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What your probability is of being correct depends on what two numbers I have. You only can guarantee that it is better than even, but it could be by a very small amount indeed. There is theoretical discussion on this, and the fact that you don't know the probability turns out to be important. (I don't remember details though.) However there is a variant of this problem where both you and I pick our numbers independently out of the same distribution. Now you can work out the probability of your being right prior to my picking my numbers. And even though your number has no actual information about mine, you turn out to be right exactly 2/3 of the time. If you disbelieve this it is easy to write a short script to test it. Again, I got this problem from a probability theorist and it really does work. In reply to RE (tilly) 2: Spooky math problem
by tilly

