作者LimSinE (r=e^theta)
看板Math
标题Re: [机统] 掷骰问题
时间Wed May 11 04:09:31 2011
※ 引述《shancool (善)》之铭言：
: Let Xn be the sum of the first n outcomes of tossing a six-side die
: repeated in an independent fashion.
: Compute
: P(Xn is dibisible by 7) as n->∞
: 先谢谢各位了!!
这题是可以用Markov chain，如原文推文
这里提供另一个作法：
设w = cos 2pi/7 + isin 2pi/7
Zi = w^(第i次点数)
则 Zi, i.i.d, E(Zi^r)= -1/6, if r=1,2,3,4,5,6; E(Zi^7)=1
设 Yn = w^Xn，则Yn = Z1Z2...Zn
E(Yn^r)=(EZ1^r)^n= (-1/6)^n, E(Yn^7)=1
因为Yn只能取值1,w,...,w^6，设P(Yn=w^k)=P(Xn=k mod 7)= P(n,k)
记
Pn = [P(n,1),...,P(n,7)]^T
A =[w^ij],1<=i,j<=7
则 APn = [(-1/6)^n,...,(-1/6)^n,1]^T
Pn = A^-1 [(-1/6)^n,...,(-1/6)^n,1]^T → P= A^-1[0,0...0,1]^T as n→无限大
故 AP = [0,...,1], 发现一解[1/7,...,1/7]^T
A 可逆，故解得P=[1/7,...,1/7]^T
所求即1/7
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