for the first one, 3 can be ahead , in the middle or after 1,2. Only in the last case you can win. Thus the prob is 1/3 For the second one, when you hold 1, you have half prob to win or loss, otherwise you still hold 1. By symmetry, the prob of win is P(draw 1 first) * 1/2 + P(draw 2 first)*1/2 = 1/3
for third one, by conditional prob, = P(draw 1 first)*(draw 2 first along 2 and 3)+P(draw 2 first)*(draw first along 1 and 3) = M/(N+M+K) * N/(N+K) + N/(N+M+K) * M/(M+K)
【在 s*******s 的大作中提到】 : for third one, by conditional prob, : = P(draw 1 first)*(draw 2 first along 2 and 3)+P(draw 2 first)*(draw first : along 1 and 3) : = M/(N+M+K) * N/(N+K) + N/(N+M+K) * M/(M+K) : : 和2
s*s
17 楼
that is irrelevant to the event of winning and shouldn't be considered in the conditional prob space.
【在 d****i 的大作中提到】 : 不对吧,楼主说没分出胜负要接着抽。比如抽了1,1或2,2的情况 : : first
c*e
18 楼
Hehe, guru from quant board comes. So does a solution.
first
【在 s*******s 的大作中提到】 : for third one, by conditional prob, : = P(draw 1 first)*(draw 2 first along 2 and 3)+P(draw 2 first)*(draw first : along 1 and 3) : = M/(N+M+K) * N/(N+K) + N/(N+M+K) * M/(M+K) : : 和2
l*e
19 楼
The third one answer: 2*M*N/((M+N+K)*(M+N+K-1))
m*k
20 楼
>for 1,2,以后输的概率是1/2 why 1/2? there are many many 1 and 2's I think no matter what I hold in hand, the P of next one is 3 is always 1/3