To decide who would play first in a human vs. AI chess match, the human contender, Garry, suggested flipping a coin. 为了决定谁在人机象棋比赛里有先走权，嘉里说，抛硬币决定。He produced a quarter from his pocket, stating heads you win, tails I win. 他从裤子里掏出一枚硬币，说正面你赢，反面我赢。The computer replied, Humans are very devious. 电脑说，人是很狡猾的。I know that your coin is not fair; the probability of heads on that coin is (2/e)^pi. 我知道你的硬币不是枚好硬币，得到正面的概率是2/e的pi次方。The computer went on to describe a method, using flips of the biased coin, which would give an exactly fair chance to both sides. 电脑接着介绍了一个方法，能利用嘉里的硬币模拟出一枚好硬币，使得得到正面和反面的概率是一样的。The expected number of coin flips required was finite. 抛硬币次数的期望值是有限的。你能想到那个方法吗？

冯诺依曼的解答：

Flip the coin twice. 抛两次硬币

If get Head followed by Tails, call the result HEAD. 假如结果是正面-反面，把这个结果定义为新的正面。

If get Tails followed by Head, call the result TAIL. 假如结果是反面-正面，把这个结果定义为新的反面。

Otherwise, twice flip again. 否则，重复上述二掷的过程。

The probability of gettting HEAD and TAIL is both x(1-x) where x = (2/e)^pi. 得到新的正面和反面的概率都是x(1-x)，x的值是(2/e)的pi次方。

In fact, the exact value of bias does not matter in this simulation, as long as the bias is not 0 or 1, we can always simulate a fair coin using a biased coin. 其实在这个方法中，只要出现正面和反面的概率不是0或者1，我们就可以用相同的方法模拟出一枚好硬币，不管原有硬币出现正面和反面的概率是多少。 In finite expected time an unbiased outcome will be achieved because p*2 + (1-p)*(2+t) = t, where p = x*(1-x) and t = expected number of flips. 在有限的抛掷次数内，一个胜负结果就会出现，因为上述式子。Thus, t = 2/p = 2 / ((2/e)^pi *(1- (2/e)^pi)) ~= 8.47 flips 经计算，抛掷次数约为8.47次。

沃顿统计社11月的题目（附中文翻译）