送交者: steven 于 July 12, 2001 01:40:46:
someone asked about the probabilty question about head and tail, here is what he said:
A的答案是第10次出现正反面的概率相同。理由是第10次抛硬币完全是一个独立的随机事件，与前几次无关。
B的答案是优先选择反面。理由是：可以把10次抛硬币中出现正面的次数看成一个随机变量，那么它将服从二项式分布。那么第10次是正面的概率为
Pr{X=10} = (10,10) * (1/2)^10 = 1/1024
其中(10,10)为组合数10选10；第10次出现反面的概率为
Pr{X=9} = (10,9) * (1/2)^9 * (1/2) = 10/1024
我个人觉得A是对的，但也很难反驳B。现在我们把问题变一下：某人一次抛出10个硬币，然后盖住一个，其余的9个恰好正面朝上，现在让另一人猜测盖住的一个。
Answer A is correct, the probabilty is remain the same.
the definition of probabilty for some event set E is:
P(E) = |E|/|U| where U is the union of all event sets.
In the coin problem U = {H, T}, and from experiment,
we know P(H) = P(T) = 1/2. and each toss is independent of previous outcome. The answer B mistaken the fact that
when Bernoulli distribution applys only to a sequence of
events, for P(X=10), the event space U'=UXUXU...XU 10 times. where UXU is cartesian product, then each event becomes a 10th tuple where
x_i \belong U. In this case, answer B is equivlent to the answer of the following question:
when I throw ten coins at once, which one, getting 10 heads, or getting 9 heads and 1 tail has higher probabilty. That is different from asking the probabilty
of head or tail for on toss.
In conclusion, answer B does not apply since B applies to a sequence of events.