这个讨论过吗？两个盒子

HF:

Here is a bookish approach, hope it can convince you or please point out where the logic fails.   Let X, Y, Z are random variables taken values in {1 2 3}, where X is the door behind which hides the precious car, Y is the door that the player picks, Z is the door that the host opens.  So we have the following:www.ddhw.com Prob( X = i, Y = j) = 1/9, for i = 1,2,3, j = 1,2,3 Prob(Z=1|X=1,Y=1)= 0, Prob(Z= 2|X=1, Y = 1) = Prob(Z=3|X=1, Y=1) = 0.5    Prob(Z=1|X=2,Y=1)=0,  Prob(Z=2|X=2,Y=1)=0, Prob(Z=3|X=2,Y=1) = 1 Prob(Z=1|X=3,Y=1)=0, Prob(Z=2|X= 3, Y=1) = 1, Prob(Z=3|X=3,Y=1)=0 .....   So, for example,  Prob(X=Y|Y=1,Z=2) = Prob(X=1|Y=1,Z=2) = Prob(X=1,Y=1,Z=2)/Prob(Y=1,Z=2) =  (0.5*1/9)/(0.5*1/9+0+1/9) = 1/3 www.ddhw.com     本贴由[HF:]最后编辑于：2007-2-22 8:13:57