Question

3 cards are chosen from 52. What is the probability that 2 cards have the same rank?

Answer 1

The solution says that the answers is \[\frac{13*\left(\begin{matrix}4 \\ 2\end{matrix}\right)*\left(\begin{matrix}48 \\1\end{matrix}\right)}{\left(\begin{matrix}52 \\ 3\end{matrix}\right)}\] I understand all the numbers except for the (48 1). I though it would've been more like (50 1) since after you pick two cards, you'd be choosing 1 from the 50 left. I am not sure why we are only picking 1 from 48 cards remaining?

Answer 2

Think about the number of ways to get specifically 2 aces.
I'll get 2 aces, of which there are 4. That's (4 2)
Then I'll choose one card that's not an ace. How many non-ace cards are there?

Answer 3

what smoothmath said. it is 48 because you cannot pick one of the same value as the other two. i guess it should say "exactly two" have the same rank

Answer 4

Ohh I see what you mean. I guess "exactly two" would have made more sense to me. Ok that helps a lot thanks :)