From a deck of $52$ cards, a $5$-card hand is dealt. Find the number of hands containing exactly one pair

combinatoricsprobability

From a deck of $52$ cards, a $5$-card hand is dealt. Find the number of hands containing exactly one pair without considering the variation of the suits?

my attempt to solve it :
split the $52$ into $13$ subset of $4$ identical items now I will choose one pair
then I will choose one from each subset and then removing duplicates?
$$ ^{13}C_1 \times (\frac{^{12}C_1 \times ^{11}C_1 \times ^{11}C_1}{3!})$$

Best Answer

Without considering the suits, my approach would be:

The pair can be one of 13 numbers, i.e. we need to pick 1 out of 13 candidates.

For the remaining 3 cards there are two possibilities.

  1. They can be 3 cards with identical number. In this case we need to pick 1 out of 12 candidates.

or

  1. They can be 3 cards with different numbers. In this case we need to pick 3 out of 12 candidates.

Which gives us:

$$\binom{13}{1} \cdot (\binom{12}{1} + \binom{12}{3}) = 13 \cdot (12 + 220) = 3016$$

EDIT

In comments it's discussed whether or not 3 cards with identical numbers represent a pair or even 3 pairs.

To me a hand like 2 2 3 3 3 contains exactly one pair and one triple. To me it's not a hand with 2 pairs nor is it a hand with 4 pairs (which has been suggested in comments).

Anyway - if a triple counts as 3 pairs the answer to the question is even simpler. Just remove the hands containing a triple.

$$\binom{13}{1} \cdot \binom{12}{3} = 13 \cdot 220 = 2860$$

However, I disagree with considering "3 of a kind as 3 pairs". Consider this example. The hand is 3 3 3 4 5. Now put your first pair on the table. So 3 3 goes to the table and the remaining hand is 3 4 5. Now put your second pair on the table. You can't ... there is no pair in 3 4 5.

So if a triple is to be considered also as pair, it must (IMO) be as exactly one pair. In that case the calculation changes to include hands like 2 2 2 3 4. So if a triple counts as 1 pair the calculation is:

$$\binom{13}{1} \cdot (\binom{12}{2} + \binom{12}{3}) = 13 \cdot (66 + 220) = 3718$$

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