m*******s
发帖数: 758 | 1 一个6面骰子，不停扔，
A等pattern: 相邻两个和=12(可以改成8或者其他数字)
B等pattern：相邻两个都是1
然后就是A的停时，B的停时，A或者B的停时，先A后B的概率. | p*****k
发帖数: 318 | 2 see
http://www.mitbbs.com/article_t/Quant/31218015.html
if it's a standard fair dice, the case of 12 is trivial, as there is
the only choice: 6+6, and no difference with 1+1 probability-wise:
E[T(A)]=E[T(B)]=42, E[T(A/B)]=21, Pr[A first]=1/2
8 is also simple since the possibilities (2+6,3+5,4+4) and 1+1
are mutually exclusive:
E[T(A)]=42/5, E[T(B)]=42, E[T(A/B)]=7, Pr[A first]=5/6
it gets complicated when the sum for A is e.g., 7, in which case;
E[T(A)]=7, E[T(B)]=42, E[T(A/B)]=287/46, Pr[A firs |
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