I need to calculate the probability to get fullhuose in a poker game, using 54 cards - included 2 jokers.
I know that I have 13 kinds of cards, but I need 2 out of 4, then I have 12 kinds and I need 1 out of 4, and then I have 2 jokers but I need 1. (beside 3744 regular fullhouses).
My calculation:
$$13\binom{4}{2}\cdot12\binom42\cdot2=11,232$$
but I know that the right score is 5616. where am I wrong?
First, $$13\binom43\cdot12\binom42\cdot2=7488\;,$$ not $11,232$; $$13\binom43\cdot12\binom42\cdot3=11,232\;.$$
In fact there are $$13\binom43\cdot12\binom42=3744$$ full houses that do not use either joker. Now we’ll count the full houses that do include at least one joker.
If both jokers are present in a hand that does not contain a pair, the hand cannot be interpreted as a full house, and if both jokers are present in a hand that contains a pair, they’ll be used to make a four of a kind, which beats a full house; thus, we may assume that the hand contains just one joker.
If the hand also contains an ‘honest’ three of a kind, the joker will be used to make four of a kind, so we may assume that the hand consists of two pairs and a joker. There are $\binom{13}2$ ways to choose the denominations of the pairs, and for each of those denominations there are $\binom42$ ways to choose $2$ cards of that denomination. Finally, there are $2$ ways to choose the joker, so there are altogether
$$\binom{13}2\cdot\binom42^2\cdot2=5616$$
full houses of this type. Note that we do not want to multiply by another factor of $2$ to account for the fact that the joker could be combined with either of the pairs to make a three of a kind: it will always be combined the higher pair. The grand total is therefore
$$3744+5616=9360\;.$$