Probability problem (F1 race)

好明顯不得。我知你條式中
P(no. 2 win)/ (1 - P(no. 1 win)) 係想表示當car no. 1贏了冠軍時car no. 2的概率﹐但有以下問題
1 1 - P(no. 1 win) 只表示car no. 1贏不到冠軍的概率
2 P(no. 2 win) 是表示當10架車一起時car no. 2贏冠軍的概率﹐它不能夠用來表示任何條件概率。
其實好簡單﹐例如car no. 1玩0西既﹐拿冠軍可能性100%。用你的式子就會得到矛盾

2008-12-23 15:50:58 補充：
其實好難。例如用相乘都不合理﹐例如假定有3架車,勝的概率如下

car no. 1 0.5,car no. 2 0.3,car no. 3 0.2

則頭一二位的結果如下

car no. 1,car no. 2 0.15
car no. 1,car no. 3 0.1
car no. 2,car no. 1 0.15
car no. 2,car no. 3 0.06
car no. 3,car no. 1 0.1
car no. 3,car no. 2 0.06

2008-12-23 15:53:57 補充：
問題
1 不mutually exclusive 2 明明car no. 1,car no. 2的概率應該大過car no. 2,car no. 1的概率。

但係一般情況下只會給每個車手勝出的概率﹐所以這類問題應該不是經典概率範疇

Assume that the 10 cars have equal chances to win the race. Therefore,
probability (no. 1 wins)
= 1/10

After no. 1 won the race, the rest 9 cars have equal chances to be the first runner up. Therefore,
probability (no. 1 wins, no. 2 be 1st runner up)
= (1/10) x (1/9)

After no. 1 won the race and no. 2 became the 1st runner up, the rest 8 cars have equal chances to be the second runner up. Therefore,
probability (no. 1 wins, no. 2 be 1st runner up, no. 3 be 3rd runner up)
= (1/10) x (1/9) x (1/8)
= 1/720
=