ok... let me try

来源: lovecorn 于 2010-03-25 22:02:28 [] [博客] [旧帖] [给我悄悄话] 阅读数 : (685 bytes)

veggiemaggie is much correct i think...

say A, B, C are the gates. one of them leading to luck(treasure/heaven) or any good things.

P(A) = 1/3 = P(B) = P(C)
and P(A_bar) = P(B_bar) = P(C_bar ) = 2/3

suppose you choose A with (mind).
And B is shown not leading to luck, noted as B_bar

Problem now is to find P(A|B_bar) and P(C|B_bar)
Let's take P(C|B_bar) = P( C & B_bar) / P(B_bar)

P(C&B_bar) = P(B_bar|C) * P(C) = 1* 1/3 = 1/3
P(B_bar) = 2/3
So now P(C|B_bar )= 1/2

Similarly, P(A|B_bar ) = 1/2 (A, C are symmetric)

so why want to switch??

Also, I probably older than 90% of people here :(




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