# The probability that a 5-card poker hand contains precisely four kinds

I want to find the probability that a 5-card poker hand contains exactly four kinds. I don't understand the difference between the following answers, yet only the first one is right. Why is the second one wrong?

For me, it equates to: choose your 4 ranks from 13 cards. For every chosen kind, choose between the 4 suits (13 C 4 * 4^4). Then, choose one of the 4 previously chosen ranks (4 choices) and for that rank, choose one of the 3 remaining suits (* 3). Where do I count duplicates? • Do you mean $4$ kings? Do you mean four ranks (so what people would normally call, one pair)? – lulu Jan 13 '17 at 17:27
• In the doubled rank you count duplicate e.g.: take 7,8,9,10 with colors heart everything and 10 spade as additional is the same as take 7,8,9 heart and 10 spade and taking 10 heart as additional. Instead you can first choose your pair and then choose colors. – ctst Jan 13 '17 at 17:32
• @lulu ranks indeed... – el-flor Jan 13 '17 at 17:33
• Given that you mean "four ranks" the pattern is $AABCD$. To count the number of such hands you need to choose $A$ ($\binom {13}1$), pick two cards of that rank, ($\binom 42$), pick three other ranks ($\binom {13}3$), pick one card from each of those ranks ($\binom 41^3$). This matches the first expression. – lulu Jan 13 '17 at 17:34
• Note: there's a typo in what I wrote. When you go to choose the three singleton ranks, $B,C,D$, you can't choose $A$ again so there are $\binom {12}3$ ways to pick them. – lulu Jan 13 '17 at 17:40