作者bora902012

給你看一個版大提供的網頁的內容,
然後你再跑跑看你寫的程式,
就可以看出是誰沒有上過機率的課了

Let Ci denote the event that the car is at door i, and Hj the event that the host opens door j . Then

P(You win the car if you switch)
= P(H3 C2) + P(H2 C3) = P(C2)P(H3|C2) + P(C3)P(H2|C3) = (1/3)·1+ (1/3)·1 = 2/3

and in similar manner we find that
P(You win the car if you don't switch) = (1/3)·p + (1/3)·(1 - p) = 1/3
看出來了嗎?
這兩項P( You win the car if XXXXXXX )的總合已經等於1,

難道另外兩個P( You don't win the car if XXXXXXX )=0嗎?

所以寫這篇文章的人不知道機率總合為1?

這個題目[2/3,1/3]或[1/2]都沒有錯,因為這兩組機率是分別代表不同的意義,
[2/3,1/3]是指在[換,不換]的條件下的條件機率,而[1/2]是指遊戲的中獎機率