只有一点我没看懂，为什么 P(B|Z)=0.5 。B“ be the event that Bob tosses more heads”，是有条件的（例如 Bob 比 Alice 多投一次的前提下？） 还是无条件的。为什么在 Z 已经发生的前提下 B 发生的概率还有 0.5 ？
题目 爱丽丝和鲍勃一共有 2n+1 枚对称的硬币.鲍勃连续抛掷了 n+1 枚硬币,而爱丽丝抛掷 n 枚硬币.证明鲍勃抛出的正面数比爱丽丝抛出的正面数多的概率为 1/2.
答案 Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2.
An alternative solution is to argue that if Alice and Bob are tied after 2n tosses, they are equally likely to win. If they are not tied, then their scores differ by at least 2, and toss 2n+1 will not change the final outcome. This argument may also be expressed algebraically by using the total probability theorem. Let B . Let X be the event that after each has tossed n of their coins, Bob has more heads than Alice, let Y be the event that under the same conditions, Alice has more heads than Bob, and let Z be the event that they have the same number of heads. Since the coins are fair, we have P(X) = P(Y ), and also P(Z) = 1 − P(X) − P(Y ). Furthermore, we see that P(B | X) = 1, P(B | Y ) = 0, P(B|Z)=0.5
Now we have, using the total probability theorem,
P(B) = P(X) · P(B | X) + P(Y ) · P(B | Y ) + P(Z) · P(B | Z)
= P(X) + 0.5· P(Z)
= ( P(X) + P(Y ) + P(Z) ) · 0.5
=0.5
as required.
1
Select Language