Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

Re:Chess: Who will win in this riveting game of Math.random(... by ekianjo on Monday January 26, 2015 @12:33AM (#8945652)

if completely random, you may prove that there is an empirical advantage to starting the game without taking in account the level of any player, because two players playing randomly should be equivalent. I'm not entirely sure, though.

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Re:Chess: Who will win in this riveting game of Math.random(... by mehrdada on Monday January 26, 2015 @02:20AM (#8945898)

This is technically correct, but does not result in a very useful information, as P(A|B) can be a very different beast than P(A).

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Re:Chess: Who will win in this riveting game of Math.random(... by ema on Monday January 26, 2015 @03:19AM (#8946003)

Imagine a really simple game where there are only 2 moves: saying 1 and saying 0. The player who first says 1 wins. So of course the player who has the first turn will in practice always win, but when both players move randomly it's 50/50.

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Re:Chess: Who will win in this riveting game of Math.random(... by jdoerrie on Monday January 26, 2015 @05:19AM (#8946328)

Not true. Assuming both players move completely random (both choose either 1 or 0 with 50% probability each) and independently (past actions don't influence the current event) player 1 has a 2:1 advantage, i.e. he will win with probability 2/3. This is because player 1 can win after an even number of rounds (including 0) with 50% probability. So his total chance of winning is 1/2 * (1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3.

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Re:Chess: Who will win in this riveting game of Math.random(... by ema on Monday January 26, 2015 @05:51AM (#8946407)

Yes you're right. Quite the embarrassing mistake I made there.

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