Timeline for How is it that you can guess if one of a pair of random numbers is larger with probability > 1/2?
Current License: CC BY-SA 2.5
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| Jan 19, 2016 at 6:11 | comment | added | Steven Landsburg | Let the two numbers be $a$ and $b$ and denote your strategy by $S$. Your probability of winning is some function $P$ of $a$, $b$ and $S$. The claim is that there exists an $S$ such that for all $a$ and $b$, $P(a,b,S)>1/2$. This is a perfectly meaningful claim and does not rely on any probability distribution for $a$ and $b$. | |
| Jul 5, 2010 at 19:46 | comment | added | Lucas K. | If the distribution isn't uniforn, then P(x>y) isn't 1/2 after revelation of x. You get a conditional probability then. One of the pitfalls of probability is to ignore a fact. | |
| Jul 5, 2010 at 17:20 | comment | added | Daniel Mehkeri | I don't agree with this either, and moreover I think the xkcd (continuous) version adds nothing and just muddies things more. Among other things the xkcd answer isn't a "strategy" since there is no finite way to compare two arbitrary real numbers (not even two computable ones). If they are compared to only finite precision, I think there is a way to defeat the strategy. | |
| Jul 5, 2010 at 5:41 | comment | added | Qiaochu Yuan | I don't think this is the only issue. As Darsh puts it, the surprise, mathematically speaking, is that the same strategy works regardless of the probability distribution on x and y. This is a nontrivial fact which is not accounted for by the observation that we should specify such a distribution. | |
| Jul 5, 2010 at 4:48 | history | edited | BlueRaja | CC BY-SA 2.5 |
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| Jul 4, 2010 at 22:59 | comment | added | Daniel Asimov | I think you mean Randall Munroe. | |
| Jul 4, 2010 at 21:04 | comment | added | Lucas K. | I always advice for these kinds of problems to simulate it (and put real money on it!!!). Then the wrong thinking becomes clear. You will see that it is rather hard to take an arbitrary number, especially when you have a computer with finite memory. | |
| Jul 4, 2010 at 21:00 | comment | added | Lucas K. | Yes, I agree with this. It is the most simple answer, but the correct one. You say 'Given that the original numbers were chosen "arbitrarily" (i.e., without using any given distribution)'. This is an inconsistent assumption, which leads to the paradox. You have to choose a distribution. In case of a die, you can choose the same probability for each side. In case of infinite set, you get troubles and you should take care that you don't assume something impossible. This is also the cause of the envelope problem paradox. | |
| Jul 4, 2010 at 20:30 | history | answered | BlueRaja | CC BY-SA 2.5 |