We've been dealth five cards: two aces one king, one five and one 9. We choose to discard the 5 and the 9 and are dealt two more cards. What is the probability that we end up with a full house?
I'm a bit overwhelmed by this problem. I'll just put my thoughts/ideas below:
We can get a full house in two distinct ways in this situation:
- draw two kings
- draw one king and an ace
There are two aces and three kings left. At the point of drawing, there are 47 cards in the deck left.
- We can draw the 2 kings in $3 \choose 2$ ways. The probability of drawing two of them then is $3 \choose 2$ $\times \dfrac{3}{47} \times \dfrac{2}{46}$. I'm not too sure of this.
- The probability of drawing the ace and the king is $2 \choose 1$$\times$ $ 3\choose 1$ $\times \dfrac{2}{47} \times \dfrac{3}{46} $ . Of this I'm even less sure.
Can someone help me figure out this problem?