Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
What kind of "hands" are you considering? There are many different kinds of poker. – Arturo Magidin Apr 16 '12 at 4:37
given 5 cards that is – Ofek Ron Apr 16 '12 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 "$13\cdot48-\text{Full Houses}$". – joriki Apr 16 '12 at 4:40
Please use LaTeX for math: $13 \cdot \binom{4}{3} \cdot 48 \cdot 49$, etc. – JeffE Apr 16 '12 at 7:10
up vote 5 down vote accepted

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.

share|cite|improve this answer
this is correct! thanks!, can you see what was i missing in my calculation though? – Ofek Ron Apr 16 '12 at 4:47
yeah i know, but then in the correction i still am missing something... – Ofek Ron Apr 16 '12 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 – Arturo Magidin Apr 16 '12 at 4:51
wow, thanks! great answer! – Ofek Ron Apr 16 '12 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 $$\frac{13\cdot 4\cdot 48\cdot 44}2=54,912\;.$$

share|cite|improve this answer
I think you're missing a factor of $2$ for the irrelevant order of the two other cards? – joriki Apr 16 '12 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... – Ofek Ron Apr 16 '12 at 4:43
You are overcounting, since you are considering the order of the remaining two cards. – Arturo Magidin Apr 16 '12 at 4:44
@joriki: Yes, I just caught that. It’ll be fixed in a moment. – Brian M. Scott Apr 16 '12 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. – joriki Apr 16 '12 at 4:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.