Puzzle. You got two envelopes with two distinct real numbers. You
chose one of them and open it. After you see the number you are allowed
to swap envelopes. You win if at the end you pick the larger number.
Find a strategy that gives you a probability more than 1/2 of winning.

Choose a real number “randomly” from some distribution with infinite support (e.g. the standard normal distribution) before opening either envelope. Open one, and swap if the number in the first envelope is smaller than the number you chose.

If the numbers in the envelopes are both higher than your chosen number, or both lower, then you gain nothing and your probability of winning is 0.5.

If the number you chose is *between* the numbers in the two envelopes, then you will choose the higher using this procedure, and your probability of winning is 1.

We don’t know anything at all about where the numbers in the envelopes came from, but we do know that they’re two distinct reals and our distribution has infinite support, therefore the probability that we chose a number in the interval between them is some p > 0, and the probability of winning the game is 0.5(1-p) + 1*p = 0.5 + 0.5p, > 0.5.

(edit: mathematical symbols did not appear in the previous comment)

Suppose that you open the envelope and see the real number x. Let f(x) ∈ [0,1] denote the probability that you swap envelopes for the observed value x. Any monotonically decreasing function f(x) will do the job.

Antony’s answer is a special case. To see this, let a be the guessed number. In this case, f(x) = 1 – u(x-a) where u(x) is the unit-step function.

Estimate x = number of penguins in the Antarctic, toss a coin to decide the sign of x, look at the number in the first envelope and keep it if it is greater than x or exchange the envelope otherwise.
Well, it sounds much less dignified than “choose a random number from some distribution” but it bears the same relevance to the question at hand, so…

More seriously, let’s a be the number in the first envelope and b the number in the second one. When both a and b are unknown we do only know that a ≠ b thus p > 0. When we learn the value of a, if a x: we have only changed our question from “is b > a” to “is b > x” and I can’t see how this can affect the probability. If a > x holds then we exchange the question “is b < a” with “is b < x” to the same effect.

In general we should'nt combine directly probabilities conditional to diffent informations.

The st petersburg paradox:
The EV of this game is infinite. Thus it is always +EV to switch, regardless of what we open. This appears to break symmetry, but that’s what we get when working with infinite EVs.

## Anthony:

Guess a number x. If the number in the first envelope is greater than x, keep it, otherwise pick the second envelope.

20 October 2019, 2:31 pm## Andrew:

Choose a real number “randomly” from some distribution with infinite support (e.g. the standard normal distribution) before opening either envelope. Open one, and swap if the number in the first envelope is smaller than the number you chose.

If the numbers in the envelopes are both higher than your chosen number, or both lower, then you gain nothing and your probability of winning is 0.5.

If the number you chose is *between* the numbers in the two envelopes, then you will choose the higher using this procedure, and your probability of winning is 1.

We don’t know anything at all about where the numbers in the envelopes came from, but we do know that they’re two distinct reals and our distribution has infinite support, therefore the probability that we chose a number in the interval between them is some p > 0, and the probability of winning the game is 0.5(1-p) + 1*p = 0.5 + 0.5p, > 0.5.

21 October 2019, 12:33 pm## Misha:

Open any of them. Always open another one. Probability 1/2.

21 October 2019, 2:01 pm## Berkan Dülek:

(edit: mathematical symbols did not appear in the previous comment)

Suppose that you open the envelope and see the real number x. Let f(x) ∈ [0,1] denote the probability that you swap envelopes for the observed value x. Any monotonically decreasing function f(x) will do the job.

Antony’s answer is a special case. To see this, let a be the guessed number. In this case, f(x) = 1 – u(x-a) where u(x) is the unit-step function.

22 October 2019, 8:36 am## Guido Bombi:

Estimate x = number of penguins in the Antarctic, toss a coin to decide the sign of x, look at the number in the first envelope and keep it if it is greater than x or exchange the envelope otherwise.

Well, it sounds much less dignified than “choose a random number from some distribution” but it bears the same relevance to the question at hand, so…

More seriously, let’s a be the number in the first envelope and b the number in the second one. When both a and b are unknown we do only know that a ≠ b thus p > 0. When we learn the value of a, if a x: we have only changed our question from “is b > a” to “is b > x” and I can’t see how this can affect the probability. If a > x holds then we exchange the question “is b < a” with “is b < x” to the same effect.

In general we should'nt combine directly probabilities conditional to diffent informations.

29 October 2019, 12:41 pm## Julien:

The st petersburg paradox:

30 October 2019, 7:11 pmThe EV of this game is infinite. Thus it is always +EV to switch, regardless of what we open. This appears to break symmetry, but that’s what we get when working with infinite EVs.

## Guido Bombi:

Let’s consider to numbers a and b, with a,b in [-L,L]: when we learn a we can assign the probability that b > a

p(b > a|a & L) = (L – a)/2L

Then we take the limit for L -> infinity

lim (L – a)/2L = 1/2 – lim a/2L = 1/2

This is how come that the knowledge of a (or x, for that) tells us nothing about b when a and b are taken from the whole R.

31 October 2019, 8:15 am