What is the probability of 3 heart and 2 distinct pairs in a hand of five cards?

Question

Given that the hand is drawn from a standard deck of 52 cards, what is the probability that it contains three hearts and two distinct pairs?

I know that there are ${52 \choose 5}$ distinct hands of five, and that each pair will contain exactly one heart, with the third heart not being in a pair.

So far, my work is that there are:

${13 \choose 3}$ ways to draw three hearts

${3 \choose 2}$ ways to select which of the two hearts are part of the two pairs

${3\choose2}$ ways to select the suit of the other card for each pair

And so the probability would be $\frac{{13 \choose 3}{3 \choose 2}^3}{{52 \choose 5}}$, about 0.003

I can't help but feel like I am leaving something out, though. Have I forgotten to include anything in the numerator?

Answer 1

This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.

There are there are ${13 \choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.

So there are ${13\choose 2}\cdot 3\cdot 3\cdot 11=$

$\frac {13\cdot 12}2\cdot 99 = 13\cdot 6\cdot 99 = 7722$ ways to do this.

So probality is $\frac {\frac {13!}{11!2!}9*11}{\frac {52!}{47!5!}} =\frac {7722}{2598960} \approx 0.003$

Which is the same answer as yours as I did $ {13\choose 2}\cdot{{13-2} \choose 1}=\frac {13!}{11!2!}\frac {11!}{10!1!}$ whereas you did ${13\choose 3}{3\choose 1}=\frac {13!}{10!3!}\frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).