一道有些难度的统计题

xyh

A pair of dice are to be rolled until all the possible outcomes (sum of the two dices) 2,3,...11,12 have occurred at least once. What is the expected number of dice rolls needed?www.ddhw.com

 

掷一对色子，投掷结果为应该2，3，4，5，6，。。。。11，12。www.ddhw.com

 

如果我们希望让所有的上述结果至少出现一次，那么我们投掷次数的期望值是多少？

（原题为英文，我的中文翻译不好，如果没有表达清楚，请斑竹修改）

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开开心心

  中文翻译不是不好，而是很烂，大家只看英文好了

HF:

First, the probability of each possibility for each roll can be calculated,  p(2)=1/36, p(3) = 2/36, ..., p(7) = 6/36, p(8) =5/36, p(9) = 4/36,..., p(12) = 1/36   At any stage, consider the set of all the outcomes that have no appeared yet, say (i1,i2,..., ik), which is a subset of (2,3,...,12) Let f((i1_,i_2,...,i_k)) be the exptected number of rolls in order for all of (i_1,i_2,...,i_k) to occur. Then we have the induction formula f( (i_1,i_2,...,i_k) ) = 1+ f( (i_2,...,i_k) ) p(i1) + f( (i_1,i_3,...,i_k)) p(i2)+... + f((i1,i2,..., i_(k-1)))p(ik) +  f( (i_1,i_2,...,i_k) )(1-p(i_1)-p(i_2)-...p(i_k)) orwww.ddhw.com f( (i_1,i_2,...,i_k) ) = [1+ f( (i_2,...,i_k) ) p(i1) + f( (i_1,i_3,...,i_k)) p(i2)+... + f((i1,i2,..., i_(k-1)))p(ik) ]/(p(i_1)+p(i_2)+...p(i_k))   For the empty set, of cause, we have f( )  = 0.   This can be implemented using a recursive function.    www.ddhw.com     本贴由[HF:]最后编辑于：2008-6-7 11:58:42

xyh

  楼上两位，闲话少说，赶紧做题！一人一个熊掌伺候

喝醉了的猫

公式好像没错，但是要这么算出来就要人命了   www.ddhw.org---

xyh

的确需要电脑计算，我没有结果 www.ddhw.com

dddwww

  用Monte Carlo方法得到结果是60左右