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Alice writes two distinct real numbers between 0 and 1 on two chits of paper and places them in two different envelopes. Bob selects one of the 2 envelopes randomly to inspect it. He then has to declare whether the number he sees is the bigger or smaller of the two. Is there any way he can expect to be correct more than half the times Alice plays this game with him?

This puzzle is from https://gurmeet.net/puzzles/bigger-or-smaller/index.html

My approach: I feel that since this is absolutely random and we don't know the distribution of Alice. I looked at this link Puzzle: Guessing the bigger number! but this deals with some fixed distribution by Alice, and that makes sense to me. But when Alice could have any strategy in mind, how does Bob expect to be correct more than 1/2 of the times? Tbh, the solution given does not make any sense to me. Why should it be 1/2 + 1/2 |p-q|?

Also, say, if it wasn't Alice, rather a uniformly random number generator, then Bob could have selected any number > 0.5 and said that to be bigger and he would have 1/2 chance only, never better. Am I correct?

Charlie
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  • If Alice can choose the numbers herself , there is not much Bob can do. She could trap Bob by writing down $0.9999$ and $0.99999$. If Bob draws $0.9999$ , he will assume this is the bigger number. The same story , if Alice chooses $0.0001$ and $0.00001$. – Peter Aug 13 '23 at 07:24
  • If the numbers are random in the interval $[0,1]$ , Bob has much better informations. The strategy for the guess is however rather trivial in this case. – Peter Aug 13 '23 at 07:26
  • @Peter Are you claiming that there is no way for Bob to do better than $50%$ when Alice chooses the numbers herself? If so, you are wrong, as demonstrated in solution to the linked question I proposed as a duplicate. – Mike Earnest Aug 13 '23 at 16:01
  • @MikeEarnest, I have checked the link but that question differs slightly from this one. I agree if there was no interval like [0,1] then Bob has slightly better chances of winning but in this case, you cannot divide like thresholds so Bob has exactly 50% chance of winning – Charlie Aug 13 '23 at 17:48
  • You are incorrect; the exact same strategy given there words here, so this is a duplicate. "You cannot divide like thresholds" I do not understand what you are saying here. – Mike Earnest Aug 13 '23 at 18:20
  • So, what I mean is that if we had range from -inf to +inf, then say we choose some random threshold as x, then if the number we see is greater than x, we say bigger else smaller. Now if both are greater than x, we have 0.5 probability of winning, if both are smaller than x, we have 0.5 probability again but if they are distributed as one smaller than x and the other larger than x, then we win for sure. So the probability is greater than 0.5 for Bob, but we cannot divide like this here. – Charlie Aug 13 '23 at 18:33
  • For this problem, then, just choose the random threshold, $x$, to be a random number between $0$ and $1$. This works for the $[0,1]$ problem, for the same reason as the $(-\infty,\infty)$ version. – Mike Earnest Aug 13 '23 at 20:00
  • Well, in (-inf, inf), we can say probability to be 1/2 for left and right sides of any random number x, but in [0,1], if we choose random number x, won't probability be x, 1-x on either sides so it won't be 1/2 right? – Charlie Aug 13 '23 at 20:18
  • Your last comment is very wrong! The probability is NOT $1/2$ for each side in the $\pm\infty$ case. I suggest that you reread the linked answer carefully, understanding that it NEVER uses the assumption of $1/2$ for each side. See https://math.stackexchange.com/questions/709984/who-discovered-this-number-guessing-paradox for more information on the history of the problem. – Mike Earnest Aug 14 '23 at 17:57

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