题看错,...被任一人(除孔明)得的概率算式如下:

来源: 于 2010-12-19 13:37:19 [旧帖] [给我悄悄话] 本文已被阅读：次

P(E)=SIGMA 4!P(132,4I0+K-1)[P{132-4I0-K+1,4(I3-I2+I2-I1+I1-I0)-3}]P(137-4I3-K)/137!   

{ 0<=K<=3,0<=I0<I1<I2<I3<=34,137-4I3-K>0 }