# How Many Three-of-a-Kinds in Standard Poker?

Given 5 cards hand,How many ways are there to have a three of a kind hand?

I think that it is $13*(4C3)*48*49-(\text{Full Houses})-(\text{4 of a Kinds})$ which is $13*4*48*49-13*12*(4C3)*(4C2)-13*48=122304-3744-624$

Can you confirm that?

Correction thanks to Arturo Magidin: $13*(4C3)*(49C2)-(\text{Full Houses})-4*(\text{4 of a Kinds})=54,912$

• What kind of "hands" are you considering? There are many different kinds of poker. Apr 16, 2012 at 4:37
• given 5 cards that is Apr 16, 2012 at 4:38
• When you write text within formulas, it gets interpreted as a juxtaposition of variable names and therefore appears italicized and incorrectly spaced. To insert text into a formula, you can use the \text command. For instance, 13\cdot48-\text{Full Houses} produces "$13\cdot48-\text{Full Houses}$". Apr 16, 2012 at 4:40
• Please use LaTeX for math: $13 \cdot \binom{4}{3} \cdot 48 \cdot 49$, etc. Apr 16, 2012 at 7:10

If your poker hands have five cards, we can count them as follows:

1. Select the rank of the 3-of-a-kind: there are $\binom{13}{1}$ ways of doing this;
2. Select the suits of the 3-of-a-kind: there are $\binom{4}{3}$ ways of doing this;
3. Select two ranks from the remaining twelve ranks; there are $\binom{12}{2}$ ways of doing this;
4. Select the suit of the higher of the two leftover cards; there are $\binom{4}{1}$ ways of doing this;
5. Select the suit of the lower rank of the two leftover cards; there are $\binom{4}{1}$ ways of doing this.

This gives $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1} = 54912.$$

Your count is off among other reasons because you are considering the order in which you pick the remaining two cards (you should have $49\times 48/2$ in the first summand instead). But the real problem is that you are counting each four-of-a-kind hand four times. For example, if you have four aces and a king, you count it once when your three-of-a-kind are the aces of hearts, diamonds, and spades; then again when they are the aces of hearts, diamonds, and clubs; then again when they are the aces of hearts, spades, and clubs; and yet again when they are the aces of diamonds, spades, and clubs. If you subtract four times the number of 4-of-a-kind hands you get the correct answer.

• this is correct! thanks!, can you see what was i missing in my calculation though? Apr 16, 2012 at 4:47
• yeah i know, but then in the correction i still am missing something... Apr 16, 2012 at 4:50
• You are counting each 4-of-a-kind four times, once for each way to select three cards from the four of a kind Apr 16, 2012 at 4:51
• wow, thanks! great answer! Apr 16, 2012 at 4:52

I’m assuming on the basis of your calculation that you’re talking about five-card hands from a standard deck.

There are $13$ ways to choose the rank of the triplet, and $4$ ways to choose which of the four suits is not reprsented in it. There are then $48$ cards of the other ranks, so there are $48$ ways to choose one of them. Finally, you want a fifth card different in rank from both the triplet and the fourth card; there are $52-8=44$ available cards. However, this counts (for instance) a fourth card of $\diamondsuit 3$ and a fifth card of $\spadesuit 4$ separately from a fourth card of $\spadesuit 4$ and a fifth card of $\diamondsuit 3$, so it counts every pair of ‘filler’ cards twice. Thus, we have to divide by $2$, getting $$\frac{13\cdot 4\cdot 48\cdot 44}2=54,912\;.$$

• I think you're missing a factor of $2$ for the irrelevant order of the two other cards? Apr 16, 2012 at 4:43
• thats not correct, what if the 4th card is the same as the triplet? then you get a four of a kind hand... Apr 16, 2012 at 4:43
• You are overcounting, since you are considering the order of the remaining two cards. Apr 16, 2012 at 4:44
• @joriki: Yes, I just caught that. It’ll be fixed in a moment. Apr 16, 2012 at 4:44
• @Ofek: "There are then 48 cards of the other ranks". That doesn't include the fourth card of the rank of the triplet. Apr 16, 2012 at 4:44