Given a standard playing card deck of 52 cards, what is the probability of being dealt a 2 pair 5 card hand consisting exactly of one pair of face cards and one pair of NOT face cards is?


There are $\binom{52}{5}$ hands, all equally likely.

For the number of hands that fit your description (the favourables), the kind of face card we have two of can be chosen in $\binom{3}{1}$ ways. For each of these ways, the actual cards can be chosen in $\binom{4}{2}$ ways. For each of these ways, the kind of non-face card we have two of can be chosen in $\binom{10}{1}$ ways, and the actual cards in $\binom{4}{2}$ ways. Finally, the useless fifth card can be chosen in $\binom{44}{1}$ ways, since we must avoid the kinds we have two of. That gives a total of $\binom{3}{1}\binom{4}{2}\binom{10}{1}\binom{4}{2}\binom{44}{1}$.

  • $\begingroup$ Why must we avoid the kinds of which we have two? $\endgroup$ – Zubin Mukerjee Nov 29 '13 at 22:43
  • $\begingroup$ Because if we get say three Jacks and two $7$'s, we don't have a "two pair" hand, we have a full house, substantially better. It has to do with poker terminology. A two pair hand is by definition a hand that has two cards of two different kinds, and a fifth card of a still different kind. $\endgroup$ – André Nicolas Nov 29 '13 at 22:47
  • $\begingroup$ Okay, thank you. I edited my solution to account for that. $\endgroup$ – Zubin Mukerjee Nov 29 '13 at 22:50

First, I'll solve the problem assuming that it doesn't matter whether or not the fifth card matches either of the pairs.

There are $3 \cdot 4 = 12$ face cards in the deck, and $52 - 12 = 40$ non-face cards.

The number of choices for the positions of the pairs is

$$\binom{5}{2} \cdot \binom{3}{2}$$

The probability of getting the desired cards in a specific order is:

$$P_0 = \frac{12}{52} \cdot \frac{3}{51} \cdot \frac{40}{50} \cdot \frac{3}{49} \cdot \frac{48}{48}$$

$$P_0 = \frac{3}{13} \cdot \frac{1}{17} \cdot \frac{4}{5} \cdot \frac{3}{49}$$

$$P_0 = \frac{3 \cdot 2 \cdot 2 \cdot 3}{13 \cdot 17 \cdot 5 \cdot 7 \cdot 7}$$

$$P_0 = \frac{36}{54145}$$

The final probability is simply the product of this and the number of choices for positions of the pairs:

$$P = P_0 \cdot \binom{5}{2} \cdot \binom{3}{2} = \frac{216}{10829}$$

If it does matter that the fifth card does not match either of the pairs, then the last term in our calculation of $P_0$ becomes $44/48$ instead of $48/48$, which gives us a final answer instead of $$ P = \frac{198}{10829}$$


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