概率趣题

baindeglace

想到一道有趣的概率题， 来考考大家。学过一点概率论的人应该很快可以做出来。
有一个均匀的骰子，6个面，各为1，2，3，4，5，6. 问平均掷多少次可以使得所有扔出的数字之和为10的倍数？

hujunhua

感觉这个问题不成立，没有答案。

wayne

楼主说是有趣的题目，而且很快能做出来，所以应该不需要很复杂的计算．所以大概是在问抛出很多次后开始统计，发现扔出的数字总和是10的倍数的概率是多少．是$1/10$

zeroieme

1. ParallelTable[NestWhile[#+{1,RandomInteger[{1,6}]}&,{0,0},#[[1]]==0\[Or]Not[Divisible[#[[2]],10]]&]//#[[1]]&,{1000000}]//Mean

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mathe

我们假设当前扔过数目之和模10为h时还需要平均仍$E_h$,容易给出递推式是一个9阶方程，解之即可

wayne

\[x_0=\left\{0,1,1,1,1,1,1,0,0,0\right\}, A = \left(
\begin{array}{cccccccccc}
0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 \\
0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 \\
0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 \\
0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 \\
\end{array}
\right)\]
则\[x_n^T = A^n\cdot x_0^T\]

投币$100$次，模为{0,1,2,3,4,5,6,7,8,9}的概率分别如下：
\[\left\{\frac{16332965587501772652417256679283619906555257213264763546370493893810792794747}{163329655875017726524172566789514455134285927618238717885767991592374285369344},\frac{302462325694477271341060308874447782269708315382052340790742985720043639175}{3024623256944772713410603088694712132116406067004420701588296140599523803136},\frac{7259095816667454512185447412912930242667074135515724677245899494042521949575}{72590958166674545121854474128673091170793745608106096838119107374388571275264},\frac{16332965587501772652417256678848797980856268718737363053350324456879182687325}{163329655875017726524172566789514455134285927618238717885767991592374285369344},\frac{65331862350007090609669026714730843137171745972067668697813908358370002233675}{653318623500070906096690267158057820537143710472954871543071966369497141477376},\frac{8166482793750886326208628339309635560150964155191490015391552212332032139561}{81664827937508863262086283394757227567142963809119358942883995796187142684672},\frac{65331862350007090609669026714730843137171745972067668697813908358370002233675}{653318623500070906096690267158057820537143710472954871543071966369497141477376},\frac{16332965587501772652417256678848797980856268718737363053350324456879182687325}{163329655875017726524172566789514455134285927618238717885767991592374285369344},\frac{7259095816667454512185447412912930242667074135515724677245899494042521949575}{72590958166674545121854474128673091170793745608106096838119107374388571275264},\frac{302462325694477271341060308874447782269708315382052340790742985720043639175}{3024623256944772713410603088694712132116406067004420701588296140599523803136}\right\}\]

1. MatrixPower[RotateRight[PadRight[ConstantArray[1/6, 6], 10], #] & /@ Range[10], 100]

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baindeglace

wayne 发表于 2018-2-8 21:17
\[x_0=\left\{0,1,1,1,1,1,1,0,0,0\right\}, A = \left(
\begin{array}{cccccccccc}
0 & \frac{1}{6} &  ...

其实不必模拟100次，把A那个矩阵写出来就可以得出结论了。考虑从1，。。。，10这10个状态的Markov链，A就是转移矩阵。它的稳定分布是（0.1,...,0.1）。于是平均10次可以使得所有的结果和为10的倍数。

更一般的，平均$n$次， 可以使得所有结果之和是$n$的倍数。

wayne

这个不是模拟，都是程序生成的结果，本来想给出一般解

mathe

\(\begin{cases}E_1=1+\frac{1}{6}(E_2+E_3+E_4+E_5+E_6+E_7)\\
E_2=1+\frac{1}{6}(E_3+E_4+E_5+E_6+E_7+E_8)\\
E_3=1+\frac{1}{6}(E_4+E_5+E_6+E_7+E_8+E_9)\\
E_4=1+\frac{1}{6}(E_5+E_6+E_7+E_8+E_9)\\
E_5=1+\frac{1}{6}(E_1+E_6+E_7+E_8+E_9)\\
E_6=1+\frac{1}{6}(E_1+E_2+E_7+E_8+E_9)\\
E_7=1+\frac{1}{6}(E_1+E_2+E_3+E_8+E_9)\\
E_8=1+\frac{1}{6}(E_1+E_2+E_3+E_4+E_9)\\
E_9=1+\frac{1}{6}(E_1+E_2+E_3+E_4+E_5)\\
\end{cases}\)
解后，计算\(1+\frac{1}{6}(E_1+E_2+E_3+E_4+E_5+E_6)\)即可

mathe

[9188594/936421, 9027468/936421, 8884862/936421, 7615596/936421, 25210/3011, 8009904/936421, 8134898/936421, 8060712/936421, 8029226/936421]
然后结果竟然是平均10次