The probability of getting a full house in a 7 card poker hand

combinatoricspokerprobability

I'm trying to calculate the probability of getting a full house in a 7 card poker hand. This is obviously the number of 7 card full house hands divided by $\binom{52}7$. To try and calculate the number of 7 card full houses, I tried this strategy:

  1. chose two ranks $\binom{13}2$
  2. get two cards of the first rank $\binom{4}2$
  3. get three cards of the second rank $\binom{4}3$
  4. chose two cards from the remaining 47 cards $\binom{47}2$

When I do this, I calculate $\frac{2,023,632}{133,784,560}$, which equals about $0.015$, or $1.5\%$. However, wikipedia says I should be getting about 2.6%. Why is my answer so wrong?

Thanks!

-Diego

Best Answer

In $7$-card poker, the best five cards are selected. Therefore, you can obtain a full house in three ways:

  • A three of a kind, a pair, and two singletons
  • A three of a kind and two pairs
  • Two three of a kinds and a singleton

Also, you failed to distinguish between the rank of the three of a kind and the rank of the pair. By selecting two additional cards at random, you created the possibility of a four of a kind, which is not permitted. The first error made your outcome too small. The second error would have led to an over count had you not made the first error, particularly since the four of a kind could have been paired with a pair or a three of a kind.

A three of a kind, a pair, and two singletons: There are $13$ ways to select the rank of the three of a kind, $\binom{4}{3}$ ways to select three of the four cards of that rank, $12$ ways to select the rank of the pair, $\binom{4}{2}$ ways to select two cards of that rank, $\binom{11}{2}$ ways to select the ranks of the two singletons, and $4$ ways to select a card from each of those ranks.

There are $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}\binom{11}{2}\binom{4}{1}^2$$ such hands.

A three of a kind and two pairs: There are $13$ ways to select the rank of the three of a kind, $\binom{4}{3}$ ways to select three of the four cards of that rank, $\binom{12}{2}$ ways to select the ranks of the two pairs, and $\binom{4}{2}$ ways to select two cards from each of those ranks.

There are $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{2}^2$$ such hands.

Two three of a kinds and a singleton: There are $\binom{13}{2}$ ways to select the ranks of the three of a kinds, $\binom{4}{3}$ ways to select three of the four cards of each of those ranks, $11$ ways to select the rank of the singleton, and $4$ ways to select a card of that rank.

There are $$\binom{13}{2}\binom{4}{3}^2\binom{11}{1}\binom{4}{1}$$

Since these three cases are mutually exclusive and exhaustive, the number of favorable cases is found by adding the three results. Dividing that sum by $\binom{52}{7}$ yields the desired probability.

Addendum: As aschepler pointed out in the comments, it is not possible to combine a full house with a straight or a flush with only seven cards, so the cases above are indeed exhaustive.

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