一道概率题切磋 6/26/2009 14:52
本题原出自我见的帖。现在我做个全面题解,欢迎指正。

两个人赌博,硬币猜正反,两个人轮流。第一个人A先猜一次,掷一次,猜中赢一次加一分,猜错不扣分。第二个人B也是猜一次,但可以选择掷N次。如果N次全是猜的那面,就得2^(N-1)分;中间有一次不是猜的那面,就不得分。两个人各轮到100次,谁得的分高,谁就赢。
假设B很聪明,可以选择最佳策略。问B最后赢的几率是多少?


下面是我的解释加答案:
The chance of head and tail follows binomial distribution, where the probability of head or tail p=1/2.

Step1:
For A:
his probability mean score = n*p = 1/2 + 1/2 + 1/2 + ..... + 1/2 = 100 * 1/2=50.

For B: his probablilty to be correct each time is also 1/2 but if he chooses to throw n times and all n times are correct - the p=1/2 ^n. When he is correct in all n times, he gets 2^(n-1) points.

Suppose he throws the numbers of n1, n2, n3, ..., n100 for 100 times.
his probability mean score = 2^(n1-1)/2^n1 + 2^(n2-1)/2^n2 + ... + 2^(n100-1)/2^n100=1/2+1/2+1/2+... + 1/2= 100*1/2= 50.
This says that statistically or in probability, B does not have advantage than A because his probability mean score is also 50.

step2:
However, if B is smart, he should choose the largest probability event each time (except when he wants to take risks meaning he chooses lower probability event), which is to throw once that has the largest p=0.5.

If this is not intuitive enough and you ask why, we need to introduce the concept of odds. If B throw once, with p=1/2, the odds of him getting 1 to him getting 0 is p/(1-p)=(1/2) /(1/2) . Or we can say his odds of getting 1 point rather than getting 0 point is 1 to 1. And if he chose to throw 2 times, p=1/4, and his odds of getting 2 points vs. getting 0 point is 1/4 over 3/4 which is 1 to 3. Although if he is correct, he gets 2 points rather than 1 point, but his odds is lower. So he should choose higher odds.

If you agree with step1 and step2 so far:
step3, 我们看看什么Critical 情况下B需要孤注一掷。(只有以下三种决定时刻)
let's consider the following deciding scenarios after the 99th throw:
1. say by the 99th time, B is leading A in 1 point.
If B chooses to throw only once。and A once, there are 4 possibilities for (A,B) combinations (1,0), (1,1), (0,1) (0,0). And 3 out 4 B is going to win; and A has 1/4 chance to win. Why not throw only once? - 所以我认为没必要孤注一掷。

2. If A is leading B for a score of m (m~=0)。
如果B想赢,这种情况下他应该孤注一掷 (meaning taking the risk). 他应该选择扔 the minimum number k that makes 2^(k-1)>=(m+1+1). 因为最后一把A may get another 1 point.
问B最后赢的几率是多少?
However, we have to know B赢的几率 = 1/2^k.

据个例子的话,如果A比B多1分,需要扔3次为了得到4分。他只有12.5%的概率赢。值得不值得呢?他反正已经是死猪了,12.5% is better than 0%.
其实即便是A领先13分,B扔5回仍然有3.125%的几率。如果硬币没有偏差上帝没有特殊偏爱的话 wink ,他们俩个如果都按1次扔,99次以后俩个人应该都在50分左右不是很远的地方。
如果竟然差别很大的话,那就是死猪。唯一区别是B死猪还可以靠投机家运气翻转p>0,而A的机会=0。

3. 那么你可能会问,其实真正最挑战性的局面是在第99把的时候,俩个人的分数是相等的怎么办. 因为这个题目没有说平了怎么办,其次不知道输了后果是什么。如果咱们假设平了就是输的话 (比如一个电视有奖活动You want to be a Millionnaire 只发给一个赢的,平了就没人得奖的话),B 应该选择扔2次,原因是:
如果B and A 都选择扔一回,B如果能赢只能是p=B对A错=1/2*1/2=1/4. 那么同样是1/4的概率,当然选择可以得俩分而不是1分了。(所以3不过是2的一个特例罢了)

为什么要孤注一掷而不是一旦落后一些分数就着急呢? 我想应该很清楚了. 从选择扔1-5次, 正确的概率迅速从50% 到25%, 12.5%, 6.25%, 和 3.125%, 每增加扔一次, 他正确的概率降低一半. 并且, 如果一直选扔一次的话, 如果十次错了, 接下来对的概率会增加, 改变策略是禁忌. 这是为什么多扔的办法应该留做最后孤注一掷 - 是赌运气.

只想赌运气不管概率的人, 以上皆可忽略.
太复杂了 我晕

实际操作可以见利就走 oops
simple at 6/26/2009 20:53 快速引用
simple :
太复杂了 我晕

实际操作可以见利就走 oops


对你怎么可能复杂呢?

如果这个游戏就是最后扔完100把, 谁赢了谁见利. 那不就非得玩完吗? 除非不想得利的才可以走. 不过确实, 不玩个100万以上, 如果只是赢千八白的就可以拒绝玩这个游戏.

应该建议出一个这样的电视节目: Who wants to be millionnaire?
不过出资人就惨了 Laughing
wildcrane at 6/26/2009 23:45 快速引用
直觉是对的,细推敲我要查书,早忘了,
好像有millionnaire这个节目,N<100 Laughing
wildcrane :
simple :
太复杂了 我晕

实际操作可以见利就走 oops


对你怎么可能复杂呢?

如果这个游戏就是最后扔完100把, 谁赢了谁见利. 那不就非得玩完吗? 除非不想得利的才可以走. 不过确实, 不玩个100万以上, 如果只是赢千八白的就可以拒绝玩这个游戏.

应该建议出一个这样的电视节目: Who wants to be millionnaire?
不过出资人就惨了 Laughing
simple at 6/29/2009 11:23 快速引用
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