一道概率题.

http://www.mitbbs.cn/article_t/Quant/25387519.html

发信人: winterlover (夫妻双双把家还~~~~), 信区: Quant
标 题: 一道概率题
发信站: BBS 未名空间站 (Fri Aug 31 14:04:08 2007)

100名乘客要上飞机,飞机上100个座位。第一个乘客不按号,随便乱坐,剩下的人都安
分守己,会对照自己的座位号(第二个去2号位,第三个去3号位,so on and so forth
),除非座位已经被占,他们才会随即挑一个,请问第100个乘客(最后一个)能坐自
己号的概率是多少?

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答案是0.5

若有n个人,令第n个人能做到的概率为P(n)
除去第一个人选到一号和n-1号的两种情况(概率分别是1/n和1/n*0.5),剩下的情况
都和有{n-1...3}个人的情况等价。 即,1/n(P(n-1)+P(n-2)...P(3))。 而P3=0.5,
故可推知,所有Pn都是0.5。
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in the case of everyone randomly picks seat, the chances are 1/100

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Set a random variable X describing the seat the first person chose. so
X could be 1,2,..., N with 1/N chance.
Set event A as the case the last guy has his right seat.
So conditional on X, total probability depending on N is
P_N(A) = 1/N [P(A|X=1) + P(A|X=2) + ... + P(A|X=N)]
and P(A|X=1)=1, P(A|X=N)=0, P(A|X=i) = P_{N-i+1}(A).
As a result, a recurrence equation is formed
P_N(A) = 1/N [ 1+P_2(A) + P_3(A) + ... + P_{N-1}(A) ]
then, after reforming the terms
P_N(A) = P_{N-1}(A)

As we know P_2(A) = 1/2,
then P_100(A) = 1/2, and P_N(A)=1/2 for any N.

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如果第一个人坐在1号椅子, 那么概率为1.
如果第一个人坐在N号椅子,概率为0,
如果第一个人坐在任何其他椅子m,
把这个人连同他做的椅子m一块儿踢出去,
然后把m号的人re-label为1.那么此后的概率为n-1个人的概率.

所以 p(n) = 1/n * 1 + 1/n * 0 + (n-2)/n * p(n-1),
p(1) = 1

thus, p(n) = 1/2


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