Probability of full house when a player already has 2 cards of the same value (ex 5 of hearts/clubs) when drawing next 3 cards (total 5)

Question

I am having trouble trying to calculate the probability of getting a full house when a player already has 2 cards in their hand. For example, if they have a 5 of hearts and 5 of clubs, then they can either draw another 5 and draw a pair, or draw a three of a kind. I am unsure how to solve this and any help would be appreciated.

Answer 1

Case 1: Draw a separate three-of-a-kind. There are 12 ranks that you do not have. Choose 1 and choose 3 of the four suits. Divide by the number of ways to draw three more cards.

Case 2: Draw one of the remaining cards of the same rank you already have, choose one of the remaining twelve ranks, and choose two of the four suits. Divide by the number of ways to draw three more cards.

Total probability: $$\dfrac{\dbinom{12}{1}\dbinom{4}{3}+\dbinom{2}{1}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{50}{3}} = \dfrac{12}{1225}$$

Answer 2

Alternative counting... Assume we have the two cards $5\heartsuit$ and $5\clubsuit$ in our hand. (The probability in the general case is the same, here it helps to fix ideas.)

Then there remain $52-2=50$ cards, and we have no information about other cards. We extract three cards, their order being important.

So there are totally $50\cdot 49\cdot 48$ cases. Each such case comes with the same probability, $1/(50\cdot 49\cdot 48)$. How many of these cases are favorable (for the full house)?

• The constellation $\{5?,X?,X?\}$ (as a set), $X\ne 5$, comes in the cases $5XX$, $X5X$, $XX5$. Here, the $?$ stays for a color as a joker, different colors are possible. There are two cases for the $5$, $52-4=48$ cases for the first $X$, and $3$ cases for the second $X$. We get $3\cdot2\cdot48\cdot3$ favorable cases with a full house $555XX$ (as a set).
• The constellation $\{X?,X?,X?\}$ (as a set), $X\ne 5$, comes (with same conventions) in with $52-4=48$ cases for the first $X$, $3$ cases for the second $X$, and $2$ cases for the third $X$, getting thus $48\cdot3\cdot2$ favorable.

The searched probability is thus $$ \frac{48(3\cdot2\cdot3+3\cdot2)}{48\cdot 49\cdot 50} =\frac{24}{49\cdot 50} =\frac{12}{1225}\approx 0.0097959183673469\dots \ . $$