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Re:Coin Puzzle: Predict the Other's Coin by datdatruth on Sunday December 30, 2012 @08:23AM (#4985602)
It doesn't, you're simply wrong. See my other comments.
Re:Coin Puzzle: Predict the Other's Coin by Derander on Sunday December 30, 2012 @08:27AM (#4985612)
Think of it like this:

What is the probability that two coins come up with the same face value? 0.5

If we use the guessing scheme where each player guesses his own guess as the result of the other player's coin, then players A and B win if their coins are the same. Their coins are the same with probability 0.5.

Thus, Expectation[game] = Expectation[Game|A & B Win] * P(A & B win) + Expectation[Game|A & B Lose] * P(A & B Lose) = 1 * .5 + -2 * .5 = -1, so C should not play.

If the players guessed randomly then the expectation of the game would be as you say it is.

Re:Coin Puzzle: Predict the Other's Coin by theonewolf on Sunday December 30, 2012 @08:55AM (#4985690)
If you want to say someone is wrong, it would be better to provide an explanation of exactly why they're wrong rather than point everyone elsewhere and make then figure out the hole in the logic.

Point out the hole directly.