# What is the probability that we end up with a full house?

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?

We can draw the 2 kings in ${3\choose 2}$ ways.

Your logic up to this point is perfect.

The probability of drawing two of them then is ${3\choose 2} \times \frac 3{47} \times \frac 2{46}$. I'm not too sure of this.

you have a little bit too much going on. you could say

$P(K,K) = \frac {3\choose 2}{47 \choose 2} = \frac 3{47} \times \frac 2{46}$, but not multiplied.

$P(A,K)$ You pick one king (3 ways) and 1 ace (2 ways).

$P(A,K)=\frac {6}{47\choose 2} = 2\times \frac 3{47}\frac 2{46}$ .

$P((K,K) or (A,K)) = 3\times\frac 3{47} \times \frac 2{46}$

• Our results seem to be different. Let's take an example to illustrate if we actually disagree (or one of us has made an error): Take the probability of drawing two kings. There are three kings left in the deck, call them $K_1,K_2,K_3$. We can then draw two of them in the following ways: $(K_1,K_2),(K_1,K_3),(K_2,K_1),(K_2,K_3),(K_3,K_1),(K_3,K_2),$ i.e. in six ways. The probability of each of these cases is $\frac{3}{47}\cdot\frac{2}{46},$ so the total probability should be $6\cdot\frac{3}{47}\cdot\frac{2}{46},$ IMO. Do you disagree? Sep 12, 2016 at 21:43
• yes, I do. As you have laid out the possibilities $(K_1,K_2)$ is different from $(K_2,K_1)$ which is fine if you want to do it that way. I would tend to think of them as being equivalent in this case. Probability is the (number of way to deal 2 cards that meet your criteria) / (total number of ways to deal 2 cards) = $\frac {3\cdot 2}{47\cdot 46}$ Sep 12, 2016 at 21:50
• Welp, I agree with you, thank you for clearing that up! (I will either delete or edit my answer.) Sep 12, 2016 at 21:54
• @Lovsovs In selection order does not matter,hence (K1,K2) is same as (K2,K1). Sep 12, 2016 at 22:23
• @DougM By that same logic, shouldn't the probability of the ace-king-scenario be $\frac{12}{47\cdot 46}$, since the number of different ways of drawing a king and an ace is $12$ ($6$ for ace first, $6$ for king first)? Sep 13, 2016 at 11:01