Two identical complete decks of cards, each with 52 cards, have been mixed together. A hand of 5 cards is picked uniformly at random from amongst all subsets of exactly 5 cards (order of cards in the hand is not important).

  1. What is the probability of choosing a hand with exact 2 identical cards (1 pair)?
  2. What is the probability of choosing a hand with exact 2 pairs of identical cards?

I tried something, but I probably go wrong somewhere (I have the oposite probability in a textbook and that one summed up with solution of 1. and 2. is not 1).

  1. Let's compute the number of hands with a single pair of identical cards. For the identical cards we have ${52 \choose 1}$. For the rest of the 4 cards we have ${51 \choose 4}$ possibilities (51, as they are not allowed to be as the first selected card), but because every card can be taken from the first or second deck, we have in total ${2^4} \cdot {51 \choose 4}$. So the required probability would be: $ \frac {{52 \choose 1} \cdot {2^4} \cdot {51 \choose 4}} {{104 \choose 5}}$, where the denumerator is the total number of possible hands out of 104 cards.
  2. Similarly for two pairs of identical pairs: $ \frac {{52 \choose 2} \cdot {2^3} \cdot {50 \choose 3}} {{104 \choose 5}}$

My problem is that each of the two probabilities is bigger than 1 (which is definitely wrong) and that they are equal (which my intuition denies to believe).

  • $\begingroup$ In the first subproblem, my error seems to be that for the rest of 4 cards (it is actually 3 cards), we have $50 \choose 3$ possibilities. The same for the subproblem 2. If someone shows me how to solve the subproblems by other ways of counting, I will accept the answer. $\endgroup$ – Andrei I Feb 19 '15 at 21:28

For case 1, first choose the value and suit of the card that is the pair ($52$). Then choose the values/suits of the three unpaired cards ($_{51}C_3$) and then choose which instance of each card goes in your hand ($2^3=8$).

For case 2, choose the value and suit of the two cards that are the pairs ($_{52}C_2$) and then choose the card from the rest of the pile ($100$).

So the probability for case 1 is

$$P_1 = \frac{(52)(_{51}C_3)(8)}{_{104}C_5},$$

and for case 2 it's

$$P_2 = \frac{(_{52}C_2)(100)}{_{104}C_5}.$$

  • $\begingroup$ This is also what I computed. My error is described in a comment. Do you know another way how to compute this? $\endgroup$ – Andrei I Feb 19 '15 at 21:47
  • $\begingroup$ There probably is a way, but I don't know if it will be clearer or more elegant. $\endgroup$ – John Feb 19 '15 at 21:51

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