## The Sleeping Beauty Problem Amplified

The Sleeping Beauty 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 tails. Regardless of her answer, if the coin was heads, the experiment ends. However, if the coin was tails, she is put back to sleep with her memory erased and awakened again on Tuesday and asked the same question. In this case, the experiment stops on Tuesday. She knows the protocol. She is awakened one morning. From her point of view, what is the probability that the coin was heads?

This problem is considered controversial. Some people, called halfers, think that when she is awakened, the probability that the coin was heads is one half. Other people, called thirders, think that when she is awakened, the probability that the coin was heads is one third.

I am a thirder, so let me explain my thinking. If it is Monday, then the probability that the coin is heads is one half. So the probability of Monday/heads is the same as Monday/tails. If the coin is tails, Sleeping Beauty can’t distinguish between Monday and Tuesday. So the probability of Monday/tails is the same as Tuesday/tails. Thus, the three cases Monday/heads, Monday/tails, and Tuesday/tails are equally probable, and the answer follows.

This problem is similar to the Monty Hall problem in some ways. The main difference is that The Monty Hall problem stopped being controversial, while Sleeping Beauty problem continues to be. The best argument that helped intuition and led to the resolution of the Monty Hall problem was increasing the number of doors. Similarly, for the Sleeping Beauty problem, we can increase the number of days she is put back to sleep when the coin is tails. Forgive me for the cruelty of this theoretical experiment.

The Sleeping Beauty Variation.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 tails. Regardless of her answer, if the coin was heads, the experiment ends. However, if the coin was tails, the experiment continues for 99 more days. Each day, she is put back to sleep with her memory erased and awakened the next day and asked the same question. She knows the protocol. She is awakened one morning. From her point of view, what is the probability that the coin was heads?

In this variation, when she is awakened, she should be almost sure that the coin was tails. I hope it will help halfers feel that, in this case, the probability of heads can’t be one half. For me, the argument is the same as before, making the probability that the coin was heads is 1 over 101.

After I wrote this essay, I discovered that Nick Bostrom made the same argument. Though, for tails, he put Sleeping Beauty to sleep one million and one times, which is about 2,740 years. He increased my one hundred days by several orders of magnitude, amplifying our point. Surely, when Sleeping Beauty awakens, she should be almost certain that the coin was tails. After Sleeping Beauty agrees to so much torture, why is the problem still controversial?

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## Theo Johnson-Freyd:

In the version of this problem I grew up with, Beauty is a PhD student in Statistics at Stanford, and the experiment is held at the Stanford Sleep Lab.

1 September 2022, 10:02 am## Tanya Khovanova's Math Blog » Blog Archive » The Sleeping Beauty Problem Amplified - New Marathi Live:

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16 September 2022, 4:57 am## JeffJo:

There is a subtle point that gets lost in all the debate. I suspect that Adam Elga saw it. He actually changed the experiment, from what he posed, in a way that let him avoid it in his solution. And the controversy that has surrounded the problem ever since, is about how (or whether) to re-integrate it back in.

Don’t believe me? Here is the problem, as Elga posed it: “Some researchers are going to put you to sleep. During the two days 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 first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”

Notes:

1) Monday and Tuesday are not mentioned.

2) The order of the mandatory and optional wakings are not mentioned.

3) I suspect that Elga was already thinking of the Monday v. Tuesday solution, since he mentioned “two days,” even tho the actual days, the fact that there are two, and the order have no significance in this formulation.

4) It is ambiguous what “when first awakened” means, but it also seems to have meaning to Elga’s solution. It could mean on Elga’s Monday; but since SB can’t know if it is Monday or Tuesday in his solution, her belief can’t be different. Or it could mean before Elga reveals some information to SB; but her lack of that information is the entire point of the question.

Anyway, the subtle point is that to SB, the “probability experiment” is not the coin flip. It is her observation of a result of the coin flip. And there are two different such results: (A) SB can be awakened and interviewed, or (B) she can not be wakened and interviewed. The subtle part is that if SB observes that she is in an (A) result, IT DOES NOT MATTER what would have happened in a (B) result. When she is interviewed, SB has definitive evidence that she is in an (A) result.

I’m going to essentially assume we wake her, but take her on a shopping spree instead of an interview. Her answer, in the interview, can’t be different because of this change. But this removes one flaw in the halfer argument: it assumes that (B) can’t happen since SB cannot observe it.

Elga avoided any consideration of (B) in his solution. He suggested two continuations of the experiment: After her first interview question is answered, we could tell her what day it is, or we could tell her the coin result. If that means we tell her it is Monday, or that the coin landed on Tails, one of the three (A) results is eliminated [(B) was already eliminated]. Since she can’t say that either of the remaining (A) results is more likely, all three must have had the same probability before the reveal. This makes the answer 1/3.

The common halfer objection is that DAY={Monday,Tuesday} is not a valid discriminator, since both “happen” to SB if the coin lands on Tails. This is why it is important to note that SB’s “experiment” is the observation of a result of the coin flip, not the coin flip itself. Yes, both a Monday and a Tuesday version of (A) will happen to her, but her observation represents only one, and not the other.

+++++

I’ll repeat a solution I’ve given before. Elga didn’t need to introduce the controversy that allows halfers to make that argument. The original problem, as posed, can be implemented with two coins. This can happen on a single day, or two days:

1) After SB is asleep, flip two coins, C1 and C2.

2A) Examine the two coins.

2B) If both coins show Tails, end step 2.

2C) Otherwise, wake SB.

2D) Ask SB the question, about coin C1 only.

2E) After she answers, put SB back to sleep with amnesia.

3) Turn coin C2 over.

4) Repeat step 2 in full.

This way, it does not matter to SB whether she is in step 2, or step 4. In sub-step A, there were four equally-likely states for the two coins: HH, HT, TH, and TT. But since she is in substep D of whichever, she knows that HH is eliminated. The answer -PROVEN BY THIS METHOD – is 1/3.

19 January 2024, 9:33 am