Quote:
Originally Posted by statmanhal
But OP asked about 4 remaining players. Multiply Bruce's result by 4.
There are 45 unknown cards, three of which are 9’s. For one of your four opponents, the probability he has two nines is C(3,2)/C(45,2) – the number of combinations of 2 nines out of 3 divided by the number of ways he can be dealt 2 cards from 45. Since any one of the four players can have the pair of nines and no more than one, you multiply this result by four.
Pr(one of 4 players has 99|3 nines left in ‘deck’ of 45)
= 4 *3/(45*44/2) = 0.01212
Using direct probabilities
Pr = 4*(3/45)*(2/44)
hmm i think i can understand your idea, so its basically:
V(5,1)*(C(3,2)/C(45,2))= 0.01515
right?
