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The Sleeping Beauty problem (Elga, 2000; see also Piccione and Rubinstein, 1997) is a philosophical dilemma related to conditional probability. It may be succinctly described as follows:
Sleeping Beauty is put to sleep, and a fair coin (say, a nickel) is tossed. If the nickel shows heads, then Beauty is interviewed on Monday only, while if the nickel shows tails, Beauty is interviewed on both Monday and Tuesday (and given an amnesia-inducing drug between the two interviews, so she does not remember the rst interview during the second). In each interview, without access to any additional information (such as the result of the coin toss, or the existence of any previous interviews, or the day of the week), Beauty is briefly woken and is asked to assess the probability that the nickel showed heads. The question is, what probability should she assign to this?
意思大概是
一个睡美人叫去被做一个实验，在周日吃个催眠药睡去。然后科学家们扔一个均匀的硬币，如果朝上，则在周一唤醒这个美人，如果朝下，则会在周一唤醒这个美人，然后再给她吃药，周二再次唤醒这个美人，值得注意的是药效很猛，导致美人无法记得是周几，也记不住自己是否曾经被唤醒。那么美人被唤醒时，会问这个美人对硬币面向上的概率进行一个估计（美人知道这个实验的设计）。
有两个答案
1. 1/2, 原因如下，一开始美人知道硬币是均匀的，被唤醒后没有增加任何信息，所以还应该1/2
2. 1/3 美人有三种情况被唤醒，tail 周二 tail 周一 head 周一， 所以概率应该是1/3
大概意思如此 具体可以参加http://en.wikipedia.org/wiki/Sleeping_Beauty_problem
大家觉得哪一种更合理呢：）
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