1

The paper, “Pick the largest number”Open Problems in Communication and Computation Springer-Verlag, 1987, p152, deals with a version of the two envelopes problem where, after seeing one number, a player has to choose whether a second number is larger or smaller. The paper says that the player can do this with greater than 50% probability if they "pick a random splitting number $T$ according to a density $f(t), f(t) > 0,$ for $t\in (-\infty, \infty)$".

The justification is that there are three outcomes:

  • The number you choose is lower than both numbers, in which case the probability of being correct is 50%
  • The number you choose is higher than both numbers, in which case the probability of being correct is 50%
  • The number you choose is in-between the two numbers, in which case the probability of being correct is 100%

My confusion is that I can't find a proof of this specific version of the two envelopes problem (the vast majority of discussion is about the doubled amount version). Specifically, I'm wondering what the mathematical justification is for sampling from a distribution and what the meaningful difference is between a player choosing a specific number and using that every time, versus them sampling from some random distribution?

RobPratt
  • 50,938
7koFnMiP
  • 111
  • 2
    So long as "The number you choose is in-between the two numbers" must have positive probability for any possible distributions for the two envelopes, your statement of the justification looks to me like a proof. But that is not the case if you use a single specific number every time: there are distributions for the numbers in the two envelopes where that specific number has zero probability of being in-between. – Henry May 20 '24 at 23:28
  • 1
    Previous questions https://math.stackexchange.com/questions/4543781/why-does-this-strategy-to-pick-the-largest-envelope-work and https://math.stackexchange.com/questions/2282445/a-two-envelopes-puzzle are more general than the "double" version – Henry May 20 '24 at 23:33
  • Ah ok I understand. I think my confusion came from how I assumed that the two envelope numbers were drawn corresponding to some distribution over all real numbers, but it's clear that that isn't necessarily the case, which is why we need to sample. Thanks! – 7koFnMiP May 21 '24 at 00:09
  • A detailed solution is also posted under a question whose title doesn't help much in finding it: Help: rules of a game whose details I don't remember!. (The table given there can be useful in deriving the formula for the probability of winning.) Also, lots of variations are discussed, and many references given, in Alexander Gnedin's article Guess the Larger Number. – r.e.s. Jun 06 '24 at 14:48
  • @7koFnMiP I think , we have not enough information. What can the other number be ? Any natural number ? In this case , it is wise to guess the other number is larger. If you see $2$ and the other number is $1$ or $3$ with probability each $1/2$ , you have no possibility to beat 50%. Without additional details , this is an ill-posed problem. – Peter Aug 21 '24 at 09:16

0 Answers0