What is the probability that you are dealt a "two pair"? (Two pairs of cards where each pair contains two cards of the same denomination, with the fifth card of a different denomination. Note that we exclude four of a kind from this definition.)

For this question, my approach is as follows (am not interested in probability but the number of ways I can select the two pairs)

There are ${13 \choose 2}$ ways to select 2 pairs out of 13 ranks.

There are ${2 \choose 1}.{4 \choose 2}$ ways to select first pair out of 2 selected ranks, out of which I select 2 cards

There are ${1 \choose 1}.{4 \choose 2}$ ways to select second pair out of 1 remaining rank, out of which I select 2 cards And ${52-4-4 \choose 1}$ ways to select one card out of remaining ranks

So Shouldn't the number of ways be

$\large{{13 \choose 2} {2 \choose 1}{4 \choose 2}{1 \choose 1}{4 \choose 2}{52-4-4 \choose 1}}$ Could someone tell me why my method is overcounting?

  • 1
    $\begingroup$ You're overcounting because of the $\binom{2}{1}$. You don't care about the order in which you get the cards, and since both the two ranks you selected will be a pair, you shouldn't be selecting the 'first pair'. For example, your current method treats $K_\heartsuit K_\clubsuit Q_\heartsuit Q_\spadesuit A_\diamondsuit$ and $Q_\heartsuit Q_\spadesuit K_\heartsuit K_\clubsuit A_\diamondsuit$ as two different cases, but it shouldn't. $\endgroup$ – stochasticboy321 Nov 22 '15 at 5:12

What you were told is right.
Once you have selected two ranks for the pairs, say $9's$ and $J's$,
all you need to do for the pairs portion is to choose $2$ cards from each rank.

You can't multiply by $\binom21$ because the order in which you choose the cards is irrelevant.


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