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I have been trying to solve this question, but in vain. Please help.

You are given two boxes with a number inside each box. The two numbers are different but you have no idea what they are. You pick one box to open; read the number inside; and then guess if the number in the other box is larger or smaller. You win if you guess correctly, and lose otherwise. Is there anyway that you can win the game with more than 50% chances no matter what the two numbers are?

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    Two-envelope paradox?2011-08-13
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    This is usually called *The two envelopes problem* and explained at length here: http://en.wikipedia.org/wiki/Two_envelopes_problem2011-08-13
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    @Didier: No, in the two-envelope problem, you have to decide whether to switch *before* seeing the number inside. If you get to see the number first, there is actually a strategy to get more than 50%.2011-08-13
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    Actually no, this isn't the two envelopes problem, you actually _can_ come up with a solution that gives (slightly) better than 50% chance of winning. Let me try to remember the solution and I will post it.2011-08-13
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    So the old answers will be trotted out once again.2011-08-13
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    @joriki, if you look at the link I gave, you will see that in some variants of the two-envelopes problem you may decide your strategy *after* seeing the number in the envelope you picked, see http://en.wikipedia.org/wiki/Two_envelopes_problem#Extensions_to_the_Problem. Nevertheless there **is** a(nother) difference: the OP does not specify that one number is the double of the other.2011-08-13
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    @Didier: Well, that's not quite fair -- it sounds as if I didn't look at the link you gave. As you say, these are variants of the problem, and not what's usually called "the two envelopes problem"; and, crucially, this one has a proper uncontroversial solution to get more than 50%, whereas the other one is more about foundational issues and people can keep banging their heads together about it forever.2011-08-13
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    @joriki: Sorry if you felt my last comment was too blunt, it was not meant as such. Two things: I DID miss the double-number vs any-numbers aspect (and I said so); and I fully agree with your banging-heads-forever verdict.2011-08-13
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    See [Who discovered this number-guessing paradox?](http://math.stackexchange.com/q/709984/16397) and especially [this previous answer](http://math.stackexchange.com/a/656426/16397). This problem is traceable to a paper of [Thomas M. Cover](https://en.wikipedia.org/wiki/Thomas_M._Cover) “[Pick the largest number](http://www-isl.stanford.edu/~cover/papers/paper73.pdf)”*Open Problems in Communication and Computation* Springer-Verlag, 1987, p152.2015-08-19

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