The Sleeping Beauty Problem
by Tanya Khovanova and Alexey Radul
This post is inspired by the following problem:
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning: What is her probability that the coin was heads?
Some people argue: asleep or awake, the probability of a fair coin being heads is one half, so her probability should be one half.
Other people, including us, argue that those people didn’t study conditional probability. On the information of the setup to the problem and the information of having awakened, the three situations “Coin was heads and it is Monday”, “Coin was tails and it is Monday”, and “Coin was tails and it is Tuesday” are symmetric and therefore equiprobable; thus the probability that the coin was tails is, on this information, two thirds.
So who is right? We are, of course. A good way to visualize probability judgements is to turn them into bets. Suppose each time Beauty wakes up she is offered the following bet: She pays $600 and gets $1000 if the coin was tails. Should she take it? If her probability of the coin being tails were one half, then obviously not; if her probability of the coin being tails were two thirds, obviously yes. So which is it? Consider the situation from her perspective as of Sunday. She can either always take this bet or always refuse it. If she always refuses, she gets nothing. If she always accepts: If the coin turns up heads, she will be asked the question once and will lose $600. If the coin turns up tails, she will be asked the question twice and will gain $800. So on average she will win, so she should take the bet. By this thought experiment, her probability of tails is clearly not one half.
To make matters more interesting, let’s try another bet. Suppose she is given the above bet just once, in advance, on Sunday. She pays $600, and she gets paid $1000 on Wednesday if the coin was tails. This has nothing to do with sleeping and awakening. If she takes the bet she loses $600 with probability one half and gains $400 otherwise. So she shouldn’t take the bet. Her probability on Sunday that the coin will come up heads is, of course, one half. The point is that just as these two bets are different bets, the sets of information Beauty has on Sunday vs at awakening are different, and lead to different conclusions. On Sunday she knows that the next time she wakes up it will be Monday, but when she then wakes up, she doesn’t know that it’s Monday.
Parting thought: The phenomenon of predictably losing information leads to the phenomenon of predictably changing one’s assessments. Suppose for some reason she decided to take that unprofitable bet on Sunday. When she wakes up during the experiment, should she feel happy or sad? From her perspective during the experiment, the odds of gaining $400 vs losing $600 are two to one, so she should be happy. Given that she knows on Sunday how she will (with complete certainty!) feel about this bet on Monday, should she take it, even given her Sunday self’s assessment that it’s a bad bet?
Share:
Bill:
Great post, but I have a bit of a quibble with the parting thought. The paradoxical nature of your last question should raise concerns about the conclusions that led to it.
If she knows she made an unprofitable bet on Sunday, she shouldn’t be any happier about it when she wakes up. Since she was going to be woken up anyway, she won’t have any additional information when she is.
Let’s use your strategy of making the probability judgement into a bet. Imagine that Sleeping Beauty makes the unprofitable bet on Sunday, risking $600 to win $1000 on the coin flip. When she is woken up, she has the option to keep that bet in its current state (which she will not be told) or to adjust it so that she is instead risking $1200 to win $2000.
If she wakes up happy with her Sunday wager, she will choose the second option, which will always either maintain or increase her wager at the same odds. If she still considers the wager unprofitable when she wakes up, she will stick with the original bet and have no interest in doubling it. Clearly, it is more profitable not to increase an unprofitable wager, and she will not.
Think of it another way. Let’s say she makes the same unprofitable wager on Sunday, but no money has yet exchanged hands when she goes to sleep. When she loses the Sunday wager, she wakes up on Monday having to pay $600. But if she wins, the payments to her (and from her) must be made over two days (or she’ll not believe she was paid at all). So when she first wakes up, she believes there is a 1/3 chance that she will have to pay $600, and a 2/3 chance that she will have to pay $300 in order to collect $500. The bet remains unprofitable for her, even as she believes there is a 2/3 chance she has won it.
26 August 2011, 12:45 pmTanya Khovanova’s Math Blog » Blog Archive » Sleeping Beauty and Mondays:
[…] we discussed in our first essay about Sleeping Beauty, she should take the bet. Indeed, if the coin was heads her loss is $550. But if the coin was tails […]
27 August 2011, 1:08 pmAlexey Radul:
Bill:
No one ever said the parting thought was a paradox.
The whole point of this post is that the information she has available
*does* change from Sunday to Monday; you are right that she gains no
information, be she loses information, because she no longer knows
what day of the week it is.
Let’s consider your first illustration. The option to double on
awakening drastically changes the situation, because she has two
chances to double if she’s winning, and only one chance to double if
she’s losing.
Specifically, two variants merit discussion: The version that
correctly corresponds to her state of mind on awakening would be the
option to double the bet each time she wakes up (with no upper limit
on the number of doublings). She should clearly always take this
option; in fact, this case is so profitable for her that it makes the
original bad bet into a good one. If the coin is heads, the bet will
get doubled once and she will lose $1200; but if the coin is tails,
the bet will get doubled twice and she will gain $1600.
The illustration you actually suggested is more subtle. The option to
double the bet at most once improves it (although with these
constants, not enough to make it good overall); and the optimal
strategy if given this option is not “always decline”. Indeed, by
choosing to double only sometimes, she can double more often when she
wins than when she loses. We can compute this: suppose her strategy
is to decline the option with probability p. Then her payoffs are:
if heads: -600p – 1200(1-p) = -600(2-p);
if tails: +400p^2 + 800(1-p^2) = 400(2-p^2).
So her overall payoff is
– 200p^2 + 300p – 200.
Differentiating with respect to p this has a unique extremum when
-400p + 300 = 0 i.e., p = 3/4.
Differentiating again, this extremum is a maximum. So her best
strategy in the game you proposed is to decline to double the bet with
probability 3/4, that is, to double it with probability 1/4. Her
overall expected payoff in this situation is then
-200(9/16) + 300(3/4) – 200 = -900/8 + 900/4 – 200 = -700/8,
which is slightly better than the expected payoff of the original bet,
namely -100. Homework: compute the maximum initial cost (instead of
$600) for which Beauty breaks even if given the option to double the
bet during the experiment, and the probability with which she should
double it in this case.
Now your second illustration. The separation into days is irrelevant:
Beauty finds a 2/3 chance that she will pay $300 and get $500 today
and also the other day of the experiment, for a total of $600 for
$1000. This looks pretty good to Beauty.
Looking forward to how one will think in a different information state
27 August 2011, 2:24 pmis counterintuitive.
Bill:
You make some excellent points, but I still disagree with your conclusion. She should not wake up happy about her bet.
If we want to structure a secondary bet to see if Sleeping Beauty is happy with her first bet after she wakes up, we have to do it carefully. Allowing her to double up twice makes it a different bet, as you point out. Allowing the mixed strategy you suggest, mathematically correct as it may be, still makes it a different bet to some degree, as she is able to take advantage of the fact that she will have – ahem – a second bite at the apple. In order to truly see if she is happy with her current bet, we must use the model I proposed.
I agree that she loses information when she wakes up, but none of the information that went into evaluating the bet. It makes no sense for her to go to sleep thinking she made a bad bet and wake up thinking it is good. Yes, she knows there is a 2/3 chance that she will be informed that she has won, but that is only because she knows she will be informed two times for each victory and one time for each loss. If the bet were repeated continuously, she would wake up each morning knowing that she has a 2/3 of learning she has won the current round, but she will also know she is on the path to bankruptcy.
It was I who called the last question paradoxical. If you disagree (or if anyone else does), perhaps you could share your answer. Given that she knows that she will feel happy about this bet on Monday, should she take it on Sunday?
1 September 2011, 3:31 pmNick:
Overconfidence usually results in embarrassment. In this case, you are wrong.
ERRONEOUS ARGUMENT AGAINST THE HALFER POSITION. Consider the following argument which I shall show to be in error. “Assume that the probability that the coin lands H is ½. Repeat the experiment a million times. On each awakening, Beauty bets one dollar, even money, that the coin was tails. If the probability was one half then she’ll break even in the long run. But she won’t break even in the long run. On each heads she’ll bet once and lose. On each tails she’ll bet twice and win twice. In the long run she’ll be about a million dollars ahead. So the assumption that the probability is one half is not correct.” In other words, the probability of T must be different to ½ if the expected return is not zero. This, however, is a bogus argument because the size of the bets are determined by the result and, unknown to SB, every time the desired outcome T occurs then the larger amount is placed. If the amount bet is x, the expected return based on ½ probabilities is then -0.5x + 0.5(2x) = x and cannot be zero. More importantly, though, the results of these bets would not be available to SB.
1 October 2011, 12:28 amNick:
And I have to comment on “Other people, including us, argue that those people didn’t study conditional probability.” Cute. You insult others for misunderstanding probability, and then misunderstand conditional probability yourself.
1 October 2011, 12:31 amBill:
Imagine that Jack, now wealthy from the sale of golden goose eggs, has taken to speculating on coin flips. Every morning, a coin is flipped and Jack has even money on the outcome. If the flip is heads, money is deducted from Jack’s bank account automatically, and if it is tails, the same amount of money is deposited.
Jack himself is far too busy to attend the coin flipping personally, but he is informed of the results by mail. If the result is heads, he is sent a letter just after the flip. If the result is tails, he is sent two identical letters just after the flip, in case he wants to keep a scrapbook of his victories. (He doesn’t.) The envelopes are all completely identical on the outside, save the postmark identifying the date the letter was sent.
Jack is a very busy man, and takes little interest in the whole endeavor. When the letters arrive, his personal assistant tosses them into a large box, where they have been gradually forming a sloppy unorganized pile over the course of several months. Jack’s accountant keeps track of his finances, so Jack never sees the money go in or out of his account.
Only today, he begins to grow curious about how the coin flipping investment is going, so he reaches into the box and pulls out an envelope randomly from the middle of the pile. It is postmarked August 24, 2011.
Now, before Jack opens the envelope, what are the odds that the coin flip on the morning of August 24, 2011 was tails?
One might be tempted to say that the odds are 50/50. It was a coin flip, after all. But while the envelope may have been selected completely at random, the date was not. Each date with a tails result is represented twice in the pile, while each heads result is only represented once. Either result being equally likely, there is a 2/3 chance August 24 was a tails day.
(If you still don’t believe it, imagine that 1,000 envelopes are mailed when the result is tails, and only one when it is heads. After 10 days, there are 4,006 envelopes in the box. Jack draws one at random and looks at the date. Do we really believe that it is equally likely that heads or tails came up on that date? There were actually more heads than tails flipped, but I’ll still bet this particular date was tails!)
That was for Nick. This is for Alexey.
In the original example above, Jack reaches into the box and pulls out an envelope randomly from the middle of the pile. It is postmarked August 24, 2011. Jack realizes, before he opens the envelope, that there is a 2/3 chance that there was a tails outcome for that date. But this does not make him feel any differently about his bet. He knows he is getting even money, and the fact that each tails outcome is represented twice in the pile doesn’t make him any more excited about it.
3 October 2011, 8:53 amJoseph:
Perhaps you should have used conditional probability to solve this problem, rather than accusing others of not studying it.
Pr(Heads|Woken Up) = Pr(Woken Up|Heads) * Pr(Heads) / Pr(Woken Up) = 0.5
Your three situations are not symmetric, in fact:
Pr(Monday and Tails) = 0.25
Pr(Tuesday and Tails) = 0.25
Pr(Monday and Heads) = 0.5
I’ll leave that to you to solve, it would appear you need the practice.
As to your betting analogy, if someone offered you the following bet:
Bet heads and win: Get $1
Bet heads and lose: Lose $2
Bet tails and win: Get $2
Bet tails and lose: Lose $1
I am sure that you would bet tails even with a fair coin.
Uneven payoffs != Uneven odds.
23 October 2011, 2:06 amJeffJo:
There is a quite simple and straightforward solution for the Sleeping Beauty problem, using conditional probability, with no need for metaphysics or a trip to a casino.
If Sleeping Beauty knew it was Monday, then she would know that the probability the coin landed heads was 1/2. If Sleeping Beauty knew it was Tuesday, then she would know that the probability the coin landed heads was 0. Thus we have these two conditional probabilities:
Pr(Heads|Monday)=1/2
Pr(Heads|Tuesday)=0
Sleeping Beauty doesn’t actually know what day it is, but she can assign a probability to that. Let M be the probability it is Monday. We don’t need to deduce a value for M yet, just can place bounds on it. Since both Monday and Tuesday are possibilities, M can’t be 0 or 1. Since no other day is possible, the probability it is Tuesday is 1-M. Finally, since being either of those two days represent independent events, we can apply the law of total probability:
Pr(Heads) = Pr(Heads|Monday)*Pr(Monday) + Pr(Heads|Tuesday)*Pr(Tuesday)
= (1/2)*M + (0)*(1-M)
= M/2.
Recalling that M is less than 1, we have now proven unequivocally that 1/2 cannot possibly be the correct answer. If you take the normal approach to deducing what M is, you will note that in repeated instances of this game, two out of every three interviews will take place on Monday, so M=2/3. That makes P(Heads)=1/3.
Joseph, your conditional probability calculation is flawed. The Pr(Heads) you include on the right-hand side should be the confidence Sleeping Beauty assigns to Heads following the rules of the Sleeping Beauty Game, which represents the information that Sleeping Beauty knows. Not the probability for an isolated coin flip. All you really proved is that Pr(Heads|Following Rules)=Pr(Heads|Following Rules), which should be obvious if you actually think about what you did by conditioning on the unity event Woken Up.
25 October 2011, 4:04 pmJoseph:
“If Sleeping Beauty knew it was Monday, then she would know that the probability the coin landed heads was 1/2.”
Nope.
“If you take the normal approach to deducing what M is, you will note that in repeated instances of this game, two out of every three interviews will take place on Monday, so M=2/3.”
Nope.
“Joseph, your conditional probability calculation is flawed.”
Nope.
Try again bro.
26 October 2011, 3:37 amJeffJo:
So, when you said “Pr(Monday and Heads) = 0.5”, you were not serious? Because if SB is told that it is Monday, that makes Pr(Monday)=1. And we all know that the definition of a conditional probability is Pr(A|B)=Pr((A and B)/Pr(B), which leads to … wait for it … P(Heads|Monday)=1/2.
But to argue for it, if SB knows that it is Monday, the experiment is no different than one where there is no second drugging. Upon awakening Monday, SB is asked for Pr(Heads),and teh game ends. You are saying this probability is not 1/2. Why?
I’d also be interested is hearing why you think an event that occurs two out of three times in indepepndent trials does not have a probability of 2/3.
Again, by using math, I have proven unequivocally that 1/2 cannot possibly be the correct answer. Using mathematics, not unsuportted claims. If you want to dispute that, try showing what you think is wrong with the math, or show the math that leads you so some other conclusion. Without that, there is no need for you to “try again, bro.”
26 October 2011, 5:46 amJoseph:
To make the situation clear:
Pr(Monday and Heads) = 1/2
Pr(Tuesday and Heads) = 0
Pr(Monday and Tails) = 1/4
Pr(Tuesday and Heads) = 1/4
Pr(Heads|Monday) = 2/3
Pr(Tails|Monday) = 1/3
Pr(Heads|Tuesday) = 0
Pr(Tails|Tuesday) = 1
Pr(Monday|Heads) = 1
Pr(Tuesday|Heads) = 0
Pr(Monday|Tails) = 1/2
Pr(Tuesday|Tails) = 1/2
Pr(Monday) = 3/4
Pr(Tuesday) = 1/4
Pr(Heads) = 1/2
Pr(Tails) = 1/2
All conditional on SB waking up and knowing the rules of the experiment.
I suggest you work through the problem again, using Bayesian inference instead of your intuition to find the probabilities; a diagram may help as well.
26 October 2011, 6:20 amJeffJo:
To make the situation both clear and proven:
Pr(Heads|Monday) = 1/2
Pr(Tails|Monday) = 1/2
Justification: The interview on Monday is independent of the coin flip.
Pr(Tuesday and Heads) = 0
Justification: The rules of the game do not allow the event.
Pr(Heads|Tuesday) = Pr(Tuesday and Heads)/Pr(Heads) = 0, regardless of Pr(Heads)
Pr(Tails|Tuesday) = 1 – Pr(Heads|Tuesday) = 1
Pr(Monday) = 2/3
Pr(Tuesday) = 1/3
Justification: By the rules of the game, in repeated instances two out of every three interviews take place on Monday.
Pr(Monday and Heads) = Pr(Heads|Monday) * Pr(Monday) = 1/2 * 2/3 = 1/3
Pr(Monday and Tails) = Pr(Tails|Monday) * Pr(Monday) = 1/2 * 2/3 = 1/3
Pr(Tuesday and Heads) = Pr(Heads|Tuesday) * Pr(Tuesday) = 0 * 1/3 = 0
Pr(Tuesday and Tails) = Pr(Tails|Tuesday) * Pr(Tuesday) = 1 * 1/3 = 1/3
Pr(Heads) = Pr(Monday and Heads) + Pr(Tuesday and Heads) = 1/3 + 0 = 1/3
Pr(Tails) = Pr(Monday and Tails) + Pr(Tuesday and Tails) = 1/3 + 1/3 = 2/3
Pr(Monday|Heads) = Pr(Monday and Heads)/Pr(Heads) = (1/3)/(1/3) = 1
Pr(Tuesday|Heads) = Pr(Tuesday and Heads)/Pr(Heads) = (0)/(1/3) = 0
Pr(Monday|Tails) = Pr(Monday and Tails)/Pr(Tails) = (1/3)/(2/3) = 1/2
Pr(Tuesday|Tails) = Pr(Tuesday and Tails)/Pr(Tails) = (1/3)/(2/3) = 1/2
Bayesian inference requires that you establish a prior, which you have not done. Now, I suggest you work through the problem again; this time, without assuming the answer you want to get, and assuming the values for the critical probabilities that will make that happen.
26 October 2011, 7:15 amJoseph:
Your justifications (except the second one) are incorrect. Remember that SB conditions on the waking up and the rules of the experiment.
Here is a diagram to help you understand (you may need to copy it into notepad if it doesn’t align correctly):
………..Monday. 0.25
26 October 2011, 8:15 am………./…….
………/……..
….Tails………
…/………….
../…………..
–………Tuesday 0.25
……………..
……………..
….Heads………
……………..
……………..
………..Monday. 0.50
JeffJo:
Your diagram is incorrect, since the events (Tails and Monday) and (Tails and Tuesday) are not independent. The two don’t happen at the same time, but by necessity in any game where one happens, so does the other. Sicne there is dependence, there is no requirement (as you tacitly assume, still without bothering to justify anything) for all three to sum to 1. Pr(Tails and Monday)=1/2 because the chances there will be is an interview on Monday and tails will have been rolled is 1/2. Same for Tuesday. Since the chances that each interview will exist are identical, the probability that SB must apply to the current interview being that one are identical. Since there are three, each has probability 1/3.
“Justification” means supplying reasons, not stating the results you think are true. Like “The interview on Monday is independent of the coin flip,” not “Pr(Monday and Heads)=1/2 because I believe it is.” And if you really believe there is something wrong with any of my justifications, you would point them out. Until you do, there really is no point in replying. All you seem to be doing is parroting what you read somewhere else, with no evidence you understand any of it.
26 October 2011, 11:10 amJeffJo:
I realized I was unclear there. The only thing you can by a diagram like you drew is the unconditional probability of the event happening. This is different than the probability that SB assigns to the event, because hers is always a conditional probability based on the rules of the game. That doesn’t just mean that a coin is flipped – it means that she has been awakened for a single interview that cannot be linked to another.
The events (Tails and a Monday Interview exists) and (Tails and a Tuesday Interview exists) are not independent; in fact, they are the same event. The events (Tails and this is a Monday interview) and (Tails and a Tuesday Interview exists) are independent, and SB must re-normalize their probabilities.
While that may sound weird, it is correct. But you don’t need to use it, because all of the probabilities can be derived from ones that don’t require such renormalization. Pr(Heads|Monday)=Pr(Tails|Monday)=1/2 because if “today is Monday” is given, the rules of the game don’t matter to the assessment of the coin flip’s probability.
26 October 2011, 2:56 pmCarl:
I’ve been reading about this fascinating problem intensively for the last few hours and, I must admit, my point-of-view has switched several times between 1/2 and 1/3.
One thing that bothers me about the simplest 1/3 argument, and I wonder if someone can clarify it for me, is this: There seems to be an assumption that each of the waking events (ie. Heads-Monday/Tails-Monday/Tails-Tuesday) are equiprobable, ie. 1/3 each. However, it seems to me that this is not the case. There is 1/2 chance of either heads or tails, but then the tails result could result in two different waking events, which would then split that probability in 1/2 again. Therefore, P(HeadsMonday) = 0.5, P(TailsMonday) = 0.25, and P(TailsTuesday) = 0.25.
So, when SB wakes up, she knows that it’s 0.75 chance that it’s Monday, and only 0.25 chance that it’s Tuesday. So when trying to work out the coin probabilities, she would go something like this:
P(Tuesday AND Heads) = 0
P(Tuesday AND Tails) = 0.25 (Because P(Tails|Tuesday)=1)
P(Monday AND Heads) = 0.5
P(Monday AND Tails) = 0.25
Therefore, there’s 0.5 chance of the coin being either Heads or Tails.
Can any of you 1/3-ers point out what’s wrong with this argument?
22 November 2011, 5:44 pmTanya Khovanova:
Carl, the tails doesn’t split later into Monday or Tuesday. It will result in both: two waking events.
22 November 2011, 9:21 pmBill:
Carl, all possible outcomes can be exhausted in two trials, one in which the outcome will be heads and the other in which it will be tails. After these two trials, she will have been woken exactly three times, one for each of the three possible states. Therefore, when she wakes, having no memory of previous wakenings, the three events are equiprobable. She does not perceive a 75% chance that it is Monday; for her, it will be Monday 2/3 of the times she wakes up.
Imagine a different scenario. Snow White takes a bite of a poisonous apple and falls into a deep slumber. Before she loses consciousness, The Wicked Queen explains the following rules, and Snow White believes her. The Wicked Queen will flip a coin. If the result is heads, she will let Snow White die. If the result is tails, she gives Snow White the antidote and awakens her. Now, Snow White awakens. Assuming the rules were followed, what are the odds that the coin flip was tails?
I don’t see how any answer other than 100% makes sense. This proves the 50/50 odds of a coin flip can be modified by conditional perception. If we were to apply your system to this puzzle, we’d run into a problem, as we couldn’t subdivide the heads probability into any possible outcomes.
23 November 2011, 9:59 amJeffJo:
“There seems to be an assumption that each of the waking events (ie. Heads-Monday/Tails-Monday/Tails-Tuesday) are equiprobable, ie. 1/3 each.” Carl, that isn’t the assumption that is being made. SB isn’t estimating the probability that a combination will occur, she is estimating the probability that today’s combination, which *has* occurred, is one of those. Since she has no evidence than can distinguish any of them, all three possibilities look the same to her, so she cannot assign probabilities that are different to any of them.
Here is a set of variations we can make that show the absurdity assigning different probabilities to them. Although I did originate some thoughts along these lines, I later discovered that Nick Bostrom also expressed some of them is his article about the problem.
1) Move the coin toss to Monday night. Since it has no effect on the experiment until Tuesday, this can’t change the problem in any way.
2) Instead of having Sleeping Beauty make a bet, say you participate in the experiment with her and she has to set the odds for you to bet against. That is, if you bet $1 on heads, what will she pay you if you end up being right? That’s a better measure of her confidence.
3) After she sets odds on Monday, tell her it is Monday, so she knows the coin hasn’t been tossed yet. Then ask her to set odds for a new bet. As Joseph so nicely pointed out, the halfer probabilities important for this new bet are Pr(Heads|Monday) = 2/3 and Pr(Tails|Monday) = 1/3. If she is a halfer, she should payoff $3 on a $1 bet that a fair coin, one she knows has yet to be flipped, will land on Heads.
4) If she feels confident in those odds, ask her if she’d be willing to sweeten the deal. Since no coin toss has been made, or conditional awakenings (or non-awakening) performed, we can lengthen the experiment without prejudice. We could make it a week long, awakening you both six more times if the result is tails, and letting you sleep all week if it is heads. The halfer arguments now say P(Tails&Monday)=1/14 and P(Heads&Monday)=1/2, so Pr(Heads|Monday)=7/8. She should pay $8 for a $1 bet on Heads. And she still knows the coin hasn’t been flipped.
5) If she agrees, ask her how changing the process affects the coin. Oh, and bet on Heads.
29 January 2012, 3:49 pmJeffJo:
Um; I mixed up heads and tails in that. With one awakening, she should pay $3 for a bet on tails, not heads. And $8 with seven awakenings. Bet on tails.
29 January 2012, 3:53 pmJeffJo:
I’m returning with some new thoughts on this problem, that more-or-less prove the result. First I’d like change the question a little. The underlying problem is unchanged, but I’ll explore variations soon: “What probabilities should Sleeping Beauty place on the three propositions ‘Awake on Monday and the coin landed Heads,’ ‘Awake on Monday and the coin landed Tails,’ and ‘Awake on Tuesday (and the coin landed Tails)’?”
The halfer argument is that, since an Awakened Sleeping Beauty (I’ll call her ASB) knew she would awake in a situation identical to all she sees, that she cannot update the probabilities assigned by Sunday Sleeping Beauty (SSB). So the probability the coin landed heads, or tails, must still be 1/2. Further, if it landed tails, it is just as likely to be Monday or Tuesday. So the three probabilities must be 1/2, 1/4, and 1/4, respectively. But now, based on my reformulation of the question, we can see a blatant contradiction: SSB’s assessment for each of these three probabilities would have been 1/2. The assertion that she cannot update has led to an actual update.
In the following variations, I’ll change the problem into a week-long experiment governed by six-sided dice. All of the probabilities can be deduced from the Principle of Indifference. It says that if the possibilities for two events’ occurrences are identical, their probabilities must be identical. Most times, the application will be obvious so I won’t explain it.
Variation 1: As the trivial example, awaken SB exactly once, on the day indicated by one die rolled on Sunday Night. Both SSB and ASB place a 1/6 probability on each event of the form “Awake on Day D and the die rolled D.” The assertion that no update is possible seems fine so far.
Variation 2: Now roll the die each morning. If the result matches the day number, awaken our intrepid sleeper. Since it is no longer certain that she will be awakened (it’s actually quite close to a 2/3 chance), if there is an ASB, by the halfer’s logic she must update the answers. Yet her answer cannot be anything except 1/6, which is exactly what SSB thinks. So, where is the update?
Variation 3: Roll two dice on Sunday. If they match, re-roll them until you get two different numbers. To see how the Principle of Indifference applies, used differently-colored dice. The red die has to have a 1/6 of ending up on any number, and so does the green one, which makes a 1/3 chance overall. Awaken Sleeping Beauty on the two days indicated, as in Variation 1, and SSB’s answer to each question (change “die rolled D” to “dice roll included D”) is now 1/3. Since she will be awakened, the halfer’s argument is that she can’t update. Yet she must, since ASB’s answers still must be 1/6.
Variation 4: Roll one die on Sunday, and awaken Sleeping Beauty on the first D days of the experiment. SSB’s probability for the event “Awake on day D” must be D/6. The halfer argument says ASB can’t update. But she can’t assign a probability of 1 to any day, like SSB did to Day 1. Still, ASB knows that, while there are 36 combinations of a day number and a die roll, only 21 of them result in her being awakened: six on Monday, five on Tuesday, etc. And the Principle of Indifference applies to these combinations, not to a day. Her answer is D/21 since there are D equally-likely combinations for each day.
And the point of these variations, is that Bayesian updating can be used to get ASB’s answer. Simply divide ASB’s probability for DAY D by the sum of all of her probabilities, which is 21/6. It is unusual that the denominator in this equation is greater than once, since the denominator must represent the sum of the probabilities of disjoint events. We don’t normally expect them to add up to more than one, yet they must in this problem.
Which points out what “new information” Sleeping Beauty has, and what the halfers are not recognizing. Events that described the same outcomes to SSB, describe different ones to ASB. In the original problem, “Awake on Monday and the coin landed Tails” and “Awake on Tuesday (and the coin landed Tails)” represent the same outcome on Sunday night, but different outcomes to Sleeping Beauty when she is awakened. This is “new information,” and allows her to update her probabilities for each event to (1/2)/(3/2)=1/3. We might see this better in:
Variation 5: Flip a coin. On Heads, use Variation 1. On Tails, use Variation 3. And here, use three differently-colored dice. Since each die has the same chance of being used, and each has a 1/6 chance of landing on a D, there are 21 equally-likely combinations of die+day. Each has a 1/21 chance of applying to ASB’s state, but twice as many correspond to Tails, as to Heads. So P(Tails)=2*P(Heads), P(Tails)=1/3, and P(Heads)=2/3.
Variation 6: Change the size of the die in Variation 6 to any N, and the results for Heads and Tails are the same. In particular, N=2 makes it the original problem, except that the awakening day for Heads is left to chance. That doesn’t matter – since the answer can’t be different if it is fixed for Monday, or fixed for Tuesday, then that answer has to be correct when it is random.
10 August 2012, 9:55 amIoannis Mariolis:
The probability of Heads should always be 1/2, since upon awakening no gain or loss of information takes place. The betting argument you have presented is erroneous. Let’s assume even bets on Heads/Tails (e.g. in case of heads you win 600 $, in case of Tails you lose 600 $). Then, if you bet on Heads your expected gain according to halfers is E(Heads)=1/2*600+1/2(-1200)=-300 , which means that in fact you have an expected loss of 300 $. According to thirders your expected gain is E(Heads)=1/3*600+2/3(-600)=-200, which means that in fact you have an expected loss of 200 $. If you are not convinced that the halfers approach is the correct one you can run a large number of trials of the experiment and you will see it is the halfers approach that gives the correct prediction about your loss. In case the bet is not even, the correct threshold that should define your strategy is to take the bet if you win at least twice than what you lose. Coincidentally, this threshold is predicted by both approaches.
13 May 2013, 4:27 amJeffJo:
. Sleeping Beauty does gain new information. On Sunday, she knows that she can be wakened once or twice. When wakened, she knows about only one. And it’s easy to demonstrate that this is new information. It requires six-sided die, and a modification to the drug, but neither affects the problem in any way except to illustrate the new information.
Assume the “wake up drug” only wakens SB for half an hour, at the end of which she falls asleep again and forgets that half hour. On Sunday Night, after Sleeping Beauty has been put to sleep, flip the coin and roll the die. If the coin lands on tails, make a note of the die result, and roll it again until you get a different number.
Obviously, there is a 1/6 probability that the first die roll is any particular number from one to six. It is not quite so obvious, but not hard to show, that the same probability applies to the second die roll, if it is required. And that there is a 1/3 probability that any particular number is included in the two rolls. Finally, if the coin flip is unknown, there is a (1/6)*(1/2)+(1/3)*(1/2) = 1/4 probability that any particular number is among however many dice rolls are required.
Wake SB up on Monday at the hour that corresponds to the first die roll (1=1PM, 2=2PM, etc.). If tails came up, wake her up on Tuesday according to the second die roll. Oh, and give her an accurate watch that doesn’t tell her what day it is.
Say SB finds herself awake at 3PM. On Sunday Night, she knew there was a 1/4 probability she would be awake at 3PM sometime during the experiment. But now she has to assign a 1/6 probability to the fact that her current awakening would be at any particular hour. So she clearly must be able to update from Sunday Night’s “I will be awake *sometime*” to “I am awake *now*.” The assertion every halfer uses to justify their intuition is wrong. And this conclusion is valid no matter what hour she finds herself awake, so it is valid even if she doesn’t know the time (this is from the Law of Total Probability, an irrefutable theorem, so don’t try to argue with it).
The likelihood ratio she uses for this update is 2/3 since (1/4)*(2/3) = (1/6). Her probability on Sunday, for a Heads-wakening disregarding hour, was 1/2. Multiply this by the likelihood ratio of 2/3, and we get P(heads|awake) = 1/3. Similarly, P(tails&Monday|awake) = P(tails&Tuesday|awake) = 1/3.
The so-called “betting argument” depends on how many times you allow a bet, and can show any result you want.
3 June 2013, 8:33 pmIoannis Mariolis:
Dear JeffJo, I have thoroughly read your post on SB’s gain of information. Although, the assigned probabilities your are using are correct, it doesn’t follow that new relevant information are provided to SB in either original or modified case. There is no need to use the modified version to make your point, since the same rationale applies also to the original case. In particular, before put to sleep SB calculates that the probability of at least one of her awakenings occurs on Monday is 1, whereas when actually awaken she calculates that the probability that it is Monday is 3/4. She is not using any new information for these calculations, she is just calculating probabilities of different events. If she was asked before put to sleep about these probabilities (e.g. P(Monday|awake)) she would make the same calculations.
17 June 2013, 9:22 amI agree that betting results depends on the times the bet is allowed. In my previous example the bet is allowed on each awakening and SB assumes that her choice is the same for both awakenings in case of a tail toss.
Best regards,
Yannis
JeffJo:
“Although, the assigned probabilities (you) are using are correct, it doesn’t follow that new relevant information (is) provided to SB in either (the) original or (the) modified case.”
Sorry to correct your English; but the issue with this problem is in the tense used to express it, and I want to make it clear that I am going slowly with it because it seems that English is not your first language.
On Sunday, SB is assessing events in a future time; but when wakened, she is assessing the present time. In order to treat the same events with the laws or probability, they need to be placed into the same frame of reference, that of *potential* time, whether it be present, future, or past. To do this, the events have to be presented in terms of “Monday” or “Tuesday;” not “today,” “I am awake,” or “I am awake at least once.”
So on Sunday, SB has two random variables: will a given time be on Monday, or Tuesday; and will it occur in a frame of reference where the coin eventually lands on Heads, or on Tails. Then she needs to form the joint probability distribution for these two random variables: they are independent, and each value occurs with probability 1/2, so each combination (Mon&Heads, Mon&Tails, Tue&Heads, and Tue&Tails) has a 1/4 chance to correspond to any specific time.
On Sunday, SB knows that each combination *can* happen*. That includes Tue&Heads. But when she is awake, she knows that three could represent the current time, and that one cannot. This is new information. She possesses it because on Sunday she knew Tue&Heads could correspond to a time, and now she knows it does not correspond to the current time.
“She is not using any new information for these calculations, …” Which is why she gets the wrong answer if she uses your methods. She has new information, and is not using it.
28 June 2013, 3:24 pmIoannis Mariolis:
Dear JeffJo, English is not my first language. However, most of the errors you have corrected are due to typos or informal syntax. I believe that my arguments are both clear and valid and you are not addressing them directly.
Quoting a passage from your reply: “In order to treat the same events with the laws or probability, they need to be placed into the same frame of reference…”
If I understand correctly by “laws or probability” you are referring to modern probability theory which is an axiomatic theory. According to this approach a sample space based on possible outcomes is defined and events are subsets of this sample space. Thus, there is no need (or room) for *potential* time and other arbitrary concepts.
You are also employing two (discrete) random variables, which are not rigorously defined, since no mapping is presented. Moreover, it is not clear, in your description of the first random variable, what you mean by the expression “given time”.
Then you go on and argue that: “On Sunday, SB knows that each combination *can* happen*. That includes Tue&Heads.” . I don’t agree with that, since if a Heads toss occurs, the experiment ends on Monday, whereas SB is aware of that fact in all cases.
Lets employ the framework of modern probability theory in order to avoid ambiguities. If the conducted random experiment described by Elga is explicitly defined, there are only two possible outcomes (Heads or Tails). Thus, its sample space is S={Heads,Tails}, and assuming a fair coin we should assign P(Heads)=P(Tails)=1/2. Then, the event “a Monday awakening occurs during the experiment” is Heads U Tails (i.e. the certain event S), whereas the event “a Tuesday awakening occurs during the experiment” is the same as the Tails outcome and its assigned probability is 1/2.
Notice that you can use the above random experiment to calculate P(Heads)=1/2 both on Sunday and whenever you are awaken. The confusing part is that it doesn’t address probabilities about what day it is upon awakening. The erroneous arguments of thirders are mainly based on such confusion, mixing (without realizing that they are actually doing that, since they don’t bother to explicitly define the corresponding random experiments and sample spaces) events of a different random experiment that can indeed address the additional uncertainty about the day upon awakening.
Finally, you are addressing my argument: “She is not using any new information for these calculations, …” and you reply: “Which is why she gets the wrong answer if she uses your methods. She has new information, and is not using it.”
Once again, new information is not equivalent to new relevant information and you have failed to demonstrate why knowing that you are awake is relevant to the coin toss. Moreover, your answer is beside the point, since if you consider my complete argument, it is clear that the word “these” refers to the calculations on the probabilities of the events “a Monday awakening occurs during the experiment” and “I am awaken and it is Monday”. The basic point is that these are different events and you don’t need new relevant information to assign different probabilities to them anyway. In my previous posts I have not suggested any method for calculating the probabilities of Heads upon awakening. Thus, it is absurd to argue “Which is why she gets the wrong answer if she uses your methods”. However, in this post I have explicitly defined the conducted random experiment and applied axioms of modern probability theory to demonstrate that (assuming a fair coin) even upon awakening SB should assign P(Heads)=1/2.
The really interesting part of the sleeping beauty problem lies in the second random experiment that should be used to assign probabilities to events regarding the day upon awakening and the corresponding conditional probabilities that it implies resulting to a very puzzling situation. I have written a paper on that which is under review on a scientific journal. If it is accepted I will post a link to it.
Best regards,
1 July 2013, 7:18 amYannis
JeffJo:
Yes, let’s employ a modern framework. Both Elga and Lewis define three possibilities, which they call H1, T1, and T2. H1 means “HEADS and it is Monday,” and the others are similar. These are disjoint events, meaning that when any one is happening, none of the others is happening. So we require all three to make a sample space.
Elga only used them for SB when she has been wakened (*A*waken is an intransitive verb, and can’t be something that is done to her), saying “the difference between your being in T1 and your being in T2 is not a difference in which possible world is actual, but rather a difference in your temporal location within the world.”
Lewis tries to use them on Sunday Night, and assigns probabilities to them. At that time it is not Monday or Tuesday, so he is quite wrong to do so. He needs to define them differently on Sunday Night. Wanting to use the same events for both systems is what requires distinction I made, and it still does.
Lewis argues that there is no new information, so no update is possible. Based on this assertion, he then changes the probabilities for T1 and T2 from 1/2 to 1/4, which is an update.
The change – which both you and Lewis use without explaining how or why – is that on Sunday Night, T1 and T2 are not disjoint. The represent the same future, but at different times (see the Elga quote). But when she is awake, they are disjoint and so require a method to distinguish them in each probability space we use. That includes Sunday.
So, on Sunday, you need to create a joint probability space for the two random variables COIN and DAY. Each has two equiprobable outcomes, and they are independent, so the probability that each of the four combinations corresponds to any specific time in the future is 1/4.
When she is awake, SB knows the combination is not H2. This is new information. She can update.
But tell me, please: In what you think is a “modern framework,” suppose the coin isn’t flipped until Tuesday Morning. And, as Lewis argues, SB is told, after a delay, that it is Monday. Your answer demands that the SB say probability this fair coin *will* land on heads tomorrow must be 2/3. How does that happen? The coin isn’t flipped yet, and has had no influence on any occurrences.
2 July 2013, 3:07 pmIoannis Mariolis:
Dear JeffJo,
many popular textbooks on probability and random variables are referring to axiomatic probability theory as modern, as opposed to the “classical theory” that treats probability as the ratio of favorable to total outcomes. As an example you can see “Probability and Random Processes with Applications to Signal Processing”, by Henry Stark and John W. Woods, Prentice Hall (2002). A central idea of this “modern” axiomatic approach is that of random experiments whose outcomes are forming a corresponding sample space. Although it is a powerful theory, it is agnostic with respect to what probabilities should be assigned to each outcome. Thus, one should employ other principles (like that of indifference) to decide what is the probability of occurrence of each outcome. Even more importantly, when a setup including some random factors is provided, it is not always obvious which random experiment should be used to model it. Therefore, a lot of controversy can arise regarding the aforementioned decisions. However, once you have defined the random experiment that you are using to model the given setup and the probabilities you assign to its outcomes straightforward mathematics can be applied and provide you with the probabilities of any event belonging to the experiment’s sample space.
As I have explained in my previous post, Elga does not define the random experiment whose outcomes forms the {H1, T1, T2} sample space, and this is the source of the resulted confusion. In order to consider T1 and T2 as disjoint events you should assume a random experiment in which T1 and T2 are two different possible outcomes (or two disjoint sets of outcomes). It is obvious that the conducted random experiment (which I have explicitly defined in my previous post) consists only of a coin toss and cannot be used to generate T1 or T2 as disjoint events. In fact, we need a random experiment in which if you are wakened on Monday and the coin is tossed Tails it is completed. Similarly, if you are wakened on Monday and the coin is tossed Heads it is also completed, and if you are wakened on Tuesday and the coin is tossed Tails it is also completed. The two later cases are consistent with the original random experiment. However, completing the experiment when “Monday and Tails” occurs is inconsistent with the original random experiment and implies that a different one should be used. The structure of this new random experiment is dictated by the change of information and the uncertainty induced by this change. There is a change in the available information. Before you are put to sleep you know that it is Sunday. Upon wakening you know that it is either Monday or Tuesday. In the context of the original random experiment this change is not relevant. Before you are put to sleep you assign P(Heads)=1/2 independently of your knowledge that it is Sunday. Even if you get additional information upon wakening that it is actually Monday, in the context of the original random experiment, you should still assign P(Heads)=1/2. It is also obvious that upon wakening you have lost no information about the setup of the conducted experiment. Thus, you can still employ it to calculate P(Heads). In the context of the original random experiment P(Sunday)=P(Monday)=P(Heads or Tails)=1 and P(Tuesday)=P(Tails)=1/2. Another piece of information that you get is that you have been wakened. However, in the context of the original random experiment P(wakened)=P(Heads or Tails)=1. Thus, none of the reported changes in information should lead you to update any of the probabilities about events of the original random experiment. On the other hand, if upon wakening you are informed that it is Tuesday you have evidence of a Tails event and you should update to P(Heads|Tuesday)=0 and P(Tails|Tuesday)=1. Thus, to summarize, in the context of the original random experiment the only relevant information that can be provided is either the result of the coin toss, or that it is Tuesday (which is actually equivalent to say that the coin was tossed Tails). None of this relevant information is provided to you upon wakening. Moreover, upon wakening you are still aware of the original setup and you can use its sample space to assign P(Heads | wakened)=1/2.
But what does the change in information implies? It implies that a new kind of uncertainty is introduced to you. Namely, you do not know if it is Monday or Tuesday. In the original setup the uncertainty regarding the day was whether a Tuesday wakening will occur or not. In this new setup the uncertainty is whether the current wakening occurs on Monday or on Tuesday. The key point is how you should model your current wakening. I argue that you should model it as a random wakening during the original experiment. This implies that you assume that the original experiment is conducted and in case of a Heads toss a Monday wakening is selected as your current state, whereas in case of a Tails toss either a Monday or a Tuesday wakening is randomly selected as your current state. Then, it is straightforward that if you consider H1, T1 and T2 as the outcomes of the above random experiment you should assign P(H1)=1/2, P(T1)=P(T2)=1/4. Notice that also in the context of this random experiment P(Heads)=P(H1)=1/2 and that P(Monday)=P(H1 or T1)=3/4 and not 2/3 as implied by Elga’s analysis.
However, the above analysis results to a puzzling situation, since it looks like that the above probabilities suggest that if you are informed that it is Monday you should update to P(H1|H1 or T1)=2/3. I believe that in this puzzle lies the true beauty of the problem and unless you have a rigorous definition of the H1, T1, T2 events which is provided by an explicitly defined random experiment it is almost impossible to understand what is going on. I am afraid I have already presented a large part of the answer and I apologize for not being at liberty to reveal the last and most interesting argument on why it is perfectly consistent with the above analysis not to update P(Heads) to 2/3 once you learn that it is Monday. However, in this unambiguous framework defined by the described random experiments the answer is just one syllogism away.
I have to insist that if you depart from the axiomatic theory of probability you are taking a very slippery road (a road which both Elga and Lewis are taking in their papers). Once the random experiments are defined, one’s temporal location is irrelevant to the probabilities assigned to them. What matters is which experiment you are using to model your uncertainty. Even on Sunday night you can use the second random experiment I have defined in order to infer the probability that it is Monday when you are wakened sometime during the original experiment. It is perfectly legal to do that and there is no update of the probabilities. You are just referring to different events of different random experiments and that is why the probabilities are different.
Therefore, your argument: “Lewis argues that there is no new information, so no update is possible. Based on this assertion, he then changes the probabilities for T1 and T2 from 1/2 to 1/4, which is an update.” is not valid.
You continue saying:
“The change – which both you and Lewis use without explaining how or why – is that on Sunday Night, T1 and T2 are not disjoint. The(y) represent the same future, but at different times (see the Elga quote). But when she is awake, they are disjoint and so require a method to distinguish them in each probability space we use. That includes Sunday.”
Let me make clear that I have no intention to defend Lewis’ arguments and I would like to ask you not to identify my views to his. I believe that it is clear from the analysis I have presented in this post that I always consider T1 and T2 as disjoints events of the second random experiment and distinguish them from the “Tails and Monday” and “Tails and Tuesday” events (which are just the “Tails” event) of the original random experiment. Thus, the probability spaces and their corresponding events are unambiguously defined by the employed random experiments. Which random experiment is used depends on the questions being asked and the setup they imply.
You continue arguing that:
“So, on Sunday, you need to create a joint probability space for the two random variables COIN and DAY. Each has two equiprobable outcomes, and they are independent, so the probability that each of the four combinations corresponds to any specific time in the future is 1/4.”
A random variable (in a formal mathematical treatment) is defined as a function whose domain is the sample space of a random experiment and its range is (usually) the set (or a subset) of the real numbers. Therefore, unless you have defined a random experiment and a corresponding sample space it is impossible to define the involved random variables. After you define the second random experiment I have suggested you can assign the following joint probabilities: P(Heads, Monday)=1/2, P(Heads, Tuesday)=0, P(Tails, Monday)=1/4, P(Tails, Tuesday)=1/4 and define the corresponding random variables (which are not independent, since the Heads toss is anti-correlated with the Tuesday wakening).
In the sleeping beauty debate, there are more than two sides. Apart from halfers and thirders there are also double halfers ( the analysis I have presented implies that I am actually a double halfer) and also those arguing that there is no contradiction and both ½ and 1/3 can be consider valid answers to different questions (implying that the description of the problem leaves room for disambiguation) . I have to admit that I resent labels and in the bottom line it is only the arguments that matter. I have studied numerous papers on the problem and although some nice arguments have been presented from all sides most of them can be dismissed as logical fallacies or based on ill defined notions (this applies to all sides). Perhaps the only line of argument I have found convincing in the published literature so far is that presented in Roger White’s paper “The generalized Sleeping Beauty problem: a challenge for thirders” in Analysis (2006), where a more general case is considered that includes a device which is consulted before each awakening and there is a fixed probability c that it suggests a wakening. In the two limits (c=0 and c=1) the probability of Heads tends to 1/3 and ½ respectively. Although, White’s approach is indirect I find it pretty convincing. I believe the reason it is so convincing is that the random experiment is explicitly defined making the calculations straightforward.
I hope I have presented a new perspective to the problem and that soon I would be at liberty to present the last piece of the puzzle. This last piece is perhaps the most interesting part revealing a pitfall that has not been addressed in any other probability paradox.
Kind regards,
4 July 2013, 9:06 amYannis
JeffJo:
Yannis:
The axiomatic components of any probability theory are the components that define the probability measure itself, not other definitions or the laws we use once we have a probability measure. Regardless of the theory used, we should get the same results if we start with the same probability measures.
Say our sample space comprises three events A, B, and C, and no others; with probabilities P(A), P(B), and P(C) which sum to 1. These events have to be disjoint, regardless of how you defined them or whether you adequately defined the experiment. Now suppose that we learn C is not possible based on some evidence E, but A and B are unrestricted by E. A universal law now says that P(A|E)=P(E|A)*P(A)/[P(E|A)*P(A)+P(E|B)*P(B)].
But this means, with the probabilities you proposed, that H1, T1, and T2 must be disjoint, and that:
P(H1|Monday) = P(Monday|H1)*P(H1)/[ P(Monday|H1)*P(H1) P(Monday|T1)*P(T1)]
= 1*(1/2)/[1*(1/2)+1*(1/4)] = 2/3.
You obviously agree this is an absurd result. But anything else, especially the rationalizations proposed by “double halfers,” is a product of rearranging probability theory through illogical steps in order to arrive at the answer you have decided, based on intuition alone, must be correct. It violates all accepted probability theories, all laws of logic, and it is wrong.
A similar argument applies to the partial generalization proposed by White. The mistake he makes is similar to, but the opposite of, the one made by many in the Two Child Problem. Say you know a woman has two children. If you ask her if the older is a boy and she says “yes,” the probability she has two boys is 1/2. But if you ask her if either is a boy and she says “yes,” the probability both are boys is 1/3. Many people will say this answer must be 1/2 by incorrectly reasoning as follows: If the boy is the older child, then the second question is the same as the first. And if the boy is the younger, it is the same as an equivalent question where you asked about the younger child. Since the boy can only be the older or the younger, and the answer is 1/2 in either case, the answer to the second question must be 1/2 also.
The error in this logic is that, if the woman has two boys, she is not referring to either one when she answers “yes” to the second question. Knowing that *either* child could be a boy is not the same as knowing that a specific child is a boy. But suppose, in a third version of the question, we ask her to tell us something about one of her children, and she tells us a story about her son John. Now the probability of two boys is again 1/2, since she has chosen a specific child. It doesn’t matter that we don’t know if John is the older, or the younger, because he can only be one or the other. The logic that was incorrect before is now correct.
White uses the event “SB is awake at least once” in his calculations, where he should use “SB is awake *today*”. While it is true that she does not know if “today” is Monday or Tuesday, it is still only one of those specific days and that is part of the information she has. Said another way, White is including the event “Awake on Monday and asleep on Tuesday” when SB is, in fact, awake on Tuesday. When her information is used correctly, the thirder logic in White’s generalization produces 1/3 regardless of what value is used for c, and the halfer logic produces an actual contradiction (the contradiction White alleges is a violation of his implied assumptions, which are wrong).
11 July 2013, 8:36 amIoannis Mariolis:
Dear JeffJo,
Thank you for taking the time to read my arguments and comment on them. However, I strongly object to your comment “…the answer you have decided, based on intuition alone…”. I believe that probability problems should not be tackled using intuition and I would welcome any 1/3 answer that is expressed through valid arguments that cannot be dismissed using methods of axiomatic probability theory. Therefore, I have reached to the “double half” result though rigorous analysis using probability theory alone and not based on any intuition. Actually, the analysis I have presented contains a highly counter intuitive result, i.e. that P(H1| Monday randomly selected as my current state)=2/3 (which although counter intuitive it is correct and at the same time it does not imply that P(Heads|Monday) is not 1/2).
You argue that: “The axiomatic components of any probability theory are the components that define the probability measure itself, not other definitions or the laws we use once we have a probability measure.”
I believe that it makes no sense to isolate a component of an axiomatic theory. In axiomatic probability theory, in particular, the three axioms are
(1) P[E] >= 0
(2) P[S] = 1
(3) P[E U F] = P[E] + P[F] if E and F are disjoint
Thus, probability is defined as a set function P[.] that assigns to every event E a number P[E] called the probability of E such that (1), (2) and (3) are satisfied. In (2) S is the sample space whose subsets define the events (like E and F). Therefore, a probability measure is not sufficient in order to apply axiomatic probability theory. We also need a sample space S (defined by the outcomes of a random experiments) and a sigma field made up by the elements of S, defining the events.
My argument is that unless we explicitly define every component needed for applying axiomatic probability theory we leave room for ambiguities and confusion. A large part of the confusion is that real life events such as “a wakening on a specific day” are not identical (and should not be treated as such) to sample space events such that “this particular wakening has been randomly selected to be my current wakening” and the subtle difference between them is visible only if the random experiment defining the sample space is explicitly defined.
In the Two Child Problem I agree only with the first part of your analysis, i. e. : “If you ask her if the older is a boy and she says “yes,” the probability she has two boys is 1/2. But if you ask her if either is a boy and she says “yes,” the probability both are boys is 1/3. Many people will say this answer must be 1/2 by incorrectly reasoning as follows: If the boy is the older child, then the second question is the same as the first. And if the boy is the younger, it is the same as an equivalent question where you asked about the younger child. Since the boy can only be the older or the younger, and the answer is 1/2 in either case, the answer to the second question must be 1/2 also.”
For the second part: “But suppose, in a third version of the question, we ask her to tell us something about one of her children, and she tells us a story about her son John. Now the probability of two boys is again 1/2, since she has chosen a specific child. It doesn’t matter that we don’t know if John is the older, or the younger, because he can only be one or the other. The logic that was incorrect before is now correct.”, I have to object. In this case you have the additional information that she randomly selected between her two children and she has selected a boy to tell a story about. Let’s denote by (1) the proposition “at least one of my two children is a boy” and by (2) the proposition “I have randomly selected one of my two children and it is a boy”.
Before you hear (1) or (2) there are 4 possible outcomes (b-b, b-g, g-b, g-g), and based on a principle of indifference you can assign equal probabilities to each outcome (1/4).
In case of (1) and (2), outcome g-g is impossible. In case of (1) and based again on a principle of indifference you can assign equal probabilities to each one of the remaining outcomes (i.e. 1/3). Therefore, in case of (1) the probability the other child is also a boy is 1/3. Namely, p(b-b|1)=1/3. However, in case of (2), you know that one child has been randomly selected by the mother to tell its story. Thus, it makes no sense to treat the three remaining outcomes as equiprobable. You can assign equal priors to the four outcomes( p(b-b)= p(b-g)=p(b-g)= p(g-g)= 1/4) and then use Bayesian inference to calculate the conditional probabilities p(b-b|2), p(b-g|2), p(g-b|2). It is twice more probable that a boy is selected by the mother telling its story when it is b-b than when it is b-g or g-b. Therefore, you should assign p(b-b|2)=1/2, p(b-g|2)=1/4, p(g-b|2)=1/4. More formally using Bayes rule, p(b-b|2)= p(2|b-b) x p(b-b)/p(2)=1 x 1/4 / p(2), p(b-g|2)= p(2|b-g) x p(b-g)/p(2)=1/2 x 1/4 / p(2), p(g-b|2)= p(2|g-b) x p(b-b)/p(2)=1/2 x 1/4 / p(2). Hence, it is obvious that p(b-b|2)= 2 x p(b-g|2)= 2 x p(g-b|2) and because p(b-b|2)+ p(b-g|2)+ p(g-b|2)=1 we get to p(b-b|2)=1/2, p(b-g|2)=1/4, p(g-b|2)=1/4.
Anyway, regardless of the above analysis I do not agree that the Two Child Problem is connected to White’s approach.
12 July 2013, 4:44 amYou argue that: White uses the event “SB is awake at least once” in his calculations, where he should use “SB is awake *today*”. As I have said in my previous post White has defined a new random experiment. Although you are free to object on his analysis on why this new random experiment is relevant, it makes no sense to invent events that have nothing to do with the defined random experiment. It is straightforward from its definition that a coin is tossed and wakening or not is randomly selected. Thus S is (H1: Heads and wakened on Monday but not Tuesday, H2: Heads and not wakened either on Monday or Tuesday, T1: Tails and wakened on Monday but not on Tuesday, T2: Tails and wakened on Monday and on Tuesday, T3: Tails and not wakened on Monday but wakened on Tuesday, T4: Tails and not wakened either on Monday or on Tuesday) with P[H1]=c/2, P[H2]=(1-c)/2, P[T1]=1/2c(1-c), P[T2]=1/2 c c, P[T3]=1/2(1-c)c, P[T4]=1/2(1-c)(1-c). The event “SB is awake at least once” (call it W) is H1 U T1 U T2 U T3 and P[W]=1/2c(3-c). SB is always aware of the generalized experiment, thus, once wakened she has information on W and she can update P[Heads|W]= (c/2)/(1/2c(3-c))=1/(3-c), which directly implies that in the limit of c=1 P[Heads|W]=1/2, whereas in the limit of c=0 P[Heads|W]=1/3.
You continue saying: “Said another way, White is including the event “Awake on Monday and asleep on Tuesday” when SB is, in fact, awake on Tuesday.” . Events are defined as subsets of a sample space and “Awake on Monday and asleep on Tuesday” is an outcome of the defined random experiment, thus an element of the sample space and it would be irrational not to include it.
You also argue that: “When her information is used correctly, the thirder logic in White’s generalization produces 1/3 regardless of what value is used for c”. But White’s argument is exactly that, i.e. “the thirder logic in his generalization produces 1/3 regardless of what value is used for c”. The only available information upon wakening that is relevant to the conducted random experiment is that she has been wakened at least once during the experiment. If you believe otherwise please define the random experiment you are using (which includes as you argue the event “SB is awake *today*”), its sample space and the probabilities assigned to its events in order to discuss without ambiguity. Can you also be more specific about what you think are White’s implied assumptions, and why they are wrong?
Best regards,
Yannis
Ioannis Mariolis:
@ JeffJo,
12 July 2013, 6:51 amI want to clarify (since it is not stressed out in my previous post) that in the Two Child Problem (the second part) I do not object to your result (I also conclude to 1/2). It is the reasoning that I find unclear and not convincing. Therefore, I have presented a more rigorous explanation.
JeffJo:
Yannis:
The laws of probability are quite clear, and incontrovertible. If SB thinks P(H1)=1/2 and P(T1)=P(T2)=1/4, and then evidence E says that T2 is impossible without saying anything that restricts H1 or T1, then she must say P(H1|E)=2/3 and P(T1|E)=1/3. If you think you have found an argument that shows otherwise – specifically that these probabilities match what they obviously have to be based on what we all agree is clearly intuition, and clearly correct – then you are rationalizing. You have said you have such an argument. I don’t need to see it to know it is wrong, and it has nothing to do with what probability theory you are using. It violates Bayes Theorem. But this answer cannot be correct, because the coin does not have to be flipped yet. The conditions of the experiment cannot change the probability it *will* flip heads in the future, as any theory of probability says if the halfers are correct.
I have shown you why 1/3 is the answer. But just to clarify it, define the experiment as follows:
1) SB will be put to sleep on Sunday night.
2) After she is asleep, a (uniformly) random time is selected from the range [12Noon,6PM], and a random day from {Monday, Tuesday}. That is, any time within either of two six-hour intervals.
3) On Monday, she will be wakened at 8AM, and put to sleep with amnesia at 8PM.
4) At some time before Tuesday 8AM, a coin will be flipped. If it lands tails, the same waking schedule will be followed on Tuesday. Otherwise, she will be left to sleep. When exactly it is flipped is irrelevant, and we can consider it to be before or after Monday’s waking period as appropriate to any argument.
5) At the time and day selected in step 2, she will be asked for her assessment of P(Heads). She must answer immediately, so she must prepare her answer in advance. The range of times guarantees that she has sufficient time to do this if she is scheduled to be awake then, but it is not guaranteed that she will be awake.
The answer she should prepare is exactly the same as the answer to the original question. But the correct answer for this version, based on the well-defined random variable of when she is asked, is quite clearly P(Heads)=1/3: There are four possible situations that could correspond to the selected time, conventionally labeled H1, H2, T1, and T2. This is a random variable in any probability theory, the four values represent disjoint events in theory, and each has a probability of 1/4 in any theory. But H2 corresponds to an asleep SB, so no answer will be given then. So, when an answer is given, P(H1|answer)=(1/4)/(3/4)=1/3.
And if you don’t like the fact that she may not be asked, get twelve professors of probability (any theories you want) whom she doesn’t know, and put them through the same sleep-amnesia schedule. But assign each of them an hour in the range of times and days defined in step 2, during which they get to interview SB. This way, each professor and SB have the same information, and getting an interview doesn’t tell either anything about day or coin. But the professors know, during their interview and based on a well-defined random variable, that P(Heads) is 1/3.
The reason my solution for the third version of the Two Child Problem (TCP) is correct, is because there is a third random variable you are avoiding. You do that by calculating its effect from the other two through conditional probability (and in fact, you try the same with SB but do it incorrectly), but it is easier to consider it directly. To make that choice independent of gender we don’t measure it as a choice between boys and girls, as you tried to argue. It is between her two children. We can conveniently label that choice E or Y, for Elder or Younger child. Thus there are eight equiprobable combinations of family+choice, not four combinations of family alone: {BBE, BBY, BGE, BGY, GBE, GBY, GGE, GGY}. In four of these the story is about a boy: {BBE, BBY, BGE, GBY}. And in half of those, the family has two boys.
White has not defined the experiment any differently than Lewis or Elga – and does not claim to – except to introduce his random wake-up machine. So I reject your claims about his argument being better. Just like the second version of the TCP which you agreed with, we do need to represent the specific values of all of the random variables. Even if the parties involved do not know those values. When representing the case where SB sleeps through Monday and is awake on Tuesday, White counts the cases where she is awake on Monday and sleeps through Tuesday. Assuming Tails, the probability SB is awake twice is c^2; that she is awake on Monday only, or on Tuesday only, is (c-c^2); and that she is never awake is (1-c)^2. When she finds herself awake, it is on a specific day she does not know. The probability that represents every “tails” possibility based on her information is [c^2+(c-c^2)]/2=c/2 because it is one day, not [c^2+2*(c-c^2)]/2=[c-(c^2)/2] which counts every time she is awake.
You see how the arguments – either yours or mine – for TCP3 apply the same way to SB by choosing randomly between waking her in a red room on Monday and a blue room on Tuesday, or a blue room on Monday and a red room on Tuesday. The color she sees tells her nothing about what day it is or about the coin, but does distinguish the days just like children were distinguished in TCP3. She sees only one color, and can dismiss the cases that may be possible where she is awake in a differently colored room.
12 July 2013, 9:36 amIoannis Mariolis:
Dear JeffJo,
You are saying about TCP:
“The reason my solution for the third version of the Two Child Problem (TCP) is correct, is because there is a third random variable you are avoiding. You do that by calculating its effect from the other two through conditional probability (and in fact, you try the same with SB but do it incorrectly), but it is easier to consider it directly. To make that choice independent of gender we don’t measure it as a choice between boys and girls, as you tried to argue. It is between her two children. We can conveniently label that choice E or Y, for Elder or Younger child.”
1) I have never argued that you are incorrect. I have only said that “It is the reasoning that I find unclear and not convincing. Therefore, I have presented a more rigorous explanation.”. And I believe that this is a fair critic since your analysis was:
“But suppose, in a third version of the question, we ask her to tell us something about one of her children, and she tells us a story about her son John. Now the probability of two boys is again 1/2, since she has chosen a specific child. It doesn’t matter that we don’t know if John is the older, or the younger, because he can only be one or the other. The logic that was incorrect before is now correct.”
Namely, all you say is that the probability is ½ because she has chosen a specific child. I believe more analysis is required (and in your next post you have provided this analysis, I agree with that and I find it satisfying).
2) I do not “avoid” using the elder – younger variable. It is just not necessary. Moreover, the analysis I have provided is more general than yours, since it uses the evidence that a boy has been selected. For example you might have a population where it is twice more probable for a mother to select the boy because they tend to have more interesting stories (I do not say they have, it is just for the shake of argument). Using my approach you can exploit this information and change your calculations to p(b-b|2)= p(2|b-b) x p(b-b)/p(2)=1 x 1/4 / p(2), p(b-g|2)= p(2|b-g) x p(b-g)/p(2)=2/3 x 1/4 / p(2), p(g-b|2)= p(2|g-b) x p(b-b)/p(2)=2/3 x 1/4 / p(2). Thus, the probability of the second child being a boy is now calculated to 3/7. Actually, in case you know that in that population mothers never choose the girl to tell a story about, the probability of the second child being a boy becomes 1/3.
3) My analysis of TCP is not related to that of SBP. In SBP all calculations derive directly from the definition of the sample space. In fact the only calculations using conditional probabilities are included only to demonstrate the counter intuitive result, which however has a perfectly rational explanation that is consistent to the theory.
Now let’s get back to the main debate on the SBP. You are arguing that the correct answer is 1/3 because: “If SB thinks P(H1)=1/2 and P(T1)=P(T2)=1/4, and then evidence E says that T2 is impossible without saying anything that restricts H1 or T1, then she must say P(H1|E)=2/3 and P(T1|E)=1/3.”
Your rationale is absolutely correct. However, I argue that you do not have evidence that T2 is impossible.
The problem is that we disagree what T2 is and what constitutes evidence about T2. You argue that once you are informed that it is Monday you should think that you have evidence that T2 is impossible. The reason I insist so much on defining the random experiment and the events of its sample space is exactly because of such subtle differences between events of a sample space and real life events. If you adopt the sample space of the random experiment I have defined the only case you can have evidence that T2 is impossible is if you are informed that it is Monday and that the coin has been tossed Heads (in that case you know that T1 is also impossible). Notice that if you know that the coin is tossed after you are put to sleep on Monday you can never get evidence on Monday that T2 is impossible. Although counter intuitive, the above results are produced because we are using a random experiment that is never actually conducted and we know it. We know that if it is Monday we have not being randomly wakened, since it has been pre-decided that in any case we will be wakened on Monday. In other words, learning that it is Monday does not mean that a Monday wakening has been randomly selected as our current state. Thus, we cannot infer just because it is Monday that if the coin is tossed Heads we are on H1 and if it is tossed Tails we are on T1. Monday has to be randomly selected in order to have an H1 or T1 event. Therefore, we still cannot rule out T2.
As you can see my argument does not violate Bayes Theorem as you have arbitrarily assumed. It is related only to what can be considered as evidence or not. I believe that the beauty of this problem lies to this pitfall and to my knowledge this is the only probability paradox that presents it.
As for your version of the SBP where you arbitrarily introduce a new random variable it is clear that it is not equivalent to the original problem, where you know that you will be interviewed at every wakening. Obviously, in your version you can update P(H1)=1/4 to P(H1|answer)=(1/4)/(3/4)=1/3. Extending your version to include the twelve professors does not change anything. Exactly because they have the additional relevant information that they are conducting an interview they should also update to 1/3.
Finally you argue that: “White has not defined the experiment any differently than Lewis or Elga – and does not claim to – except to introduce his random wake-up machine.”
Lewis and Elga refer to the conducted experiment which can be directly modeled by a random experiment consisting of a coin toss having a sample space S = {Heads, Tails}. When they try to argue about what happens upon wakening they introduce H1, T2, T2 events silently assuming that they are events of the original sample space, which is obviously impossible. White on the other hand proposes a different version of the conducted experiment and the random experiment that can be used to model it is straightforward (Swhite={ H1: Heads and wakened on Monday but not Tuesday, H2: Heads and not wakened either on Monday or Tuesday, T1: Tails and wakened on Monday but not on Tuesday, T2: Tails and wakened on Monday and on Tuesday, T3: Tails and not wakened on Monday but wakened on Tuesday, T4: Tails and not wakened either on Monday or on Tuesday}). This is why it is easy to follow his arguments.
Can you please explain in: “When representing the case where SB sleeps through Monday and is awake on Tuesday, White counts the cases where she is awake on Monday and sleeps through Tuesday.”, what you mean by “representing” and “counts” and why you think White does that?
Please also explain what you mean by: “The probability that represents every “tails” possibility based on her information is [c^2+(c-c^2)]/2=c/2”. Do you mean P[tails| wakened]? Because I have already demonstrated in my previous post that P[heads | wakened]=1/(3-c) implying that P[tails | wakened]=(2-c)/(3-c).
As I have argued in my previous post modeling White’s experiment is straightforward. I have presented the corresponding random experiment and assigned probability values to its outcomes. If you object please be specific about which event and what value you are referring to. In the context of White’s random experiment, being wakened provides you with the information that H2 and T4 (see above) cannot occur during the conducted experiment and this is the information you are using to update probability of heads or tails.
The only connection I see to the TCP is the following: if we replace wakenings with gender (by letting Boy (B) represent wakening and Girl (G) represent being asleep), in case of tails there are four possible outcomes BB, BG, GB, GG, with P[BB]=c^2, P[BG]=P[GB]=c(1-c), and P[GG]=(1-c)^2. If you are informed that the coin tossed tails and that at least one wakening has occurred, you know that GG cannot occur and you can update P[BB] to c^2/(c^2+2c(1-c)), i.e P[BB|tails, wakened at least once]= c/(2-c). However, if you are actually wakened and informed that the coin tossed tails you should update P[BB] to c which is the probability of the other wakening to occur. Therefore, I don’t see anything incompatible with White’s analysis and TCP and I believe that in the arguments you present based on the color of the room, the color is wrongly connected to the calculation of heads or tails probabilities.
Best regards,
15 July 2013, 7:17 amYannis
JeffJo:
The reason I brought up the TCP was to compare it to White’s analysis. The point is, you must account for two distinct events (which is what we accomplish by ordering them) when you assign prior probabilities. You did that when you assigned probabilities of 1/4 for each family type. It is not whether you use that order in your solution, which is all you addressed in your counterargument.
The two waking events SB may experience are distinct events, yet White does not account for that when he assigns probabilities to them. So his numbers are wrong.
> 3) My analysis of TCP is not related to that of SBP. In SBP all calculations derive directly from the definition of the sample space.
It is when you cite White, who mixes the definitions in his sample space between “SB will be awake on Tuesday” and “SB is awake today, and it is Tuesday.”
The way I define them, there are four elements in the sample space: an observation opportunity (not a realized observation, just the opportunity) can correspond to {H1, H2, T1, T2}. The a priori chance a random observation opportunity will correspond to any of them is 1/4. When SB is wakened, she knows that she has a random observation opportunity, and that it is not H2. This represents new information, and she can update:
P(H|H1 or T1 or T2) = P(H&(H1 or T1 or T2))/P(H1 or T1 or T2)
= P(H1)/[P(H1)+P(T1)+P(T2)] = (1/4)/(1/4 + 1/4 + 1/4)
= 1/3.
White’s analysis is wrong first (technically) because it uses 1/2, not 1/4. That doesn’t matter to the result, but the second error does. He says T, T1 and T2 are three names for the same event; which in the TCP is the same as saying OneBoy, ElderBoy, and YougerBoy are the same event. Which was my point.
> Now let’s get back to the main debate on the SBP. You are arguing that the correct answer is 1/3 because: “If SB thinks P(H1)=1/2 and P(T1)=P(T2)=1/4, and then evidence E says that T2 is impossible without saying anything that restricts H1 or T1, then she must say P(H1|E)=2/3 and P(T1|E)=1/3.” The problem is that we disagree what T2 is and what constitutes evidence about T2. You argue that once you are informed that it is Monday you should think that you have evidence that T2 is impossible.
You are misinterpreting what I call T2. Being impossible doesn’t mean “SB can’t be awake on Tuesday after Tails,” it means “Today is not Tuesday after Tails.” If SB is informed it is Monday, my T2 is impossible. Go back and read what I wrote on 2 July 2013, 3:07 pm about how Lewis defines his events in such a way as to make them impossible when evaluated on Sunday.
It just isn’t possible to define atomic (“indivisible”) events that work for both SSB (“Sunday Sleeping Beauty”) and ASB (“Awake SB”). So let me start a new analysis, where I define them consistently for both, but add an additional set for ASB:
SSB and ASB:
H1 = Heads and, by the end of the experiment, SB will have been awake on Monday.
H2 = Heads and, by the end of the experiment, SB will have been awake on Tuesday.
T1 = Tails and, by the end of the experiment, SB will have been awake on Monday.
T2 = Tails and, by the end of the experiment, SB will have been awake on Tuesday.
ASB only:
D1 = Today is Monday.
D2 = Today is Tuesday.
Sunday Probabilities P0:
P0(H1)=P0(T1)=P0(T2) = 1/2. They add up to more than 1 because T1 and T2 are not disjoint.
P0(H2)=0.
P0(D1) and P0(D2) are meaningless.
In-experiment probabilities P1 (halfer, or thirder):
P1(H1) = 1/2, or 1/3
P1(T1) = P1(T2) = 1/4, or 1/3
P1(D1) = P1(H1)+P1(T1) = 1/2+1/4=3/4, or 1/3+1/3=2/3
P1(D2) = P1(T2) = 1/4, or 1/3
[Note that halfers do, indeed, update P0(T1)=1/2 to P1(T1)=1/4, demonstrating “new information.”]
P1(H1|D1) = P1(H1&D1)/P(D1) = P(H1)/P(D1) = (1/2)/(3/4) = 2/3,
or (1/3)/(2/3) = 1/2
The halfer’s paradox here is that, given D1 and a coin flip that ASB knows will happen tonight, how can she say its probability is anything other than 50/50? Worse, say we alter the experiment: After she is told it is Monday, and only on Monday, she chooses how many more days she will be wakened if the coin flip will come up tails. By the halfer argument, the probability is now a function of her choice. For a coin she knows has not been flipped. How does the coin know what she will choose?
> The reason I insist so much on defining the random experiment and the events of its sample space is exactly because of such subtle differences between events of a sample space and real life events.
You haven’t done that yet. I just did.
> Lewis and Elga refer to the conducted experiment which can be directly modeled by a random experiment consisting of a coin toss having a sample space S = {Heads, Tails}. When they try to argue about what happens upon wakening they introduce H1, T2, T2 events silently assuming that they are events of the original sample space, which is obviously impossible.
Actually, they define the events in the H1, T1, and T2 space, and silently assume they can use them on Sunday. Lewis says H1 means, and I quote, “HEADS and it’s Monday.” Both analyses should also define H2, which can happen (it just won’t be observed by SB if it does happen, which is not the same thing as not happening and the ultimate halfer error). However, Elga’s analysis solves the problem without needing to refer to H2, where Lewis’ needs to. By omitting it, he gets the wrong answer.
> Can you please explain in: “When representing the case where SB sleeps through Monday and is awake on Tuesday, White counts the cases where she is awake on Monday and sleeps through Tuesday.”, what you mean by “representing” and “counts” and why you think White does that?
There are four distinct states that can happen. Since the decision to wake her or not is addressed two different times, their probabilities sum to 2. If you want to make them sum to 1, you have to introduce the concept of a randomly-selected observation opportunity, which has a 1/4 chance of corresponding to any situation. White considers the two disjoint events T1 and T2 to be the same.
+++++
Or, try a simple variation that is meant to illustrate how the concept of an observation opportunity works. SB is always wakened on both days, but in three out of the four possibilities it is in a room that is painted red. On Tuesday after Heads, it is in a room that is painted blue.
When SB finds herself awake in a red room, her information is identical to the original problem. The amnesia drug means that it only matters *that* she knows one possibility cannot correspond to her current situation, not what (or how, or if) she would know about any other situation if she were in it.
Yet the solution to this problem is trivial: P(Heads|Red)=1/3, P(Tails|Red)=2/3, P(Heads|Blue)=1, and P(Tails|Blue)=0.
Jeff
11 October 2013, 2:59 pmIoannis Mariolis:
Dear JeffJo,
You are arguing:
>You are misinterpreting what I call T2. Being impossible doesn’t mean “SB can’t be awake on Tuesday after Tails,” it means “Today is not Tuesday after Tails.” If SB is informed it is Monday, my T2 is impossible. Go back and read what I wrote on 2 July 2013, 3:07 pm about how Lewis defines his events in such a way as to make them impossible when evaluated on Sunday.
Well, this is exactly why I have said that ” The problem is that we disagree what T2 is and what constitutes evidence about T2″. I am not misinterpreting what you call T2, I am just saying that it is different from the event I have defined as T2. Moreover, the way I have defined T2 (in the context of an explicitly defined random experiment) it is a valid event of a sample space (in fact it is an outcome of the defined random experiment).
You also argue that:
>It just isn’t possible to define atomic (”indivisible”) events that work for both SSB (”Sunday Sleeping Beauty”) and ASB (”Awake SB”). So let me start a new analysis, where I define them consistently for both, but add an additional set for ASB:…
In axiomatic probability theory there is no such thing as atomic events. Defining real life events in order to use them to calculate probabilities not only is not admissible but it is also not safe.
In your analysis you have defined T1 as “T1 = Tails and, by the end of the experiment, SB will have been awake on Monday” well this is just Tails since she will always wake up on Monday. Also for T2 you say “T2 = Tails and, by the end of the experiment, SB will have been awake on Tuesday”, which is also just Tails, since when Tails she will always have been awake on Tuesday by the end of the experiment. Notice that also H1 actually means Heads. Thus, as you have defined your events halfers argue that also P1(H1)=P1(T1)=P1(T2)=1/2.
I quote myself saying:
“The reason I insist so much on defining the random experiment and the events of its sample space is exactly because of such subtle differences between events of a sample space and real life events.”
and your response:
>You haven’t done that yet. I just did.
I have defined both random experiments (one corresponding to the experimental setup using only Heads/Tails outcomes and the other taking into account current status upon awaking), in my post on 4 July 2013, 9:06 am. On the other hand in your analysis there is only the original random experiment implied for P0 probabilities, but you have not defined the corresponding random experiment for P1 probabilities and its sample space.
Finally on your red/blue variation:
I believe that it is not equivalent to the original problem, where SB is NOT “always wakened on both days”. As I have explained in the second random experiment I have defined, current status upon awakening is part of the random experiment, thus adding a new potential outcome such as “Tails and blue” drastically alters its sample space.
Best regards,
17 October 2013, 2:36 amYannis
JeffJo:
yannis
> The way I have defined T2 (in the context of an explicitly defined random experiment) it is a valid event of a sample space (in fact it is an outcome of the defined random experiment).
No, it isn’t valid for this one. To be valid, it has to be a disjoint event from T1, otherwise you are defining it differently for Sunday Sleep Beauty (SSB) and Awake Sleeping Beauty (ASB).
>In axiomatic probability theory there is no such thing as atomic events.
I overstated it – all I meant was events that are not divided further for the current analysis. The problem “axiomatic theory” has is with “absolutely atomic” events. You can always introduce more random variables that make such a claim absurd.
> In your analysis you have defined T1 as “T1 = Tails and, by the end of the experiment, SB will have been awake on Monday” well this is just Tails …
Right – the point was that it isn’t “the coin *did* land on heads, or “the coin *will* land on heads,” when the tense doesn’t matter and is in fact ambiguous when you have to apply it to both SSB and ASB. It also allows the proper application of the events you ignored, the ones that only ASB can use because they do require tense – present tense.
> I have defined both random experiments (one corresponding to the experimental setup using only Heads/Tails outcomes and the other taking into account current status upon awaking), in my post on 4 July 2013, 9:06 am.
You mean the one where you said “It is obvious that the conducted random experiment (which I have explicitly defined in my previous post) consists only of a coin toss and cannot be used to generate T1 or T2 as disjoint events.” You explained the deficiency yourself. Please repeat it if you meant something else.
> On the other hand in your analysis there is only the original random experiment implied for P0 probabilities, but you have not defined the corresponding random experiment for P1 probabilities and its sample space.
I was merely comparing the results of the two arguments, without deriving them directly. The point was that the halfer argument produced an absurd result. To define both, I need to add another random variable (see “no atomic events” above) that allows SSB to distinguish Monday from Tuesday, since ASB needs to. You’ve resisted that, so I didn’t try.
> Finally on your red/blue variation:
> I believe that it is not equivalent to the original problem, where SB is NOT “always wakened on both days”. As I have explained in the second random experiment I have defined, current status upon awakening is part of the random experiment, thus adding a new potential outcome such as “Tails and blue” drastically alters its sample space.
Then your “second random experiment” is changing the problem, because ASB’s information is identical in the two.
25 October 2013, 3:05 pmIoannis Mariolis:
Dear JeffJo,
It looks like we approach probability problems very differently. Thus, instead of addressing separately each one of your remarks I will briefly present my perspective hoping it could help building some common ground. My approach is based on explicitly defining a random experiment to model the problem at hand, and constructing a corresponding sample space of possible outcomes. According to this approach there can be more than one valid model, whereas the model is chosen based on the questions you need to answer, i.e. the outcomes you are interested in. Thus, if you need to know the probability of Heads given that SB has been awakened, the following Random Experiment (RE) can be used: A coin is tossed and there are only two outcomes Heads and Tails. Thus, S={Heads,Tails} and assuming a fair coin P(Heads)=P(Tails)=1/2. It is valid to use this model because we know that no matter what is the result of the coin toss, SB will be awakened at least once.
Now let us change to a first person perspective, and ask what should SB think of the probabilities of Head once she is awakened. Well, nothing has changed. SB still knows that her awakening is a certain event she can still use the above random experiment and calculate P(Heads|awaken)=P(Heads)=1/2.
However, the interesting part with the first person perspective in this case is that you can start asking different questions regarding what day of the week it is upon awakening. The original RE cannot account for this kind of uncertainty. Therefore, you should use a different RE to model the augmented problem. The new RE should be defined in such a way that the current day that SB is awakened has been randomly selected, otherwise there is no sense in assigning probabilities (other than 1 or 0) to it. Thus, in order for SB to answer this new question she should model it like this (quoting myself from my post on 04 July 2013):
“… the original experiment is conducted and in case of a Heads toss a Monday wakening is selected as your current state, whereas in case of a Tails toss either a Monday or a Tuesday wakening is randomly selected as your current state. Then, it is straightforward that if you consider H1, T1 and T2 as the outcomes of the above random experiment you should assign P(H1)=1/2, P(T1)=P(T2)=1/4. Notice that also in the context of this random experiment P(Heads)=P(H1)=1/2 and that P(Monday)=P(H1 or T1)=3/4 and not 2/3 as implied by Elga’s analysis.”
In case it is not clear by the above description:
H1: Heads and Monday current wakening.
T1: Tails and Monday randomly selected as current wakening
T2: Tails and Tuesday randomly selected as current wakening
and as you can see T1 and T2 are disjoint events.
The new RE is consistent with the original one, but it is also more general since it can address SB’s uncertainty on the day upon awakening.
Both Elga and Lewis are mixing events of these two random experiments resulting to absurd calculations.
This is a synopsis of my approach which is typical textbook approach on applying axiomatic probability theory on this kind of problems. However, a very important detail (and the most critical part of the confusion) is that the consistency of the two REs is not obvious, since one can be easily confused about what they imply for P(Heads|Monday). The confusion derives by the difference in the definition of the “Monday” event in the two REs. In the original RE the event “Monday” is defined as “you are awaken and it is Monday”, but in the new RE it is defined as “Monday has been randomly selected as your current awakening day”.
Thus, if you learn that it is Monday you only have evidence of the “Monday” event of the original RE and not for the new RE event (because you know that your awakening has not been actually randomly selected, you have only modeled it as such in order to account for your uncertainty). Thus, in the context of the new RE, once you learn that it is Monday you cannot update P(Heads|Monday) to 2/3.
To sum up, my approach is based on the widely applied and well tested axiomatic probability theory.The REs I am employing are consistent with each other when they are used for answering the same questions. The second RE is more general than the first, since it can account for SB’s uncertainty on the day of the week upon awakening. Elga and Lewis mix events of both random experiments producing absurd results. Moreover, in my approach there is no need to differentiate between Sunday Sleeping Beauty and Awake Sleeping beauty. Both REs I used start with a SSB and result to an ASB. The fact that ASB can account for different uncertainties than SSB is addressed by using a new random experiment, and not by merely changing the employed random variables.
From your arguments it is not clear if you are objecting to the entire approach I am using or to how I apply it in the SB case. For example in your previous post you conclude saying:
>Then your “second random experiment” is changing the problem, because ASB’s information is identical in the two.
What do you mean by “changing the problem”? What is the problem according to you? Is it the calculation of the probabilities of Heads upon awakening? As I have explained both REs are addressing this problem. However, when you start calculating probabilities regarding the day of the week upon awakening you automatically augment the problem in such a way that the original RE becomes obsolete and you have to use the second RE in order to be valid. Thus, if you adopt my approach the issue is not whether the problem changes or not, but whether the corresponding RE is modified accordingly to address the new problem.
If you dismiss my entire approach I expect some arguments on why you think axiomatic probability theory (APT) should not be applied in the SB case. On the other hand if you object on the way I am applying APT on the SB problem, I am expecting some arguments addressing the formulation of the two REs I have presented and the selection of the values assigned to the calculated probabilities.
Kind regards,
29 October 2013, 7:58 amYannis
JeffJo:
> My approach is based on explicitly defining a random experiment to model the problem at hand, and constructing a corresponding sample space of possible outcomes.
To do so, you have to define events for *both* SSB and SSB that distinguish {Awake,Tails,Monday} from {Awake,Tails,Tuesday}. Any model utilizing “awake at least once” or “today is” fails to do this.
> Now let us change to a first person perspective, and ask what should SB think of the probabilities of Head once she is awakened. Well, nothing has changed.
This is an incorrect statement. Something has indeed changed. An isomorphism might help: say I have two coins; one is a fair coin, and one has “tails” on both sides. I mix them up, and hold out in each of my closed hands. You pick one – it’s tails. What are the chances the other is heads?
What you are saying about SB is equivalent to saying “since I know there will be at least one tails, I have learned nothing about the two coins, and the probability must still be 1/2.” But you have learned something. You learned about one specific coin, but don’t know which specific coin it is. This is not the same information as the “at least one will be tails” you knew before. Seeing a tails represents three of the four possible outcomes, one of which includes a heads. The answer is 1/3.
Similarly, SB has learned about one specific day, which is different than “I will be awake at least once.” So something has changed – she has information about a specific day, but not the other specific day.
> The new RE should be defined in such a way that the current day that SB is awakened has been randomly selected…
Almost correct – its backwards. Whether she is awake is determined by the randomly-determined-from-two-possibilities day.
> … the original experiment is conducted and in case of a Heads toss a Monday wakening is selected as your current state, whereas in case of a Tails toss either a Monday or a Tuesday wakening is randomly selected as your current state.
This is incorrect, as I said before. The day of your “current state” is not selected by being awake, it is *OBSERVED*. It is randomly selected to be either Monday or Tuesday regardless of the coin, and the combination of the coin+day outcomes tells you whether SB will be awake.
There are four possible outcomes of the experiment, not three. One is not observed by SB, but that does not make it a non-outcome. You should assign P(H1)=P(H2)=P(T1)=P(T2)=1/4, and now it is straightforward to see that the answer is 1/3. [And my coin experiment is really the same problem – make the “heads” a magical Medusa head, so that you will be turned to stone before realizing what side of the coin shows. Knowing you wouldn’t have observed “heads” does not affect your knowledge when you observe something else.]
> … Then, it is straightforward that if you consider H1, T1 and T2 as the outcomes of the above random experiment you should assign P(H1)=1/2, P(T1)=P(T2)=1/4
No, it isn’t. On Sunday, SB would (assuming you change your definitions of these events so she can apply them) say P(H1)=P(T1)=P(T2)=1/2. Knowing she is awake, you changed them to P(H1)=1/2, P(T1)=P(T2)=1/4. You updated the probabilities, which you can only do if you consider being awake to represent “new information,” invalidating the assumption you and Lewis both make.
> In case it is not clear by the above description:
> H1: Heads and Monday current wakening.
> T1: Tails and Monday randomly selected as current wakening
> T2: Tails and Tuesday randomly selected as current wakening
Yes, I saw your definitions. I contest that you can include the word “current” in the definition of the event as used by SSB. I contest your assignment of probabilities to them, WHICH ARE BASED ON SSB’s PRIOR ASSIGNMENTS. You didn’t define events for SSB properly, but used them to determine ASB’s probabilities based on the incorrect assumption “well, nothing has changed.”
> Both Elga and Lewis are mixing events of these two random experiments resulting to absurd calculations.
Yet you used Lewis’s “absurd calculations” when you assigned probabilities.
Let me propose a new, unambiguous RE: Quadruplets are used for SB. They are to be wakened in separate rooms on *both* days, still using the same drugs. The four events H1, H2, T1, and T2 are assigned to them randomly, but none know which they got. *One* interview is conducted on Monday, based on the coin and whoever was assigned H1 or T1. If the coin was tails, the SB who got T2 is interviewed on Tuesday. They are allowed to discuss their intended answers in advance, and are instructed to come to a consensus.
Please note that whoever is awake, but not being interviewed, at each time does not affect the knowledge of the SB who is interviewed then, or possibly on the other day. The information an interviewee has is not affected, in any way, by her sisters. Even if some, or all, are left to sleep while she is interviewed.
The probability SB1 was assigned H1 is 1/4. When she is awake, but before the interview, the probability it is Monday and she was assigned H1 is 1/8. The probability it is Monday, she was assigned H1, and the coin was Heads is 1/16. Since the quadruplets are four distinct people, the probability it is Monday, the coin was Heads, and an interview will take place is 1/4. I’ll call this event IH1; it is the event you called H1.
The same probability applies to IT1 and IT2; or your T1 and T2. The only difference for Tuesday after Heads is that an interview won’t take place; the probability it is Tuesday, the coin was heads, and an interview won’t occur is 1/4. This is an event you need to define in your RE, but you didn’t. I’ll call it NH2 (“N” for “no interview).
How is this version different than the original problem? You will want to say it is, but you can’t find a reason. By mis-defining the events, you included the probabilities for both IH1 and NH2 in your H1. That’s wrong.
The answer is P(Heads|IH1 or IT1 or IT2)=(1/4)/(1/4+1/4+1/4)=1/3. You get the wrong answer because you use P(H1)=1/2, when it is 1/4.
> To sum up, my approach is based on the widely applied and well tested axiomatic probability theory.
Sure – but applied incorrectly. Mine is based on a correct application of it. It has nothgn to do with APT itself, just your use of it.
2 November 2013, 9:49 amJeffJo:
Let me add something to that, since I know how you are going to misinterpret it. You described the analysis of your “second RE,” but you never created the steps that go into it. They are: Flip a coin. If it comes up heads, wake SB on Monday and interview her. If it comes up tails, flip another coin. If the second coin comes up heads, wake SB on Monday and interview her. If it comes up tails, wake SB on Tuesday and interview her. This is exactly what you described, and the answer is indeed P(first=heads|awake)=1/2.
What you didn’t do, was relate this to the original problem in any way. One obstacle may have been that, in trying to put it in the first person, you were concentrating on one awakening only. So you created an RE with only one awakening. But by writing out the steps, we can easily see that you did not represent the original problem. It follows these same steps, but without the second coin. SB is wakened both days after the “first” coin lands tails. So by being correct for your RE, P(heads)=1/2 cannot be right for the SB problem, which is different if the coin lands tails.
What I was saying is that your implication that it does represent the original SB problem is equivalent to saying there is no new information to be gained by being awake, which is true if there is only one new awakening. But there are two in some cases. Because there could be two, SB needs to take into account which day “today” is, and so has new information.
And your second RE proves it.
2 November 2013, 10:57 amIoannis Mariolis:
Dear JeffJo,
I have read both of your previous posts and although some interesting arguments are presented in the first one, I believe that they are not as fundamental as those in the second, where I think you have pinpointed the key part of our disagreement. Thus, I will only address the issues raised on that second post.
As perhaps you already know, probabilities are not universal truths. Take for instance the Monty Hall problem. After the door opening by the host, in case of a switch, the winning probabilities for the contestant becomes 2/3. However, if another person that has just entered the set (a passer by) and is unaware of the previous events (i.e he does not know which door has been originally chosen) is asked to choose one of the two closed doors he should use a different model and assign 1/2 probability for each door to be the winning one. I believe that you do not have any objection to the above application of APT to the Monty Hall problem and that you will justify the need for different models by the fact that there is a difference in the available information.
However, it is also perfectly valid to use different models in case the same information is available, but different questions are asked. Of course, in that case (unlike the case where available information differs) the employed models must be consistent with each other. Thus, I argue that either a change of available information or a shift in what kind of questions we need to answer, can justify the use of different models to tackle the “same” problem. Of course one can argue that when new questions are asked the problem itself changes.
In order to avoid confusion, let’s call a well posed probability problem in problem A, a ‘main problem’ of A. In case A is the SB problem posed by Elga, the main problem is “When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”, given of course the described in A experiment regarding the awakenings and the drug.
Since you know all along that you will be awakened at least once, when you are first awakened you gain no relevant information on the coin toss result. Thus, you can use the simple RE of a coin toss to model the main problem and assign 1/2 probability to Heads (and this is the only part of my analysis that is in agreement with the one of Lewis and the other mainstream halfers’ arguments). However, the confusion arises, when in order to find some arguments on the change or not in information, or other even more absurd mambo jumbo, halfers and thirders, depart from the main problem and start referring to probabilities on what day of the week it is upon awakening, or conditional probabilities of heads given Monday, etc. When, they do that, although they are referring to the same SB problem, they augment its main problem. Thus, a different RE is needed to model the augmented problem (which, as I have already said above should be also consistent with the original RE/model).
Let’s get to what I believe is the core of our disagreement. Your main objection is that we cannot use the second RE to model the SB main problem (and if I understand correctly, you also object to using the second RE to model the augmented problem as well). Let me at this point remind that according to my approach the second RE is not even needed for addressing the SB main problem anyway. However, the second RE is needed to explain the confusion in halfers’ and thirders’ arguments. Therefore, I will try to demonstrate why the second RE can be used to model the SB augmented problem. Of course, it is straightforward that in that case it can also be used to address the main problem as well.
In the beginning of your last post you are arguing:
“Let me add something to that, since I know how you are going to misinterpret it. You described the analysis of your “second RE,” but you never created the steps that go into it. They are: Flip a coin. If it comes up heads, wake SB on Monday and interview her. If it comes up tails, flip another coin. If the second coin comes up heads, wake SB on Monday and interview her. If it comes up tails, wake SB on Tuesday and interview her. This is exactly what you described, and the answer is indeed P(first=heads|awake)=1/2.”
This, is exactly my point. The second RE can and should be used to model a situation like the one you have described, which is actually an equivalent situation to the one that SB finds herself upon awakening. Of coarse, SB is always aware that even in case the first coin is tossed Tails, there is no actual second coin flip, but her lack of information on the day of the week allows for modeling her uncertainty as a second coin flip.
It is perfectly valid for her to do that. It is like the passer by in the Monty Hall problem, who can model his situation as if a coin is tossed and the winning door is selected. Even though, knowing how the game show works, he is aware that this is not how the winning door has been actually selected by the show production. He knows that it has been actually selected randomly out of three and not two doors and that the host (who knows where the prize is) has opened one of the two doors that has not been originally selected by the contestant and which does not contain the prize. The passer by just doesn’t know which door has been originally selected by the contestant and it is this uncertainty that makes him use the coin toss model.
Moreover, exactly because SB always knows that her awakening is not actually the result of a second coin flip, she can infer that upon learning the day of the week she cannot use the new RE to update her belief on the result of the first coin toss. On the other hand, SB can use the second RE upon awakening to calculate that the probability of Monday is 3/4 and of Tuesday 1/4.
You have continued arguing:
“What you didn’t do, was relate this to the original problem in any way. One obstacle may have been that, in trying to put it in the first person, you were concentrating on one awakening only. So you created an RE with only one awakening. But by writing out the steps, we can easily see that you did not represent the original problem. It follows these same steps, but without the second coin. SB is wakened both days after the “first” coin lands tails. So by being correct for your RE, P(heads)=1/2 cannot be right for the SB problem, which is different if the coin lands tails.”
Yes, I have created an RE with only one awakening, and this is the only way SB can model her uncertainty on the day of the week upon awakening, but also this is what Elga’s analysis imply. Elga (who is using a Bayesian Update (BU) approach and not an APT approach like I am) is using three predicaments (not events of a defined RE’s sample space) upon awakening:
H1: Heads and it is Monday
T1: Tails and it is Monday
T2: Tails and it is Tuesday
Apparently these real life predicaments are mutually exclusive and collectively exhaustive. Based on that, his argument on 1/3 is that P(T1)=P(T2) and P(H1)=P(T1), thus P(H1)=P(T1)=P(T2)=1/3. Thus, the mutually exclusive property is crucial in his analysis.
Therefore, if you shift from BU to an APT approach you should be able to use an RE that defines H1,T1 and T2 as mutually exclusive and collectively exhaustive events of its sample space (notice that my second RE does that). It is clear that the simple coin toss RE does not satisfy the above demands since it allows for both awakenings to occur in the same trial of the random experiment.
As you can see it is actually Elga’s analysis that begs for the second RE. Elga’s error is not that he is using the mutually exclusive events to augment the main problem (even though because he is using the BU approach he is not aware that he is actually doing that), since it is perfectly valid to do this. His error is that in his proof of P(H1)=P(T1) he is updating the probability of Heads using non existing evidence of an “H1 or T1” event. This is because he fails to identify what actually constitutes evidence for “H1 or T1” in case H1,T1 or T2 are defined as mutually exclusive events.
I cannot see how can someone accept Elga’s arguments and dismiss the RE that implicitly derives by these arguments. If you believe that you can define any other RE that is consistent with the described experiment and defines H1,T1 and T2 as mutually exclusive and collectively exhaustive events of its sample space I would be very interested in hearing about it.
In you last post you say that:
“What I was saying is that your implication that it does represent the original SB problem …”,
where ‘it’ I believe refers to the second RE I am using.
I believe that by ‘original SB problem’ you are referring to the SB main problem and by “represent” you mean that “it can be validly used to model”. Then, yes I believe that as the simple coin toss RE represents the SB main problem, so does the second RE, which however also represents the SB augmented problem, whereas both REs are consistent with each other.
Then, you continue saying:
“…is equivalent to saying there is no new information to be gained by being awake, which is true if there is only one new awakening. But there are two in some cases. Because there could be two, SB needs to take into account which day “today” is, and so has new information.”
It is clear by my analysis that there is no new information relevant to the coin toss result gained by the fact that you have been awakened. This is true in any case, whatever RE you (correctly) employ to model your situation. Actually, as it is clear by my analysis, even learning upon awakening that today is Monday does not constitute information that is relevant to the coin toss result. Only if you learn that it is Tuesday you have gained relevant information and you can update P(Heads)=0 and P(Tails)=1 in both REs. I cannot see what you mean by “…SB needs to take into account which day “today” is, and so has new information”. When SB awakes she does not know which day “today” is. Therefore, she cannot take it into account. But even so, how can taking into account something you cannot possibly know provides new information? Let alone that even if somehow you learn what day is “today”, it is not always information relevant to the coin toss result as I previously explained. Can you please explicitly describe what is the new information that according to you is gained upon awakening and why you think it is relevant to the coin toss result? Or even more rigorously how this new information could be used to update P(Heads)=1/2 to P(Heads|new relevant information)=1/3, since P(Heads|awakened)=1/3 is not straightforward and cannot be employed without some kind of proof?
Please don’t let the last part on gained information shift the focus of this post which is on the validity or not of employing the second RE for modeling SB’s situation upon awakening. I hope you can see my points on why I believe not only that the employed RE is valid, but also why it is necessary for
a) allowing SB to calculate probabilities on what day of the week it is upon awakening
b) clearing the confusion created by Elga’s analysis that is based on H1, T1, T2 mutually exclusive events
Best regards,
4 November 2013, 10:20 amYannis
JeffJo:
Yannis
You said:
> As perhaps you already know, probabilities are not universal truths.
This is a common misinterpretation of (well, maybe I should say “emphasis on the wrong consequences of”) two facts. (1) Kolmogorov never defines probability, just the properties it has in his mathematical system; and (2) You can “update” probabilities based on a change in information. Well, (1) Kolmogorov didn’t need to, since he wasn’t applying it to a specific physical process, and (2) What is changing is the entire probability space, as it applies to an incompletely-described physical process. Although the two spaces are related, they are different. In your Monty Hall example, you are comparing a sample space with three locations for the prize (one is eliminated but still a factor), to one with two.
The fact is that, although we can’t rigorously define the concept of probability so that it applies to all of the physical processes we would ever model, it does have an accepted meaning (“universal truth”) related to the relative likelihoods of events within the same physical system. The possible variations you describe are due to considering different probability spaces in different systems. If this were not true, we wouldn’t be arguing about Sleeping Beauty right now; any Kolmogorov-consistent measure would be a correct answer. That is, I could choose any Q>=0, and claim that P(Heads)=Q/(1+Q), P(Tails)=1/(1+Q) is a correct answer.
It is because we must base probabilities on the knowledge that exists, that we can conclude P(Heads)=1/3. In that knowledge set, Q represents the confidence ASB has in whether she is in the Heads path, or the Tails path. Since she can’t distinguish any of the situations in either path from each other, and there are half as many in the Heads path as in the Tails path, Q=1/2.
> Since you know all along that you will be awakened at least once, when you are first awakened you gain no relevant information on the coin toss result.
Yes, we do. We know that there are three situations that *could* apply to SB right now, that exactly one *does*, and that only one of them includes “Heads.” Before, we knew that exactly one, or exactly two, of these situations would occur to SB, depending on the coin. This is a different set of information. In fact, that’s why you claim we need two RE’s. If the information were the same, we wouldn’t.
But we also know that we have no information – I’m not calling it new or old – that would allow SB to consider any one of them differently in a model of the physical process she knows is responsible for her current plight. So she can’t assign different probabilities to any of them. Elga’s definitions apply only to what you called the second RE, not the first. Lewis’ did, but not Elga’s.
> The second RE can and should be used to model a situation like the one you have described, which is actually an equivalent situation to the one that SB finds herself upon awakening.
No, it isn’t equivalent. SB knew she would be in *both* situations after tails, but the current situation she is in can only be one of them. In what I described, you know that you are in the only one possible, but not if it is the heads path or the tails path.
You are still tacitly assuming there is no information content in the fact that SB is awake today. As long as you make that assumption, however you hide it, you will get an answer that depends on that assumption. That answer will be that P(Heads) is the value it had in the system you claim you are not modeling, but in fact are; the one from which “no information changed.” You will get P(H1)=1/2 as a result of assuming that (1) P(heads)=1/2 and that (2) there is no new information. You will then take this as confirmation (“It is clear by my analysis that there is no new information relevant to the coin toss result gained by the fact that you have been awakened”) of your assumption that there is no new information.
This is circular logic. Whether or not the result is correct, it is still a fallacy. But as I showed you, and you ignored, it is not correct. “No new information” applies to T1 and T2 as well. That means P(T1) and P(T2) should be the same as P(tails), which is absurd since the three disjoint possibilities would add up to a probability of 3/2.
The only way we can logically deduce that there is no new information, is to perform an analysis that does not pre-suppose a result you want to get. You can’t start with P(heads)=1/2. You have to start with P(heads) as an unknown, and derive it somehow from current, in-the-RE knowledge. Which is exactly what Elga did, in valid APT. His events are clearly, and properly, defined in ASB’s world only. Not SSB’s. The probabilities he uses are based on the only information she has, which is that she cannot distinguish H1, T1, and T2. *AS* *BOTH* *YOU* *AND* *ELGA* *DEFINE *THEM*.
+++++
But we aren’t going to resolve this by arguing about the information content. You are going to keep finding a new solution that expresses a probability that is based on the assumption – not fact – of “no new information”, and I am going to keep finding an illogical consequence of that solution. As long as you keep ignoring those consequences by expressing yet another such solution, we will get nowhere.
So again, I suggest the physical process with the Quadruplets, which you avoided. What answer do you think they should prepare in advance? If it is not 1/2, how is the information possessed by the one being interviewed different than ASB’s in the problem where you claim 1/2 is correct.?
Or you could address how your assumption of “no new information” is contradicted by the fact that for SSB, P(T1)=1/2, but for ASB, you claim P(T1)=1/4. Yes, I know these events aren’t defined properly. THAT’S THE POINT – you can’t use SSB’s knowledge of P(Heads)=1/2 for your second RE. The statement “There is no new information so…” does not belong in the discussion.
+++++
> I cannot see what you mean by “…SB needs to take into account which day “today” is, and so has new information”. When SB awakes she does not know which day “today” is. Therefore, she cannot take it into account. …
This is where the example you dismissed before, about the Two Child Problem, applies. Dismissing it here is another way you tacitly assume there is “no new information.”
Say I flip two identical quarters, and show you one. You see that it landed heads. What are the chances both landed heads? There are two potential solutions: (1) You know that *AT LEAST ONE* quarter landed on heads, so that one of three possible cases is two heads; or (2) you know that *THIS SPECIFIC* quarter landed heads, so one of two possible cases is two heads. But to get the second answer, 1/2, you need to take into account which quarter you see. If you “do not know which quarter it is, therefore (as you argue above) you cannot take it into account” and the answer should be 1/3.
Now say I flip a penny and a dime, and show you one. You see that it is the penny, and that it landed heads. The chances that both landed heads are clearly 1/2. But there is no difference in the information content you have, relative to the other coin, in these two experiments. I didn’t have to show you the penny, or a coin that landed heads. I merely showed you a coin; it’s even possible I know a way to distinguish the two quarters that you don’t, rendering the two experiments identical.
Knowing it is the penny in the second experiment is not what is relevant to the other coin’s state. It is the fact that it is a specific coin that matters. Ordered information is always relevant when you know information about just one specific member in the order, whether or not you know its place that order. And that is exactly the situation ASB finds herself in, if tails was flipped. She knows she is awake on a specific day, but not which. Her information about the days is the same as the information about the quarters in this example, where it matters that it is a specific coin. ASB needs to take into account that today is a specific day, and that is different information than “it is one of the two days.”
> … But even so, how can taking into account something you cannot possibly know provides new information?
You use the probability that it is each of the two specific possibilities you know it could be, in the Law of Total Probability. You may not even know those probabilities, and still be able to do this.
Call the quarter that stopped moving first Q1, and the one that stopped last Q2. You have no idea which of these quarters is the one I showed you, or how I might have used that fact in deciding which to show you. But you do know that P(show=Q1)+P(show=Q2)=1. And you also know that P(Q2=Heads|show=Q1) = P(Q1=heads|show=Q2) = 1/2. These two facts comprise your knowledge of the situation, and using them in the Law of Total Probability produces P(both=heads|show=heads)=1/2.
In the SB problem, ASB doesn’t know if it is Monday or Tuesday. She knows she is awake today, but does not know if she was/will be awake the other day. She doesn’t need to “learn what day is ‘today’” to use this information, she just does the same thing I did with the quarters (and in this case, it is easier since P(T1)=P(T2)).
6 November 2013, 3:59 pmIoannis Mariolis:
Dear JeffJo,
You are saying: “You will then take this as confirmation (“It is clear by my analysis that there is no new information relevant to the coin toss result gained by the fact that you have been awakened”) of your assumption that there is no new information”.
If I understand correctly, you think that I am using the RE results to confirm that there is no new information provided to SB upon awakening. I am not doing such thing. I believe that there is no need to confirm that no new information is provided upon awakening, since it is straightforward. SB knew all along that she would be awakened at least once during the experiment and her awakening just confirms that knowledge. It is as if upon awakening someone told SB that it is not Thursday. Well, this is no new information since SB knows it cannot be Thursday. It is the thirders that are using the alleged change in probability to confirm the assumed change in information, but I have not seen any convincing argument on that matter. Saying that SB’s awakening provides to her more information than learning it is not Thursday is absurd and totally unjustified, thus leaving thirders only with circular arguments. We are talking about information, thus, it should be apparent exactly what this new information is before we can use it to update probabilities. When SB gets the information that “she is awake now”, since she doesn’t know to which day of the experiment ‘now’ corresponds, it is equivalent to the information that she is awake a day during the experiment. But even in Sunday she knew that no matter the result of the coin toss she will be awaken at least one day during the experiment.
On the contrary, since Elga is based on wrong application of Bayesian update (regarding what happens if SB learns that the day of the week is Monday) to infer that upon awakening P(Heads) should be updated to 1/3, thirders assume a mysterious change in available information must be the reason. However, my arguments against Elga’s analysis are not based on any ‘no new information assumption’, since I am using the second RE that directly applies upon SB’s situation when awakened and I am using well defined events of its sample space to dismiss Elga’s erroneous proof.
To sum up, information has an existence on its own, beyond probabilities and its application in their calculations. First you have to determine what is the new information and then you can use it (if it is relevant) to update probabilities. It is clear to me that no new information is available and that at best what you are implying is that somehow upon awakening SB can exploit the available information (that she would be awaken during the experiment) differently than the way she could on Sunday, which is also absurd. Thus, if you want to continue arguing in favor of new information, you should first provide an explicit definition of what is this new information in the form of a fact, then explain why it is different from what SB knew on Sunday and why it is relevant with the coin toss and then demonstrate how it can be used to update P(Heads).
What is the new fact that SB learns upon awakening? According to you :”We know that there are three situations that *could* apply to SB right now, that exactly one *does*, and that only one of them includes “Heads.” Before, we knew that exactly one, or exactly two, of these situations would occur to SB, depending on the coin.”.
Why according to you the fact that ‘wakening in the middle of the experiment will bring SB to one of these three situations’ constitutes new information for SB, since she already knows this fact on Sunday also? I believe you are confusing ‘new information’ with ‘new situation’ and I expand on that while responding to the rest of your argument:”…This is a different set of information. In fact, that’s why you claim we need two RE’s. If the information were the same, we wouldn’t.”
As I have clearly explained in my previous post the need for using a different RE does not always arise from a change in available relevant information, but also from the introduction of new uncertainties. My thesis is that when SB wakes up she has not gained or lost any of the information she had on Sunday night. According to my understanding the change in her situation only allows for the introduction of a new uncertainty, i.e. whether it is Monday or Tuesday. Not knowing which of the two days of the week it is upon awakening by no means provides any new information (or deletes any of the information that was available to SB on Sunday night). Therefore, the second RE is necessary only if SB wants to model her uncertainty on the day of the week upon awakening (I am using it however because it is relevant to Elga’s arguments). Thus, although SB’s situation has changed, her available information hasn’t. However, her new situation introduces more uncertainties that can only be addressed if we augment the probability space using the second RE.
You are also arguing:
“This is where the example you dismissed before, about the Two Child Problem, applies. Dismissing it here is another way you tacitly assume there is “no new information.”
Say I flip two identical quarters, and show you one. You see that it landed heads. What are the chances both landed heads? There are two potential solutions: (1) You know that *AT LEAST ONE* quarter landed on heads, so that one of three possible cases is two heads; or (2) you know that *THIS SPECIFIC* quarter landed heads, so one of two possible cases is two heads. But to get the second answer, 1/2, you need to take into account which quarter you see. If you “do not know which quarter it is, therefore (as you argue above) you cannot take it into account” and the answer should be 1/3.”
Let me make clear from the beginning that I consider the above example totally irrelevant to SB’s situation upon awakening or to the existence or not of new information. Nevertheless, I will comment on your example trying to analyze what is going on with these quarters. First of all, the distinction between *AT LEAST ONE* quarter and *THIS SPECIFIC* quarter is not informative enough to dictate how the probability of the undisclosed result should be calculated. However, some assumptions can be employed based on what common sense suggests and this is why the two cases are treated differently. Thus, we usually interpret *AT LEAST ONE* quarter as if the result of both tosses is checked and as long as at least one of the quarters landed Heads the truth of the proposition is validated (under this assumption the 1/3 result is correct). On the other hand, *THIS SPECIFIC* quarter is usually interpreted in a way that suggests that the obtained information is firmly restricted to the result of a specific toss (under this assumption the 1/2 result is correct). However, if someone checked only one of the tosses and it happened to be Heads, he could correctly inform you that *AT LEAST ONE* quarter tossed Heads. But then, if you knew that he only checked one you should calculate P(Heads of the other quarter)=1/2. Similarly, if someone checked both quarters and on purpose picked one that it is Heads, he could correctly inform you that *THIS SPECIFIC* quarter landed Heads. But then, if you know what he did you should calculate P(Heads of the other quarter)=1/3. It is upon the selection procedure and not the phrasing that calculations should be based. The important factor is whether the toss results are independent of one another or not. In case only one toss result is checked it is clear that the other toss result is not correlated to this result at all (thus, it all comes down to a single toss of a fair coin and P(Heads) is apparently 1/2) . However, if both toss results are checked and the quarter selection is based on both results then the results become correlated and learning what one of them is provides information that can be used to update the probability for the other (to 1/3 in the case of the two Quarters).
In the penny/dime variation nothing changes. Before calculating probabilities, we still need to know how ‘showing the penny result’ has been decided. The penny selection could have been random, or it could be predefined or even decided after the toss, it doesn’t matter as long as it is not based on the result of the other coin toss. If we know that, we can safely calculate P(dime=Heads)=1/2. Since you state that “I didn’t have to show you the penny, or a coin that landed heads” it looks like you are randomly selecting one of the coins without bothering about the toss results. Thus, the above calculation makes perfect sense. However, the way the penny is selected is the reason it does make sense to calculate P(dime=Heads)=1/2 and not the mere fact that ‘we are shown the penny which is a “specific” coin’.
You are arguing that:
“Knowing it is the penny in the second experiment is not what is relevant to the other coin’s state. It is the fact that it is a specific coin that matters”.
I believe it is clear from my analysis above that ironically ‘specific’ is not specific enough and that what is directly relevant to the probability calculation is whether the result you are informed about is independent or not from the other toss result.
Let me now explain why I can’t see the analogy between your example and SB’s situation. You argue that:” And that is exactly the situation ASB finds herself in, if tails was flipped. She knows she is awake on a specific day, but not which. Her information about the days is the same as the information about the quarters in this example, where it matters that it is a specific coin. ASB needs to take into account that today is a specific day, and that is different information than “it is one of the two days” “.
In your quarters example the information is that a specific quarter has landed Heads. In SB’s case the information is that a specific day has been selected for her current awakening. This directly implies that the two quarters correspond to Monday and Tuesday respectively and that Heads correspond to ‘this specific day has been selected as your current awakening’ (which also suggests that Tails correspond to ‘this specific day has NOT been selected as your current awakening’). However, unlike the two quarters case which can both land Heads (or Tails), in SB case the two days are anti-correlated since if one is selected as current day the other obviously isn’t. This means that as soon as SB learns that this specific day is selected (translating to Heads in your analogy), she knows that the other day is not selected (translating to Tails for the other quarter in your analogy).
Can you please explain what you mean by: “In the SB problem, ASB doesn’t know if it is Monday or Tuesday. She knows she is awake today, but does not know if she was/will be awake the other day. She doesn’t need to “learn what day is ‘today’” to use this information, she just does the same thing I did with the quarters (and in this case, it is easier since P(T1)=P(T2)).”
What did you do with the quarters? You simply assumed independence between the coin toss results (and ‘specific’ is used for justifying exactly that). How is the quarters example useful in SB’s situation, since in the quarters case you are provided with actual new information (i.e. that ‘a specific quarter landed Heads’), whereas in SB’s case you learn that you have been ‘awakened on a specific day’ which you always new that it would happen. Moreover, in SB’s case unlike in the quarters’ case ‘current awakening on Monday’ is anti-correlated with ‘current awakening on Tuesday’, which translates in the quarters example that a Heads-Heads or Tails –Tails result is excluded.
Best regards,
8 November 2013, 8:08 amYannis
JeffJo:
> If I understand correctly, you think that I am using the RE results to confirm that there is no new information provided to SB upon awakening. I am not doing such thing. I believe that there is no need to confirm that no new information is provided upon awakening, since it is straightforward.
I’m not saying that is your purpose, I’m saying you find enough confirmation in the fact that P(heads) doesn’t change that you refuse to reconsider how it is not straightforward. So all of your conclusions are circular.
When I solve the problem, I do not assume the information is the same, or that it has changed. I only assume that I can’t quantify any such change, present or not, at the start. So I don’t assume there is any information carryover from “before,” as you do. I can only base probabilities on the current information, whatever that may be. I have tried to explain how, after we find that we do end up changing the probabilities from this starting point, we can see where the change in information came from – and that we should have expected it (i.e., it is not “straightforward”) from the start. I have also tried to get you to stop making the not-straightforward assumption that keeps producing the same circular results.
> It is the thirders that are using the alleged change in probability to confirm the assumed change in information, but I have not seen any convincing argument on that matter.
No, they don’t. They do not need to assume that any probabilities of observable events changed. They also do not need to assume that any stayed the same. They only need to assume that they have a new probability space, well-defined in APT. This new probability space has two well-defined random variables: today is Monday or Tuesday, and by the end of the experiment the coin will have landed Heads or Tails.
ASB does not need to assume that any probabilities changed, or stayed the same, because she does not need to assume any values for these probabilities (let alone values based on “before” information that may, or may not, may have changed). She only knows that, given “Today is Monday,” she has no information that allows her to assign different probabilities to Monday&Heads or Monday&Tails. She doesn’t assign probabilities to them – “changed” or “the same” as whatever state your “no new information” compares them to – she only applies the Principle of Indifference to them in her current probability space.
She also knows that, given “The coin will have landed Tails,” she has no information that allows her to assign different probabilities to Monday&Tails or Tuesday&Tails. So once again, she applies the Principle of Indifference – still not requiring any assumptions about information that may or may not have changed.
But now, taking these three well-defined-in-APT events into account, and still not having compared them any other probability space, she knows that they form a partition of her current sample space. Since they all must have the same probability in her current probability space, and APT says the probabilities of a partition must sum to 1, each is now *CALCULATED* to be 1/3. This is not based on a change in information. It is also not based on “new information” in any way. It is based on her current information alone.
And I have never seen anybody argue that there is a wrong assumption in that. Lewis certainly didn’t – he even agreed with the PoI as Elga applied it. He disagreed with a premise he incorrectly attributed to Elga; that P(Heads|Monday)=1/2. That isn’t what Elga used as a premise.
Elga said that for a coin which is going to be flipped on Tuesday, that P(Heads|Monday) ought to be the same as P(Heads|Sunday) (which is 1/2). Yes, this has poorly defined events, but it was an ancillary deduction, not a premise. The premise was P(Heads|Monday)=P(Tails|Monday), which has well-defined events, from which we can deduce that P(Heads|Monday)=1/2, just like P(Heads|Sunday), and for the same reasons. That’s what Elga said. Lewis disagreed with the incorrectly-attributed premise P(Heads|Monday)= P(Heads|Sunday), but never addressed whether P(Heads|Monday)=P(Tails|Monday) should be accepted or not.
> Saying that SB’s awakening provides to her more information than learning it is not Thursday is absurd and totally unjustified, thus leaving thirders only with circular arguments.
Since no Thirder says anything at all about Thursday, this is a non sequitur. All they say is that ASB knows that it is Monday, or that it is Tuesday, and this knowledge set contains no information about the other day (see more below). She is not comparing these events to anything SSB would have considered as part of her probability space on Sunday, making for circular arguments. Only Halfers do that.
> On the contrary, since Elga is based on wrong application of Bayesian update …
Elga did not apply a Bayesian update, so he could not possibly have applied one incorrectly. All Elga said is that being in H1 is subjectively equivalent to being in T1, and that being in T1 is subjectively equivalent to being in T2.
> … (regarding what happens if SB learns that the day of the week is Monday) …
What you refer to here may be where Elga states that conditional probabilities exist for events that are well-defined in a valid APT-based system, and a condition C in that same system. These statements are not based on any form of Bayesian updating. They are from the direct application of the definition of conditional probability, P(A|C)=P(A&C)/P(C). It’s even a degenerate case of that definition, where C=A or B for disjoint events A and B. All he used it for was to show, without “updating,” that P(A|A or B) = P(A)/(P(A or B). Do you think that is a misapplication of anything? From that, since we also know P(A|A or B)=P(B|A or B), we can deduce P(A)=P(B). Do you think that is a misapplication of anything?
(Later, about my quarters experiment)
> However, if someone checked only one of the tosses and it happened to be Heads, he could correctly inform you that *AT LEAST ONE* quarter tossed Heads. But then, if you knew that he only checked one you should calculate P(Heads of the other quarter)=1/2.
Exactly. This is exactly the situation SB finds herself in. The information she has is based on checking only one day, and knowing that only one day has been checked. My point was that it doesn’t matter whether you can tell which of the two quarters/days was checked, just that you know that one, and only one, was. Thank you for expressing it in terms you can accept.
When you use “she knows she will be awake at least once” in place of “only one day has been checked in the current information state,” you are (1) basing it on the definitions in RE1, not RE2, and (2) checking both days. White’s calculation, which you claimed to accept, does that by saying P(W)=1-P(~W1)*P(~W2). If P(W) was properly accounted for using the “only one checked” model, he’d correctly get 1/3 for the probability in his experiment.
Try it this way: Four coins are flipped Sunday night. The quarter defines the sleep schedule. ASB will be told the result of either the penny, or the dime, but not both and not which coin it is. The nickel determines if ASB is told about the penny on Monday and (potentially) the dime on Tuesday, or vice versa. When ASB is told a penny-or-dime result, what is the probability both landed the same? As you just argued, if she knows that only one was checked, she should say 1/2. If she thinks both were, she should say 1/3.
Or to see it another way, stop dismissing my variation where SB is always wakened and drugged; but she is not interviewed in T2. You dismissed it on the basis of checking two days in the variation, but one-or-two in the original. But her information content in either case includes the fact that only one day is checked, so how – or if – the other is checked is irrelevant. The two experiments are equivalent.
Or consider the isomorphism of the quadruplets, who know that only one of them will be wakened in the experiment and they must come to a collective agreement on the answer. The only difference from the original problem is that the one combination that is checked is determined by checking only one, instead of drugging the information about the checks out of SB.
12 November 2013, 3:55 pmIoannis Mariolis:
Dear JeffJo,
I disagree with most of the comments on my quotes, which you are using selectively and sometimes out of context. (I don’t think this is done on purpose. I just think that because of our different perspectives we give emphasis to different things). Although I don’t think that determining, whether you call what Elga is using Bayesian updating or not, or whether he refers to new information or not, is crucial for our cases, in order to restore the truth on these matters:
Elga wrote on his 2000 paper “Self-locating belief and the Sleeping Beauty
problem”:
A.”In contrast (assuming that you update your beliefs rationally), when you are awakened
on Monday, you count your current temporal location as relevant to the truth of H:
your credence in H, conditional on its being Monday, is 1/2, but your credence in H, conditional
on its being Tuesday, is 0″
B.”At the start of the experiment, you had credence 1/2 in H. But you were also certain
that upon being awakened on Monday you would have credence 1/3 in H—even though
you were certain that you would receive no new information and suffer no cognitive
mishaps during the intervening time.”
As you can see by A., in contrast to how you are interpreting what he is doing, it is Elga himself that calls what he is doing “updating of beliefs”. Moreover, in his proof he had stated:
“But your credence that the coin will land Heads (after learning that it is Monday) ought to be the
same as the conditional credence P(H1|H1 or T1). So P(H1|H1 or T1) = 1=2, and hence
P(H1) = P(T1).” . Well, this directly implies that according to Elga, once you learn that it is Monday you should use P(H1|H1 or T1) to update your credence on H1 based on this new evidence (which as I have demonstrated with my second RE it is wrong. SB cannot use Monday as evidence for an “H1 or T1” event, since this event implies a random selection of the awakening day, which SB knows that never happens). Technically speaking, Elga is not directly applying Bayesian update, since he knows that SB does not learn upon awakening that it is Monday. However, what is important is that his proof is based on Bayesian update and the evidence involved in that hypothetical updating (i.e. the updating based on the hypothesis that SB learns that it is Monday) does not justify updating SB’s credence on P(H1) using the P(H1|H1 or T1) conditional probability.
Moreover, as you can see by B. it is also Elga that is referring to the reception or not of “new information”. Although, he is actually suggesting that in fact no new information is received, thus a new
variety of counterexample to Bas Van Fraassen’s ‘Reflection Principle’ has been provided. Despite of your decision to not call new or old the “different set of information”, the fact remains that an agent had some information on Sunday and if she has a different set of information on Monday she either received new information or lost some of the original.
However, back to your comments:
I said: “However, if someone checked only one of the tosses and it happened to be Heads, he could correctly inform you that *AT LEAST ONE* quarter tossed Heads. But then, if you knew that he only checked one you should calculate P(Heads of the other quarter)=1/2.”
You comment:”Exactly. This is exactly the situation SB finds herself in. The information she has is based on checking only one day, and knowing that only one day has been checked. My point was that it doesn’t matter whether you can tell which of the two quarters/days was checked, just that you know that one, and only one, was. Thank you for expressing it in terms you can accept.”
You have not considered the beginning of my analysis on the two quarters example where I state that: “Let me make clear from the beginning that I consider the above example totally irrelevant to SB’s situation upon awakening or to the existence or not of new information” and I have argued why I do so in:
“Let me now explain why I can’t see the analogy between your example and SB’s situation. You argue that:” And that is exactly the situation ASB finds herself in, if tails was flipped. She knows she is awake on a specific day, but not which. Her information about the days is the same as the information about the quarters in this example, where it matters that it is a specific coin. ASB needs to take into account that today is a specific day, and that is different information than “it is one of the two days”.
In your quarters example the information is that a specific quarter has landed Heads. In SB’s case the information is that a specific day has been selected for her current awakening. This directly implies that the two quarters correspond to Monday and Tuesday respectively and that Heads correspond to ‘this specific day has been selected as your current awakening’ (which also suggests that Tails correspond to ‘this specific day has NOT been selected as your current awakening’). However, unlike the two quarters case which can both land Heads (or Tails), in SB case the two days are anti-correlated since if one is selected as current day the other obviously isn’t. This means that as soon as SB learns that this specific day is selected (translating to Heads in your analogy), she knows that the other day is not selected (translating to Tails for the other quarter in your analogy).”.
The above, is a key argument on our disagreement and you are not addressing it at all.
Therefore, I cannot understand why you insist that I should accept your erroneous analogy.
What your analogy implies is that upon awakening what you check for is whether this specific day has been selected for your current awakening or not. But this directly implies that the other day hasn’t been selected (in case of course there is another day, which you do not know anyway since the experimenter’s coin could have been tossed Heads). Thus, it is as if you are checking both quarters and the only admissible results are Heads-Tails, Tails-Heads. Therefore, the SB and the Quarters situation are not equivalent (even in case the experimenter’s coin has been tossed Tails).
I believe that you should clarify what exactly do you mean by “check”. As I said before your Quarters example implies that by “check” you mean that:”you check for whether this specific day has been selected for your current awakening or not.”. As you can see if you use my interpretation upon awakening SB actually checks on both days. If by check you mean something else please define it explicitly and demonstrate how it is related to the Quarters example.
You are arguing:
“When you use “she knows she will be awake at least once” in place of “only one day has been checked in the current information state,” you are (1) basing it on the definitions in RE1, not RE2, and (2) checking both days”
Exactly. The “she knows she will be awake at least once” argument I used was to show why I believe that no new information is received upon awakening and it does (as it should) apply to the RE1 setup, which is therefore valid even after awakening. I am using RE2 not because it is necessary for calculating P(Heads) upon awakening, but because Elga’s analysis is implicitly based on that. Moreover, as I argue above, according to what is implied for the meaning of “check” by your misleading analogy, both days are checked upon awakening.
I am looking forward to your response.
Kind regards,
Yannis
PS. I will address your 4 coins an quadruplets examples in a different post, after concluding our discussion on the above issues and the Quarters example.
13 November 2013, 6:10 amJeffJo:
> As you can see by A., in contrast to how you are interpreting what he is doing, it is Elga himself that calls what he is doing “updating of beliefs”.
That statement was made after he had derived the result P(H|Awake)=1/3 totally within the APT-valid probability model of an Awake Sleeping Beauty (ASB), without referring to such updates. It was used to describe the change that must have occurred when you compare ASB’s model to SSB’s. So you are the one taking it out of context.
> “But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1|H1 or T1). So P(H1|H1 or T1) = 1/2, and hence P(H1) = P(T1).” . Well, this directly implies that according to Elga, once you learn that it is Monday you should use P(H1|H1 or T1) to update your credence on H1 based on this new evidence …
No, it does not say anything about updating. It enumerates two potential conditional probabilities within the ASB model (not an update from SSB to ASB, as you seem to imply), and says (1) that they must have the same value, and (2) that they represent all of the possibilities, so (3) their value must be 1/2. It doesn’t use Bayes Theorem to evaluate a probability from one model to another, which is the usual definition of Bayesian Updating.
What it does use, is the definition of a conditional probability: P(A|B)=P(A&B)/P(B). Not an update. And really it is doing so only to address P(A&B). The paragraph works just as well if written: “if (upon awakening) you were to consider the event MONDAY, that would amount to considering either H1 or T1. Your credence for HEADS&MONDAY should be the product of your credence for MONDAY and your credence that a fair coin, to be tossed independently of the event MONDAY, will land Heads. It is irrelevant that your ability to observe the event TUESDAY depends on COIN. Your credence in HEADS&MONDAY ought to be the same as your credence in TAILS&MONDAY. Hence P(H1) = P(T1).”
> which as I have demonstrated with my second RE it is wrong.
Only by assuming there was “no new information,” which itself is wrong. So any conclusion from it are invalid.
> SB cannot use Monday as evidence for an “H1 or T1″ event, since this event implies a random selection of the awakening day, which SB knows that never happens).
And here is where we get to the gist. “Random” does not mean “selected in an arbitrary manner,” or whatever you mean by “implies a random selection.” A random variable is any quantity that is unpredictable in an instance of a process. Since the experimenters deny ASB the knowledge of what day it is, in her world DAY is a just as much a random variable as COIN. Depending on how you want to utilize them, there are three or four events she can enumerate: T1, T2, H1, and maybe H2 (it happens, but she can’t observe it – whether that is relevant depends on how you utilize the fact that she is observing one specific event when she is asked for a probability).
> … what is important is that his proof is based on Bayesian update …
No, it isn’t.
> Moreover, as you can see by B. it is also Elga that is referring to the reception or not of “new information”.
He is deducing that “new information” exists from what the fact that he determined a probability change without assuming anything about “new information.”
> the fact remains that an agent had some information on Sunday and if she has a different set of information on Monday she either received new information or lost some of the original.
My point exactly. Lewis, White, and all halfers (including you) assume there is no new information relevant to COIN, and use that to assign probabilities in her RE. If there is such “new information,” or even if the possibility exists that there is, you can’t assign a value to it directly. Which you do.
> However, unlike the two quarters case which can both land Heads (or Tails), in SB case the two days are anti-correlated since if one is selected as current day the other obviously isn’t.
And you are comparing different things. If one quarter is shown to you, then obviously the other isn’t. Both systems have this property you describe, and it is irrelevant because it is treating the wrong quantity as the random variable. “Selection” isn’t what is random in my point, it is the value of whatever is “selected.” That has no analogy for SB, because the day isn’t selected. It is random (unknown and unpredictable) to SB, but not the experimenters.
> The above, is a key argument on our disagreement and you are not addressing it at all.
It is an artificial point that has no bearing on our disagreement.
> Exactly. The “she knows she will be awake at least once” argument I used was to show why I believe that no new information is received upon awakening …
And you are dismissing the fact that “it is only one day” is new information based on invalid reasoning.
+++++
Again, the problem can be solved without reference to “new information” or “updating.” It works with, and without, White’s “wake up machine.”
1) There are two random variables, COIN={HEADS,TAILS} and DAY={MONDAY,TUESDAY}.
14 November 2013, 3:36 pm2) There are three disjoint, observable events, {HEADS&MONDAY, TAILS&MONDAY, TAILS&TUESDAY}.
3) TAILS&MONDAY is subjectively equivalent to HEADS&MONDAY. The PoI says they must have the same probability.
4) TAILS&MONDAY is subjectively equivalent to TAILS&TUESDAY. The PoI says they must have the same probability.
5) By the transitive property, all three events must have the same probability. By axiom, their probabilities must sum to 1. Therefore, their probabilities must each be 1/3.
6) The event HEADS is equivalent to the event HEADS&MONDAY. Its probability is 1/3.
Ioannis Mariolis:
Dear JeffJo,
Elga wrote:“But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1|H1 or T1). So P(H1|H1 or T1) = 1/2, and hence P(H1) = P(T1).” .
I have commented: “Well, this directly implies that according to Elga, once you learn that it is Monday you should use P(H1|H1 or T1) to update your credence on H1 based on this new evidence”,
and you replied:”No, it does not say anything about updating. It enumerates two potential conditional probabilities within the ASB model (not an update from SSB to ASB, as you seem to imply), and says (1) that they must have the same value, and (2) that they represent all of the possibilities, so (3) their value must be 1/2. It doesn’t use Bayes Theorem to evaluate a probability from one model to another, which is the usual definition of Bayesian Updating.”
Can you provide any references to the usual definition of Bayesian Updating? I cite the
Bayes’ theorem
(Mathematics & Measurements / Statistics) Statistics the fundamental result which expresses the conditional probability P(E/A) of an event E given an event A as P(A/E).P(E)/P(A); more generally, where En is one of a set of values Ei which partition the sample space, P(En/A) = P(A/En)P(En)/Σ P(A/Ei)P(Ei). This enables prior estimates of probability to be continually revised in the light of observations.
It is clear from the above definition that Bayesian updating is using Bayes’ theorem to revise prior probabilities in light of new observations. Thus, the “from one model to another” constraint you are introducing is not implied. Nevertheless, I don’t think this is crucial to your point. I agree that Elga compares P(Heads|Monday) to P(H1|H1 or T1), so he indeed does as you say: “enumerates two potential conditional probabilities within the ASB model (not an update from SSB to ASB, as you seem to imply), and says (1) that they must have the same value, and (2) that they represent all of the possibilities, so (3) their value must be 1/2.”
But I do not imply “an update from SSB to ASB” as you think. All I say is that by doing the above Elga directly implies that the prior probability of P(H1) should be updated to 1/2 (using your terminology within the ASB model, using mine within the RE2 framework) based on the evidence that today is Monday, using the conditional probability P(H1|H1 or T1) . Then, he is using this result (in conjunction to the fact that H1, T1, T2 are mutually exclusive and collectively exhaustive events) to prove that the prior probability for H1 should be 1/3.
My argument against Elga is that even if SB gets evidence that today is Monday she cannot use the conditional probability P(H1|H1 or T1) to update the prior probability of P(H1). In case you define (as Elga’s does) the H1, T1, T2 events as mutually exclusive and collectively exhaustive you directly imply that a T1 event should occur, without a T2 event ever occurring. But in case you learn that it is Monday and Tails, if you consider that as a T1 event, you should also believe that if tomorrow you learn that it is Tuesday and Heads you should also consider it a T2 event. This line of thought is incompatible with the mutually exclusive property of T1, T2. Thus, it should be dismissed and Monday should not be considered evidence for an ‘H1 or T1’ event.
However, if you still object to calling what Elga is implying Bayesian update, the main issue remains and is the following:
evidence of Monday is not equivalent to evidence of an “H1 or T1” event , thus P(Heads|Monday) and P(H1|H1 or T1) must not necessarily have the same value.
Best regards,
15 November 2013, 5:08 amYannis
JeffJo:
> Elga wrote:“But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H1|H1 or T1). So P(H1|H1 or T1) = 1/2, and hence P(H1) = P(T1).”
Yes, he did. I explained how that is not an update, it is merely the definition of a conditional probability so he can address the union of the two events HEADS and MONDAY.
> Can you provide any references to the usual definition of Bayesian Updating?
From Wikipedia: “Bayesian inference (Redirected from Bayesian update): In statistics, Bayesian inference is a method of inference in which Bayes’ rule is used to update the probability estimate for a hypothesis as additional evidence is acquired.”
Bayes’ rule involves actual evidence that re-defines the sample space. Elga merely divided the unchanging sample space into cases that he could interpret directly. Bayes’ Rule involves exchanging the “evidence” and “consequence” in the conditional probability to achieve the result. Elga did not do that.
> I cite the (Mathematics & Measurements / Statistics) Statistics the fundamental result which expresses the conditional probability P(E/A) of an event E given an event A as P(A/E)*P(E)/P(A); …
Yep – see the exchange of evidence and consequence? Did Elga do that?
> I don’t think this is crucial to your point.
I agree – but it is showing how you are being argumentative, instead of addressing the point.
> All I say is that by doing the above Elga directly implies that the prior probability of P(H1) should be updated to 1/2 (using your terminology within the ASB model, using mine within the RE2 framework) based on the evidence that today is Monday, using the conditional probability P(H1|H1 or T1).
No, he isn’t. He is saying that there are only two possibilities given “H1 or T1”, that these two possibilities are subjectively equivalent, and so their prior probabilities must be equal. His point is the last part of what you quoted, “hence P(H1) = P(T1).” He comments *INCIDENTALLY* that this means the posterior probabilities are 1/2, if such evidence were obtained. And arguably he shouldn’t have, but that was not his point and he likely wasn’t expecting your level of nitpicking when he stated an obvious, yet incidental, fact.
> My argument against Elga is that even if SB gets evidence that today is Monday she cannot use the conditional probability P(H1|H1 or T1) to update the prior probability of P(H1).
Yes, she can. Elga is not implying “that a T1 event should occur, without a T2 event ever occurring.” He is implying that the observation of the T1 event occurs without evidence of an observation of the T2 event, because the knowledge of any other observation is denied to SB. This whole problem is a temporal paradox that you have to get past, not use in contradictory manners for different facets of the problem. Which you do, when you say P(T1)+P(T2)=P(TAILS). This implies exactly what you said shouldn’t be implied.
> evidence of Monday is not equivalent to evidence of an “H1 or T1″ event , thus P(Heads|Monday) and P(H1|H1 or T1) must not necessarily have the same value.
I’m not talking about evidence, and Elga only did so as an easy way to express the real concept. Which is that H1 is subjectively equivalent to T1, because they are differentiated only by a binary process whose two outcomes are themselves subjectively equivalent.
+++++
Again, the problem can be solved without reference to “new information” or “updating.” It works with, and without, White’s “wake up machine.”
1) There are two random variables, COIN={HEADS,TAILS} and DAY={MONDAY,TUESDAY}.
16 November 2013, 7:07 am2) There are three disjoint, observable events, {HEADS&MONDAY, TAILS&MONDAY, TAILS&TUESDAY}.
3) TAILS&MONDAY is subjectively equivalent to HEADS&MONDAY. The PoI says they must have the same probability.
4) TAILS&MONDAY is subjectively equivalent to TAILS&TUESDAY. The PoI says they must have the same probability.
5) By the transitive property, all three events must have the same probability. By axiom, their probabilities must sum to 1. Therefore, their probabilities must each be 1/3.
6) The event HEADS is equivalent to the event HEADS&MONDAY. Its probability is 1/3.
Ioannis Mariolis:
Dear JeffJo,
If T1 and T2 events are defined (as Elga does) as mutually exclusive they cannot be both outcomes of the same trial of the RE that can be used to model SB’ situation upon awakening (whatever random experiment one uses, even if you dismiss the one I have proposed). This implies that learning it is Monday does not provide evidence that an ‘H1 or T1’ event has occurred, for the reasons I have presented in detail in previous posts. However, if you do not like the evidence centered terminology I use, or the fact that my analysis also refers to why you shouldn’t update P(H1) from 1/3 to 1/2 once you learn that it is Monday, but instead your prior should be 1/2 on the first place and your posterior should also remain 1/2 when you learn that it is Monday, feel free not to take it into account. The key argument is that according to my analysis the event “SB wakes up and it is Monday” is not equivalent to the event “H1 or T1” and Elga’s assignment of 1/2 to P(H1|H1 or T1) is invalid. In fact according to the model I propose (which is based on the fact that T1 and T2 are mutually exclusive), P(H1|H1 or T1)=2/3 (since P(H1)=1/2 and P(T1)=1/4).
Regarding the solution you propose based on PoI, I suggest examining how is Elga applying PoI (correctly according to me) in case of T1 and T2:
“If (upon first awakening) you were to learn that the toss outcome is Tails, that would
amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively
just like being in T2, and since exactly the same propositions are true whether you are in
T1 or T2, even a highly restricted principle of indifference yields that you ought then to
have equal credence in each. But your credence that you are in T1, after learning that the
toss outcome is Tails, ought to be the same as the conditional credence P(T1jT1 or T2), and
likewise for T2. So P(T1jT1 or T2) = P(T2jT1 or T2), and hence P(T1) = P(T2).”
Thus, premise number 4) of your proposed solution is correct. But if you apply PoI to number 3) you get:
If (upon first awakening) you were to learn that it is Monday, that would
amount to your learning that you are in either H1 or T1. Since being in H1 is subjectively
just like being in T1, and since exactly the same propositions are true whether you are in
H1 or T1, even a highly restricted principle of indifference yields that you ought then to
have equal credence in each. But your credence that you are in H1, after learning that it is Monday, ought to be the same as the conditional credence P(H1|H1 or T1), and likewise for T1. So P(H1|H1 or T1) = P(T1|H1 or T1), and hence P(H1) = P(T1).”
And I am surprised why Elga is not using the above (erroneous according to me) proof, since he already did it (correctly ) for proving P(T1)=P(T2). However, even if he did so, the key issue remains. He still has to use the (invalid according to my analysis) premise that “your credence that you are in H1, after learning that it is Monday, ought to be the same as the conditional credence P(H1|H1 or T1)”.
I admit that if one uses Elga’s approach and refer to H1 and T1 as predicaments it almost comes natural not to ever question this premise. However, if you start to ask what random experiment can result to T1, T2 events that are mutually exclusive then it is directly implied that ‘H1 or T1’ and Monday suggest different things.
Elga also proposed a second argument in favor of 1/3:”1/3, of course! Imagine the experiment repeated many times. Then in the long run, about 1/3 of the wakings would be Heads-wakings — wakings that happen
on trials in which the coin lands Heads. So on any particular waking, you should have
credence 1/3 that that waking is a Heads-waking, and hence have credence 1/3 in the
coin’s landing Heads on that trial. This consideration remains in force in the present
circumstance, in which the experiment is performed just once.”
As you can see even though in his first argument he defined T1 and T2 as mutually exclusive events, in this argument in every trial he counts both awakenings which means that he treats T1 and T2 as if they are not mutually exclusive events (and this is exactly the problem with both of his arguments, although he (correctly) defines T1 and T2 as mutually exclusive events, he keeps using premises that violate this definition).
Kind regards,
16 November 2013, 9:10 amYannis
JeffJo:
Yannis:
> If T1 and T2 events are defined (as Elga does) as mutually exclusive they cannot be both outcomes of the same trial of the RE that can be used to model SB’ situation upon awakening (whatever random experiment one uses, even if you dismiss the one I have proposed).
Elga defines them as observations occurring on that day, and with that coin result. Literally, T1 means “TAILS and it is Monday.” He does not define them as the coin result coinciding with an instance of the day, as you incorrectly say he does. The “trial” is a single observation, taken in isolation from the other possible observations. T1 and T2 can be, and are, disjoint events in an RE based on observation.
For example, take away the coin, and drug-and-wake SB all five days from Monday to Friday. When interviewed, SB’s credence that it is Wednesday is 20% (Elga’s definition), but 100% that an interview will take place on Wednesday (what you claim is his definition). In the actual experiment, SB’s confidence is 1/3 that this interview is a T1 interview, but 1/2 that a T1 interview will take place.
> In fact according to the model I propose (which is based on the fact that T1 and T2 are mutually exclusive),…
It is based on both the observation of, and the occurrence of, T1 and T2 being mutually exclusive. In the SB experiment, only their observation is mutually exclusive, so your RE is an invalid comparison.
> Regarding the solution you propose based on PoI, I suggest examining how is Elga applying PoI (correctly according to me) in case of T1 and T2:
So you are treating his T1 and T2 as mutually exclusive here.
> And I am surprised why Elga is not using the above (erroneous according to me) proof, since he already did it (correctly ) for proving P(T1)=P(T2). However, even if he did so, the key issue remains. He still has to use the (invalid according to my analysis) premise that “your credence that you are in H1, after learning that it is Monday, ought to be the same as the conditional credence P(H1|H1 or T1)”.
And that’s why he added the other argument, even though he implied that the one you posted applies. As I described before, the only difference between H1 and T1 is based on a binary process that is completely independent of the (hypothetical) event “H1 or T1,” and whose two outcomes are themselves subjectively equivalent. So H1 and T1 are themselves subjectively equivalent. He lost focus and described that probability, which is not his relevant point. The argument you gave for him, and called incorrect, is correct.
17 November 2013, 9:40 amIoannis Mariolis:
Dear JeffJo,
You argue that:
>Elga defines them (T1 and T2 events) as observations occurring on that day, and with that coin result. Literally, T1 means “TAILS and it is Monday.” He does not define them as the coin result coinciding with an instance of the day, as you incorrectly say he does.
I do not understand what you mean by “the coin result coinciding with an instance of the day”, can you please clarify? I also don’t see how relating the events to observations can be either meaningful or helpful? What do you mean by ‘observation’ and ‘possible observation’? SB never observes the toss result or the day of the week.
You continue saying:
>The “trial” is a single observation, taken in isolation from the other possible observations. T1 and T2 can be, and are, disjoint events in an RE based on observation.
A ‘trial’ of an RE is a very specific concept related to the outcomes of the RE’s sample space. I find the phrase “an RE based on observation” confusing. In one sense all REs are based on observations of their outcomes. Can you please explicitly define the RE you are referring to, in which T1 and T2 are disjoint events of its sample space? (I don’t think it can be different from the one I propose).
Then, you argue:
>For example, take away the coin, and drug-and-wake SB all five days from Monday to Friday. When interviewed, SB’s credence that it is Wednesday is 20% (Elga’s definition), but 100% that an interview will take place on Wednesday (what you claim is his definition).
Excellent simplification. If you take away the coin toss my point becomes even more clear. I never claimed that Elga’s definition was similar to the ‘an interview will take place on Wednesday’ definition. On the contrary I insisted that he has defined T1 and T2 in a similar manner to ‘SB’s credence that it is Wednesday is 20%’ of your example! This is why T1 and T2 can be considered disjoint events and he can correctly use this fact in his proof. I argue that his proof is incorrect because his premise that ‘H1 or T1’ is equivalent to ‘Monday’ suggests that T1 and T2 are not disjoint events.
Notice that in your example (if M1 is defined as ‘Monday has been selected as current day out of five possible days’) P(M1)=P(T1)=…=P(F1)=1/5 and that P(M1)+P(T1)+…+P(F1)=1, with M1,T1,…,F1 being mutually exclusive and collectively exhaustive events. Also notice that in that case in order for SB to model her situation upon awakening she has to use an RE that she knows that it never actually takes place (like in the original SB problem). She knows that each wakening is determined, however, every time she wakes up she is using this assumed RE to model her situation and calculate the probability of Monday, Tuesday, etc to 1/5. In order to calculate the probability that ‘an interview will take place on Wednesday’ she should use a different RE with M,T,..,F events and P(M)=P(T)=…=P(F)=1. These events are obviously not mutually exclusive. Thus, if you are trying to prove something about M1,T1,…,F1 events you cannot use M,T,…F events (as Elga does implicitly in his proof).
>In the actual experiment, SB’s confidence is 1/3 that this interview is a T1 interview, but 1/2 that a T1 interview will take place.
If you accept the RE I propose SB’s confidence is 1/4 that this interview is a T1 interview, and 1/2 that an interview will take place on ‘a Monday that the coin was tossed tails’, which by your definitions it is equivalent to ‘a T1 interview will take place’, but in the RE context it is not. It makes no sense to refer to the occurrence of a T1 event during the entire duration of the experiment. In the simple example you described above it is like mixing an M1 event with an M event. It is like you are saying that ‘SB’s confidence is 100% that an M1 awakening will take place’, which is absurd. The only thing that makes sense is to say that ‘SB’s confidence is 100% that an M awakening will take place’.
>It is based on both the observation of, and the occurrence of, T1 and T2 being mutually exclusive. In the SB experiment, only their observation is mutually exclusive, so your RE is an invalid comparison.
I am not sure how you define ‘observation’ and ‘occurrence’. However, when I speak of ‘occurrence’ I mean it in the context of the corresponding RE. Thus, since T1 and T2 events are disjoint by definition, T1 and T2 occurrences are mutually exclusive.
>So you are treating his T1 and T2 as mutually exclusive here.
Yes and so does he in accordance to his definitions.
>And that’s why he added the other argument, even though he implied that the one you posted applies. As I described before, the only difference between H1 and T1 is based on a binary process that is completely independent of the (hypothetical) event “H1 or T1,” and whose two outcomes are themselves subjectively equivalent. So H1 and T1 are themselves subjectively equivalent. He lost focus and described that probability, which is not his relevant point. The argument you gave for him, and called incorrect, is correct.
Elga did not loose focus, he just applied PoI rigorously. If one adopts your line of thought he could also argue that since H1, T1 and T2 are subjectively equivalent, P(H1)=P(T1)=P(T2)=1/3. It is wrong to assume this, because H1 will certainly occur in case of Heads, whereas in case of Tails either T1 or T2 will occur. Thus, although T1 and T2 are interchangeable, H1 and T1 (or H1 and T2) are not. That’s why the only correct application of PoI is the one Elga does for T1 and T2 and it breaks down if you try to apply it to H1 and T1 because learning that it is Monday is not equivalent to learning that a ‘H1 or T1’ event occurred.
Best regards,
26 November 2013, 8:15 amYannis
JeffJo:
> I do not understand what you mean by “the coin result coinciding with an instance of the day”, can you please clarify?
That’s what the example of simply waking SB five days in a row was for. To ASB, the observation of a day is, itself, an event. It is disjoint from the observation of any other day, even if from an external POV one event assures the other will occur.
> I also don’t see how relating the events to observations can be either meaningful or helpful?
And I believe it is because you don’t want to try. To ASB, in a valid RE, {H1,T1,T2} are disjoint events. As I’ve described.
> A ‘trial’ of an RE is a very specific concept related to the outcomes of the RE’s sample space.
And this problem requires an unusual RE, because it is based on ASB concept of “now.”
> If you take away the coin toss my point becomes even more clear. I never claimed that Elga’s definition was similar to the ‘an interview will take place on Wednesday’ definition.
You implied it when you said T1 and T2 cannot be both disjoint, and products of the same RE.
In my simplification, all five days are only “outcomes of the same trial of the RE,” as you put it in your previous reply, from the experiment director’s POV. But the correct RE is in ASB’s POV; observing an interview in her “now.” What the director sees is irrelevant.
> Notice that in your example (if M1 is defined as ‘Monday has been selected as current day out of five possible days’) P(M1)=P(T1)=…=P(F1)=1/5 and that P(M1)+P(T1)+…+P(F1)=1, with M1,T1,…,F1 being mutually exclusive and collectively exhaustive events.
Yep – but this is from ASB *OBSERVATIONAL* point of view. In this RE, what happens on other days are different trials even though they are “outcomes of the same trial of the RE” in the director’s POV.
> Thus, if you are trying to prove something about M1,T1,…,F1 events you cannot use M,T,…F events (as Elga does implicitly in his proof).
And my entire point is that you can, because the RE is ASB’s observation.
> If you accept the RE I propose …
I don’t. It is from the director’s POV, and artificially forces T2 to not happen if T1 does. In ASB’s RE, they occur in different trials.
JeffJo
26 November 2013, 4:12 pmJeffJo:
Yannis
Let me just reiterate one point: a TRIAL of the RE we need for this problem is not the three-day experiment the director sees. It is a single day that ASB sees because she is awake. This is correct because it is the only point-of-view (POV) ASB can have from which to answer the question. Any other day, even if it is in the same experiment from the director’s POV, is a different trial to ASB. Most of your arguments have ignored this fact, and so are invalid. In particular, the RE you suggested – which correlates what must be different trials to ASB – cannot be used to answer the question.
26 November 2013, 4:35 pmIoannis Mariolis:
Dear JeffJo,
I will begin with your last post since it is very crucial to resolve this misunderstanding. You said:
>Let me just reiterate one point: a TRIAL of the RE we need for this problem is not the three-day experiment the director sees. It is a single day that ASB sees because she is awake.
I completely agree that “a TRIAL of the RE we need for this problem is not the three-day experiment the director sees”. In fact, when you are referring to ASB I refer to RE2 and by its definition it is clear that a TRIAL of RE2 ends with the selection of a single day where SB is awake. On the other hand, when you refer to SSB I refer to RE1 which spans all three days. In fact, my whole approach is based on the above point.
>Most of your arguments have ignored this fact, and so are invalid.
None of my arguments ignores this fact. I always clarify which RE I am referring to and treat TRIALS accordingly.
>In particular, the RE you suggested – which correlates what must be different trials to ASB – cannot be used to answer the question.
I have suggested two REs one that can be used to model SB’s situation upon awakening (RE2) and the trivial one (RE1) that can be used to model SB’s understanding of the whole experiment (which is always valid not only on Sunday. Even when she wakes up she can still use RE1 to calculate that her credence on ‘a Monday interview will occur during the experiment’ should be 100%.). RE2 defines T1 and T2 as disjoint events and does not correlate different TRIALS. In an RE2 TRIAL either a T1 or a T2 event can occur (but not both) and I have never suggested otherwise.
Taking the above under consideration lets get back to your 4:12pm post.
I said: “I also don’t see how relating the events to observations can be either meaningful or helpful?” and you replied:
>And I believe it is because you don’t want to try. To ASB, in a valid RE, {H1,T1,T2} are disjoint events. As I’ve described.
It must be clear by now that I always argued that to ASB, in a valid RE, {H1,T1,T2} are disjoint events. However, I still can’t see what ‘observations’ have to do with this.
I said: “If you take away the coin toss my point becomes even more clear. I never claimed that Elga’s definition was similar to the ‘an interview will take place on Wednesday’ definition.” and you replied:
>You implied it when you said T1 and T2 cannot be both disjoint, and products of the same RE.
I never said that. On the contrary I have proposed RE2 where T1 and T2 are both disjoint events, and products of the same RE. My entire line of arguments is actually based on this fact!
You continue:
>In my simplification, all five days are only “outcomes of the same trial of the RE,” as you put it in your previous reply, from the experiment director’s POV. But the correct RE is in ASB’s POV; observing an interview in her “now.” What the director sees is irrelevant.
I can’t see why you think I ever said or suggested any of the above. I never used the experiment director’s POV. I am explicitly defining the REs I am using and this is why I do not need to refer to POVs. RE2 is a random experiment that can be used to model ASB’s situation. Thus, ASB can use it to calculate her credence on what day of the week it is ‘now’. Moreover, ASB can still use RE1 to to calculate her credence on ‘a Monday awakening will occur during the experiment’. In my approach, the focus is not on the POV but on what you are modelling. You can say that when ASB is using RE1 she adopts SSB’s (or experiment director’s) POV but this is besides the point. If I try to use your terminology I could say that when ASB uses RE2 to calculate her credence on what day of the week it is ‘now’ she is using her own POV and not the director’s or SSB’s and clearly in that case the days of the week are NOT “outcomes of the same trial of the RE”.
>Yep – but this is from ASB *OBSERVATIONAL* point of view. In this RE, what happens on other days are different trials even though they are “outcomes of the same trial of the RE” in the director’s POV.
Exactly! We do not disagree on that. In fact it is what I tried to demonstrate by using the different REs and their corresponding sample spaces. In the {M1, T1,…,F1} sample space the events are disjoint (you call that “ASB *OBSERVATIONAL* point of view”), whereas in the {M,T,…,F} sample space they are not and they can all occur in the same TRIAL (you call that “the RE” in the director’s POV”).
I said: “If you accept the RE I propose” and you responded:
>I don’t. It is from the director’s POV, and artificially forces T2 to not happen if T1 does. In ASB’s RE, they occur in different trials.
I have explained in this post that RE2 is from ASB’s POV and its T1 and T2 events occur in different TRIALS.
I also said: “Thus, if you are trying to prove something about M1,T1,…,F1 events you cannot use M,T,…F events (as Elga does implicitly in his proof)” and you replied:
>And my entire point is that you can, because the RE is ASB’s observation.
But the RE that is ASB’s observation is the one using the {M1,T1,…,F1 } sample space. It is absurd to mix it with M,T,…F events which correspond to a different RE (the director’s POV according to your terminology).
What about the correct application of PoI? Have you accepted my arguments or you will address them on a different post?
Best regards,
27 November 2013, 5:03 amYannis
JeffJo:
Yannis:
Let me try coming at this from a different direction. The twist that makes the Sleeping Beauty Problem controversial is that it is really two problems in one: an external problem, as seen by Sunday Sleeping Beauty (SSB) or the director of the experiment, and an internal problem, as seen by Awake Sleeping Beauty (ASB). We can directly apply probabilities, based on intuitive statements like P(H)=P(T)=1/2, only to the external problem. Internally, the occurrences are derived from combinations of the external occurrences, and so may or may not have intuitive values.
But how to derive the internal probabilities, whether or not they are the same, is controversial. To distinguish these POVs, I’ll call their probabilities Pe(*) and Pi(*). Errors are made in the literature when the author assigns internal probabilities based on the external POV. It started with Lewis, who literally assumed Pi(C=H)=Pi(C=T)=1/2 in order to “prove” that Pi(C=H)=Pi(C=T)=1/2. You can’t prove anything by assuming the answer, whether or not it is correct.
The biggest difference between the two problems is that externally, the two-day experiment constitutes one trial. So T1 and T2 represent the same outcome. But internally, the T1 and T2 that were part of the same trial externally must be considered different trials. SB can only observe (this is what I mean by that word) that one, and only one, day applies to her at the moment. As a result, the number of trials is potentially different, and this famously leads to another controversy about how to test the answer with frequencies.
When you addressed this difference with a new RE, you claimed you did it to better define the events. But in reality what you did was reduce both the internal and external experiments to utilizing the same trial. That in turn allowed you to associate both the external, and internal, definitions to the same, single trial. That RE is:
RE2: Choose a coin result C from {H,T}. If C=H, set day D=1; otherwise, choose a day D from {1,2}. Wake SB on day D and ask her for Pi(C=1).
This RE still has to represent what we know to be the same true from *either* POV in the original problem, except what is affected by the number of trials. (I.e., by altering the number of trials you changed the absolute values of the external probabilities.)
So, the relative values of the external probabilities must still be the same for RE3 to represent the original problem. In the original problem, Pe(H1)=Pe(T1). You treated both POVs as the same, with P(H1) = 2*P(T1) both internally and externally. Literally, if the original problem were to be repeated many times, an external observer expects to see just as many H1 interviews as T1 interviews, irrespective of the frequency controversy I mentioned above. But RE2 has twice as many external H1s as T1s, so it cannot be a correct representation of the original problem. In its place, I propose:
RE3: Choose a coin result C from {H,T}. Choose a day D from {1,2}. Assign X=S (sleep) if the result is H2; otherwise assign X=W (wake). Wake SB according to X on day D and ask her for Pi(C=1).
I’ll keep Pe(*) and Pi(*)separate. It can’t be wrong to do so, and it may prove important.
You assumed Pe(C=H)=Pe(C=T)=1/2, and Pe(D=1)=Pe(D=2)=1/2. So will I. Now is it straightforward to derive Pe(H1W)=Pe(C=H)*Pe(D=1)=1/4; and the same for Pe(T1W), Pe(T2W), and Pe(H2S). Other combinations have probability 0. And the important thing is that these values maintain the relative values all of the external probabilities. They don’t maintain absolute values, because RE3 alters the definition (and the number) of “trials.” But by having all the same relative values, it is externally equivalent to the original. By having the internal events determined from the external ones the same way as the original, it is also internally equivalent.
But now we can see that Pi(H1)=Pe(H1|X=W)=Pe(H1W)/[Pe(H1W)+Pe(T1W)+Pe(T2W)]=1/3.
The only conceivable objection to RE3, is that an interview is no longer certain. It isn’t a valid objection, but that never stops people in this kind of controversy. This objection was the purpose of my quadruplet experiment, which extends RE3 in such a way to guarantee an interview (the same way the original did – there might be two) while maintaining the same information as in RE3 for any one of the girls. That RE was:
RE4: Secretly, and randomly, label four quadruplets H1,H2,T1, and T2. Choose a coin result C from {H,T}. On day 1 wake and interview quadruplet H1 or T1 according to C, and on day 2 wake and interview quadruplet T2 if C=T. Have them determine P(H|I get interviewed).
For any quadruplet who is wakened, her situation is identical to that in RE3 since she was randomly assigned day 1 or day 2. And I showed that RE3 is identical in principle to the original problem, while RE2 is not. RE4 is also shown to be identical by having the quadruplets calculate an answer jointly. That answer must be the same regardless of which quadruplet is being interviewed, so it is the same as the original SB’s answer must be.
+++++
> If one adopts your line of thought he could also argue that since H1, T1 and T2 are subjectively equivalent, ..
No, one can’t; and no, they aren’t. You are taking those statements out of context (yes, I repeated them out of context, but I assumed that context was understood when I did so).
T1 is subjectively equivalent to H1 under the condition that {D=1} has occurred, because with that condition the only difference is the subjectively equivalent {C=T} or {C=H}. But we can use this fact to derive the equation P(T1)=P(H1) without any such conditions.
T1 is subjectively equivalent to T2 under the condition that {C=T} has occurred, because with that condition the only difference is the subjectively equivalent {D=1} or {D=2}. But we can use this fact to derive the equation P(T1)=P(T2) without any such conditions.
The transitive property does not apply to these two subjective equivalences, because the conditions that are needed to make them subjectively equivalent are different. The transitive property does apply to the two equations because they are not conditional.
> It is wrong to assume this, because H1 will certainly occur in case of Heads, whereas in case of Tails either T1 or T2 will occur. Thus, although T1 and T2 are interchangeable, H1 and T1 (or H1 and T2) are not.
And this line of reasoning is fallacious. First, you are again arguing about internal events based on the external POV. You are relying on the temporal difference between Monday and Tuesday; “temporal” is not an internal property. This is the difference I am calling “observational.” Since SB can’t distinguish the days, she can only observe that “today” is either Monday, or Tuesday, but not both. This is subjectively equivalent to knowing the coin will be Heads, or Tails, but not both.
The equivalent internal statement is “H1 must surely apply to this interview, if Heads applies to it; whereas in if Tails applies either T1 or T2 could have. Thus, although T1 and T2 are interchangeable, H1 and T1 (or H1 and T2) are not.”
Second, you could just as easily say that “T2 must surely apply to this interview, if Tuesday applies to it, whereas in if Monday applies either H1 or T1 could have. Thus, although H1 and T1 are interchangeable, T1 and T2 (or H1 and T2) are not.” Since this says just the opposite of what the same line of reasoning said before, there must be a fallacy somewhere. That fallacy is that, where you concluded that “H1 and T1 are not interchangeable,” all you can actually deduce is that “This statement does not imply whether other pairs of events are interchangeable or not.”
> That’s why the only correct application of PoI is the one Elga does for T1 and T2 and it breaks down if you try to apply it to H1 and T1 because learning that it is Monday is not equivalent to learning that a ‘H1 or T1′ event occurred.
It is correct for H1 and T1. It does not “break down,” because learning *INTERNALLY* that it *IS* Monday, is equivalent to learning that a {H1 or T1} event occurred for this *INTERNAL* trial. What occurs for other trials is irrelevant.
JeffJo
27 November 2013, 3:47 pmJason:
Wow, I cannot believe the back and forth that JeffJo and Yannis went through!
In regards to the original SB problem, learning “internally” that it is Monday is not actually new information! If the coin is heads, she will be awoken Monday with probability 1. If the coin is tails, she will be awoken Monday and Tuesday with probability 1. This is a completely sensible probability model for the event that she will be awoken Monday. Thus, P( H | awake on Monday) = 1/2 by bayes theorem. If she does not know which day it is, which is the original intent of the problem, then her subjective beliefs regarding Heads or Tails should still be 1/2 since she has no new information on which to update her beliefs. Being awoken is not an event that is able to provide information via bayes theorem. As you see, I am basically using the axiomatic approach to define a valid probability model for the events combined with bayesian updating. The other way to think about this is that the actual information she is using to update with is that she has been awoken at least once. This is a completely valid approach using conditional probability.
Where the frequency argument gets into trouble is that it is inadvertently assessing the long-run frequency of (# times heads and awoken) / # times awoken = 1/3, which is not the same probability that the question is asking for. Of all the times she is awoken, the proportion of times it will be when the coin is heads is subtly different than your beliefs as to whether or not the original coin flip was heads. These are actually two very different probability spaces. It’s irrelevant to the original question of the problem whether she thinks it is more likely to being questioned when the toss was heads or tails.
In regard to really old post from JeffJo on 10/26/11 7:15, the argument appears to be flawed. The events Monday and Tuesday are not the only information you have to condition on, you also have to condition on being awake. When you argue that
13 April 2014, 4:23 amP(H) = P(H|M) * P(M) + P(H|T)*P(T), I believe it is improper to argue that P(M) is her belief that it is Monday. Instead, P(M) should be the probability that she is awoken on Monday. Thus,
P(H) = P(H| awoken on Monday) * P(awoken on Monday) = 1/2.
You have to build your probability model forward in time to make sure you are modeling the appropriate sequence of random events. Start with the coin toss and then what will happen as a result of the coin toss. Otherwise you will get confused as to what are random events, so like ending up thinking that P(M)=2/3. You should have a prior probability for heads or tails, then build a probability model for the events that will occur under each of those two scenarios, this is the likelihood. The implied likelihood you have created doesn’t make sense. I believe the formulation of what constitutes the probability model for the data, the likelihood, is key to solving the paradox.
JeffJo:
And I cannot believe the amount of denial people will exhibit when trying to defend their intuition, which is widely known to be suspect in the case of conditional probability.
Rather than go into your unexplained, and incorrect, definition of “new information,” and your arbitrary shifting between different probability models, let me describe a simple analogy: The experiment is essentially the same; except SB is always wakened. In the cases where she was wakened in the original, she will be led into a room that is painted blue for her interview. On Tuesday after Heads, she is led into a room that is painted red. Once she is awake:
1) What is the probability she will be led into a red room? That is, if we call that event Red, what is P(Red)?
2) What is the probability she will be led into a blue room? That is, what is P(Blue)?
3) If she is led into a red room, what is the probability of Heads? That is, what is P(Heads|Red)?
4) If she is led into a blue room, what is the probability of Heads? That is, what is P(Heads|Blue)?
5) Do Red and Blue represent a partition of the sample space?
If the answer to #5 is “yes” – and it obviously is – then the Law of Total Probability applies. That means the P(Heads) must equal P(Heads|Red)*P(Red)+P(Heads|Blue)*P(Blue). Four of these five terms are pretty easy to assign values to: P(Red)=1/4, P(Blue)=3/4, P(Heads|Red)=1, and P(Heads)=1/2. That means (1)*(1/4)+P(Heads|Blue)*(3/4 )=1/2, so P(Heads|Blue)=1/3. Since this changed, being led into a a blue room, or a red room, clearly represents “new information.” My final question is, what part of the procedure makes SB’s information “new” in one case, but not the other? Specifically, why does it matter what could happen in the situation that SB knows IS NOT HER CURRENT SITUATION that affects her answer?
My point is that it does not matter what happens in the “other” situations. There are initially four equally-likely situations, and all that matters to SB is that she has both necessary and sufficient conditions to define a subset of those four that represent her current situation. The two problems are identical.
And please note that this does not involve a “frequency argument.” Getting the same answer does not mean I use the same argument. I agree about building the model forward in time, but you have to include every possibility that exists, not just those that represent what SB sees in the cases of interest.
16 April 2014, 6:13 amLouis Wilbur:
From https://en.wikipedia.org/wiki/Sleeping_Beauty_problem we have the problem:
“–Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be wakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be wakened and interviewed on Monday only. If the coin comes up tails, she will be wakened and interviewed on Monday and Tuesday. In either case, she will be wakened on Wednesday without interview and the experiment ends. Any time Sleeping Beauty is wakened and interviewed, she is asked, “What is your belief now for the proposition that the coin landed heads?”–”
If the experiment is repeated one thousand times then, on average, she will be awakened and asked about her belief fifteen hundred times during the experiments. On average, five hundred of those fifteen hundred times she is asked (i.e. one third of the times) the coin will have landed heads and so the proposition “the coin landed heads” will be true. On average, one thousand of those fifteen hundred times she is asked (i.e. two thirds of the times) the coin will have landed tails and so the proposition “the coin landed heads” will be false.
So sleeping beauty knows several facts. She knows that in repeated trials of the experiment, on average, one out of every three times she is awakened the proposition “the coin landed heads” will be true. She knows she was just awakened. So it is reasonable for her to give one third as an answer. She also knows that in repeated trials of the experiment, on average, one out of every two times that the experiment is run the proposition “the coin landed heads” will be true. She knows the experiment is being run. So it is reasonable for her to give one half as an answer. Her answer to the question should be: My belief is that, on average, the proposition is true one out of every two times the experiment is run and, on average, one out of every three times I am awakened and asked my belief about it.
8 August 2014, 7:48 pmIoannis Mariolis:
Dear JeffJo you have posted:
19 August 2014, 9:31 am“The experiment is essentially the same; except SB is always wakened. In the cases where she was wakened in the original, she will be led into a room that is painted blue for her interview. On Tuesday after Heads, she is led into a room that is painted red. Once she is awake:”
Then, you pose 5 questions on certain probabilities.
To answer your questions:
1) What is the probability she will be led into a red room? That is, if we call that event Red, what is P(Red)?
P(Red)=P(Heads)=1/2
2) What is the probability she will be led into a blue room? That is, what is P(Blue)?
P(Blue)=1
3) If she is led into a red room, what is the probability of Heads? That is, what is P(Heads|Red)?
P(Heads|Red)=1
4) If she is led into a blue room, what is the probability of Heads? That is, what is P(Heads|Blue)?
P(Heads|Blue)= P(Heads)=1/2
5) Do Red and Blue represent a partition of the sample space?
No they do not.
In order to define a sample space you have to define a random experiment whose outcomes constitute this space. In the experiment you have defined (which is an extension of RE1), the only random element is the original coin toss. Thus, the probabilities in question are nothing like the ones you have calculated (the correct values are given in my answers). The confusion arises by the fact that you are actually referring to a different random experiment you have not defined (and is actually an extension of RE2). Thus, like in RE2, once awakened, SB models her situation as the outcome of a random experiment in which her current state is the result of a random procedure on top of the original coin toss (notice that such random procedure does not take place and SB is aware of this fact). Therefore, it could be either Heads and Blue with probability ¼, or Heads and Red also with probability ¼, or Tails and Blue with probability ½. Thus, P(Red)=1/4 and P(Blue)=3/4 (notice that in the context of this random experiment Red and Blue represent a partition of the sample space). Of course, in the context of this random experiment the probabilities you have calculated are correct, including P(Heads|Blue)=1/3. The subtle part is that once SB is awakened in a blue room she cannot update P(Heads) to 1/3, because she knows that a Blue event (as defined by the later random experiment) has not actually occurred. She knows that (apart from the role of the coin toss) her awakening on that blue room was the result of a deterministic procedure. Thus, no evidence on a Blue event (as defined by the later random experiment) has been produced. On the contrary only evidence of a Blue event as defined by the former random experiment has been produced hence P(Heads|Blue)= P(Heads)=1/2. The same applies for the Red room, she can only use the conditional probabilities of the first random experiment to update. Hence, once awakened in a red room P(Heads|Red)=1.
Best regards,
Yannis
JeffJo:
>> 1) What is the probability she will be led into a red room? That is, if we call
>> that event Red, what is P(Red)?
> P(Red)=P(Heads)=1/2
Incorrect. In three out of four possible situations, she will be led into a blue room. In one, she will be led into a red room. So P(Red)=1/4.
What you ignore, is that there are two different days that occur after a “heads” result. You consider “Room=Red” to be true on some Mondays when she is led into a blue room. Yes, she will be led into a red room the following day, but when she observes “Room=Blue,” it is not true FOR THAT OBSERVATION regardless of whether it is true on another day.
Similarly, in the original experiment, she will be awakened in one of three possible situations. When she finds herself awake, each is equally likely to be her current situation. Your error is considering a different situation to be her current situation, just because it is guaranteed to happen. The error is that it ISN’T happening.
20 August 2014, 4:30 pmJeffJo:
Let me clarify that. Your answer of P(Red)= 1/2 is based on the event “Red” meaning “SB will be led into a red room in this line of history.” My answer of P(Red)= 1/4 is based on the event “Red” meaning “SB observes that she is led into a red room.”
SB has to base her assessment on what she sees, not what fortune tellers may know. She only sees that she is led into a red room, not what may happen on another day.
20 August 2014, 4:40 pmIoannis Mariolis:
Dear JeffJo,
21 August 2014, 4:58 amIn my previous post I have presented the probabilities for both definitions of the event Red. It is clear that we both agree that in case we define Red as the event “SB will be led into a red room in this line of history” (call it Red1) P(Red1)= 1/2 and that if we define the event Red as “SB observes (at her current awakening I may add) that she is led into a red room” (call it Red2), that P(Red2)=1/4. My argument is that you can’t define the events unless you have already defined a random experiment whose outcomes constitute the sample space these events belong to. This is why I have described two random experiments, one corresponding to the first interpretation of event Red (Red1) and the other to the second interpretation (Red2). The subtle point is (and to my opinion the main reason for all the controversy around SB problem) that although the first random experiment actually takes place (there is a coin flip whose outcome is random), the second random experiment never actually takes place. It is used by SB to model her uncertainty on which stage of the actual experiment she is at her current awakening. So it is perfectly consistent to model her ignorance upon her current state as if the stage she is at has been randomly selected and assign P(Red)=1/4 and P(Blue)=3/4 for her current awakening. However, she knows that such random selection never actually takes place. Thus, if she is actually led on a Blue room she cannot update to P(Heads|Blue2)=1/3 but she still can use P(Heads|Blue1)=1/2.
Best regards,
Yannis
JeffJo:
Just found this blog again. But in the intervening time. I found a definitive proof of the answer.
First, consider a meaningless change: SB will be assigned, but not told, a random coin result C from the set {Heads,Tails}, and a random day D from the set {Monday,Tuesday}. She will be put to sleep on Sunday night, and the coin will be flipped. She will be wakened at least once, and maybe twice, during the next two days. She will he left asleep only on day D, and only if the actual coin flip was C. Anytime she is wakened, she is asked for her assessment of the probability that the coin result was C.
The change is meaningless because it is the original question when D=Tuesday and C=Heads, and the other combinations are symmetric variations that must have the same answer.
Then, use four volunteers and only one coin. Accomplish the random assignments mentioned earlier by randomly distributing the four possible combinations of {C,D} among them. On each day, three of the volunteers will be wakened, but only one of those three will have been assigned the C that matches the actual flip. So each can confidently answer 1/3 to the question asked.
15 March 2018, 3:05 pmGofer:
This problem is easily solved using two applications of the law of total probability. The answer is heads with 1/3.
Let C be the coin flip {h,t}, and D the day {mon,tue}. Then
(1) y := Pr(C=h) = Pr(C=h|D=mon)*Pr(D=mon) + Pr(C=h|D=tue)*Pr(D=tue) = 1/2*x + 0*(1-x) = x/2
(2) x := Pr(D=mon) = Pr(D=mon|C=h)*y + Pr(D=mon|C=t)*Pr(C=t) = 1*y + 1/2*(1-y) = x/4 + 1/2
(1) and (2) defines an equation system in x and y with solution 2/3 and 1/3 respectively. QED.
“:=” means “defined as” – introducing a definition.
Thus, there is no need for convoluted explanations betting on outcomes, or sidetracking with modified problems. All the information needed for solving the problem at hand is readily available in the problem description.
23 October 2021, 5:13 pmGofer:
I just wanted to add a few concluding remarks.
Some argue that Beauty, upon awakening, learns no new information, and therefore the priori probabilities of the initial coin flip are valid. While it is true that B learns no new information, that is NOT what determines the posterior probabilities; they are changed by the simple act of waking up – thus entering phase two of the experiment, the first one being the coin flip. For example, consider the same rules but for flipping the coin a second time on tails to determine whether Beauty should be awoken on Monday or Tuesday. Now the answer is heads with 1/2. The equations remain the same, but Pr(C=h|D=mon) evaluates to 2/3 instead of 1/2.
(1) and (2) work for all possible scenarios with respect to awakening Beauty. For Pr(C=h|D=mon):=1 however, signifying Beauty is never awoken on Monday for tails, they evaluate to x=y, meaning Monday and heads are synonymous, thus C taking on its priori probabilities, 1/2.
23 October 2021, 6:55 pmJeffJo:
Gofer, you are still asserting facts that Halfer’s dispute.
It will surprise many people, but nobody is using the correct wording of the problem. They are including elements that Adam Elga introduced in order to explain his thirder solution simply. The two wordings have the same answer, of course, but the objections Halfers raise to his solution all relate to the parts he introduced.
Here is the actual wording from Elga’s paper, with two edits that remove irrelevant information: “Some researchers are going to put you to sleep. During the [time] that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are [awake], to what degree ought you believe that the outcome of the coin toss is Heads?”
In the original wording, my “[time]” was two days. But it didn’t actiually say the wakings would occur on different days, and no order was specified. My “[awake]” was “first wakened,” which is a bit ambiguous. Since no order is specified, it can’t mean “Monday, but not Tuesday.” And since Elga’s solution involved telling SB information after her first answer, it seems obvious that it means “before podssibly learning something.”
There is another way of implementing this, where the “new information” becomes obvious, and the Halfer’s methods work against them.
Once SB is asleep, flip two coins, a Quarter and a Nickel. If either is showing Tails, (1) wake her, (2) ask her the question, and after she answers, (3) put her back to sleep with amnesia. Then, TURN THE NICKEL OVER, and repeat the same three steps. The question is “to what degree ought you believe that the Quarter is showing Heads?”
Note that the Quarter never changes, so what it is currently showing is the same as the outcome of its flip. But even though the Nickel might not be showing the what its flip-outcome was, the prior distribution for the two coins is found by Halfer techniques. Each of {HH, HT, TH, TT}, where the Quarter is listed first, has a probability of 1/4.
But the “new information” that SB has is that HH is eliminated. She wouldn’t be awake, and facing the question, it both coins afre currently showing Heads. The remaining three combonations each update to a conditional probability of 1/3. Since only one of them has the Quarter showing Heads, the answer to the question is 1/3.
8 December 2021, 4:11 pmTanya Khovanova's Math Blog » Blog Archive » Sleeping Beauty Wants the Prize Money:
[…] already blogged about the Sleeping beauty problem three times: The Sleeping Beauty Problem, Sleeping Beauty Meets Monty Hall, and Sleeping Beauty and Mondays. The posts were more than ten […]
30 August 2022, 12:48 pmTanya Khovanova's Math Blog » Blog Archive » Sleeping Beauty Wants the Prize Money - New Marathi Live:
[…] already blogged about the Sleeping beauty problem three times: The Sleeping Beauty Problem, Sleeping Beauty Meets Monty Hall, and Sleeping Beauty and Mondays. The posts were more than ten […]
16 September 2022, 5:28 amTh.:
Test. Test.
13 July 2023, 11:53 amWith all those pages of text and way more than necessary of technical terms, jeffjo is still not convinced. I would like to try but first i must make sure that he is reading this message or my attempt would be in vain.
All those explanation against jeffjo, even correct, but does not hit the nail in the head and fail to convince him.