I am trying to calculate the probability of getting a full house on a standard 5-card deal.
I am comfortable with how combinations, permutations and the fundamental principle of counting, but I am not sure how this problem works.
I understand that we need to avoid situations such as having a 4 of a kind or a flush, so can someone help me out ?
I would deeply appreciate it if you could give me another example like this as well.
Best Answer
You'll need to count how many different combinations of cards will result in a full house.
There are $13 \times 12$ ways to choose the particular two values of the cards that will make up your hand (i.e. kings over eights).
For each of these combinations, there are $_{4} C_{3} = 4$ combinations for the three of a kind, and $_{4} C_{2} = 6$ combinations for the pair.
Overall, there are $_{52} C_{5} = 2598960$ card combinations.
Hence the probability is $$\frac{13 \times 12 \times 4 \times 6}{2598960} \approx 0.00144.$$