For each of k=0,1,2,3,4, find the probability that a poker hand (5 cards) contains just k aces.?
Can anybody explain how to do this? You can just do one of k= 0,1,2,3,4...I will do the rest of course. I just want someone to explain it to me.
Thank you in advance
- siamese_scytheLv 78 years agoBest Answer
There are C(52,5) ways to choose a poker hand, where C(n,k) is binomial coefficient.
There are C(48,5) ways to choose five non-ace cards.
There are C(4,1) * C(48,4) ways to choose an ace and 4 non-ace cards.
There are C(4,2) * C(48,3) ways to choose two aces and 3 non-ace cards.
- Bail OutLv 68 years ago
P(poker hand has exactly k aces) = 4ck*48c(5-k)/52c5
P(poker hand has exactly 2 aces) = 4c2*48c3/52c5 = 0.03993