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How many 7-card hands can be dealt having exactly 3 of a kind?

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?

13 is for the possibilities of the rank of the 3 of a kind, ${4\choose 3}$ is for the selection of 3 cards from that rank, and ${48\choose 4}$ is for the selection of the remaining 4 cards from the 48 cards that are not in the selected rank.

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    $\begingroup$ Doesn't look right....you choose three of a kind, yes, but then you choose four cards randomly (not matching the rank you previously selected). But if you started with $3$ Aces, the four cards you draw might all be Kings! You need to specify that the four cards you draw all have different ranks. $\endgroup$
    – lulu
    Commented Feb 22, 2017 at 2:11
  • $\begingroup$ How did you get this answer? Also, please put the question into the main text itself, and show us the logic behind the answer. $\endgroup$
    – The Count
    Commented Feb 22, 2017 at 2:15
  • $\begingroup$ in the unusual interpretation of two three-of-a-kinds being allowed (which i don't think it should be) you will have overcounted as well with this $\endgroup$
    – JMoravitz
    Commented Feb 22, 2017 at 2:23
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    $\begingroup$ What does this mean? Three cards with the same face value and then four more which all have different face values from each other and from the three of a kind? $\endgroup$
    – user49640
    Commented Feb 22, 2017 at 2:26
  • $\begingroup$ You have to be specific about what you mean by "exactly 3 of a kind." Do you mean that the best poker hand you have is "3 of a kind?" Or are you allowed to have a flush, or a full house? $\endgroup$
    – Thomas Andrews
    Commented Feb 22, 2017 at 2:33

3 Answers 3

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$13\binom 43\binom{48}4$ counts the ways to select one from thirteen values of three from four suits and four from the remaining forty-eight cards not of that value.   That is a triple and four other cards not of the triple's value.

This does not exclude pairs, quadruples, nor a second triple within those four cards.   This over counts the cases of two three of a kind.

Use the Principle of Inclusion and Exclusion to avert this.

Further, what does "exactly three of a kind" mean if not "one triple and four singletons".   In that case you need to count ways to select: a value and three suits for the triple, and three other distinct values and a suit for each of the singletons.

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You're not paying attention to whether any of the four cards you pick from the remaining 48 have the same face value.

You need to pick your five different face values. There are $\binom{13}{5}$ ways to do that.

Then you need to pick which of the five face values will be the three. There are five ways to do that.

Then you need to decide, for each face value of which you only have one card, which suit that card will be. There are four ways to do that for each face value, so $4^4$ ways in all.

Finally, you need to decide which three suits will be your three of a kind. There are $\binom{4}{3}=4$ ways to do that.

So overall, there are $\binom{13}{5} \times 5 \times 4^5 = 6589440$ ways to pick your hand.

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  • $\begingroup$ Since it's a 7-card hand, did you actually mean to say ${13\choose 7}\times{7}\times{4^5}$? $\endgroup$
    – Leanne
    Commented Feb 22, 2017 at 3:09
  • $\begingroup$ No. Let's say the hand is AAAKJ94. Then the first step in my approach was to select the face values A, K, J, 9, 4. There are five of them. $\endgroup$
    – user49640
    Commented Feb 22, 2017 at 3:11
  • $\begingroup$ Oh, I see now. Thanks. It's been quite some time since I did some serious combinatorics. $\endgroup$
    – Leanne
    Commented Feb 22, 2017 at 3:13
  • $\begingroup$ Sorry, I've corrected the final answer. $\endgroup$
    – user49640
    Commented Feb 22, 2017 at 3:16
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The first part is correct, but you need to exclude the possibility of having a second triple within the other four cards.

C(exactly one triple) = C(triple) - C(pair triples)

C(triple)

Choose a rank for "triple": C(13,1)

Choose three cards for "triple": C(4,3)

Choose four cards not "triple": C(48,4)

C(pair triples)

Choose a rank for "each" triple: C(13,2)

Choose three cards for "first" triple: C(4,3)

Choose three cards for "second" triple: C(4,3)

Choose one card not from "pair triples": C(44,1)

*52 cards in deck; first triple = 4 cards, second triple = 4 cards, remaining cards to choose that are different from the pair of triples: 52-4-4=44

C(triple) = C(13,1)*C(4,3)*C(48,4) = 13 x 778,320

C(pair triples) = C(13,2)*C(4,3)^2*C(44,1) = 78 x 704

C(exactly one triple) = 10,118,160 − 54,912 = 10,063,248 ways

If you want to know the probability: just divide the above, by the number of ways to choose 7 cards from a deck of 52 = C(52,7) = 133,784,560

P(exactly one triple) = Actual Outcomes / Total Outcomes = 10,063,248 / 133,784,560 = 0.075219.. ~7.5%

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  • $\begingroup$ Please use MathJax to format your answer. $\endgroup$
    – an4s
    Commented Jul 25, 2018 at 19:19
  • $\begingroup$ This is confusing to follow. You have C(triple) but don't define your notation C(). I presume you mean combinations. Please learn MathJax to format your questions properly, see: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Daniel Buck
    Commented Jul 25, 2018 at 19:25

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