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Printjmc
counting and probability senior
Problem
We have a standard deck of 52 cards, with 4 cards in each of 13 ranks. We call a 5-card poker hand a full house if the hand has 3 cards of one rank and 2 cards of another rank (such as 33355 or AAAKK). What is the probability that five cards chosen at random form a full house?
Solution
The total number of outcomes is just the number of ways to choose 5 cards from a set of 52, which is . Notice that in this count, we don't care about the order in which the cards are chosen.
To count the number of successful outcomes, we turn to constructive counting, thinking about how we'd construct a full house.
To form a full house, we have to choose:
A rank for the 3 cards. This can be done in 13 ways.
3 of the 4 cards of that rank. This can be done in ways.
A rank for the other 2 cards. This can be done in 12 ways (since we can't choose the rank that we chose in (a)).
2 of the 4 cards of that rank. This can be done in ways.
Again, note that in each of the steps in our constructive count, we don't care about the order in which the cards are chosen.
So there are full houses. Thus, the probability is
To count the number of successful outcomes, we turn to constructive counting, thinking about how we'd construct a full house.
To form a full house, we have to choose:
A rank for the 3 cards. This can be done in 13 ways.
3 of the 4 cards of that rank. This can be done in ways.
A rank for the other 2 cards. This can be done in 12 ways (since we can't choose the rank that we chose in (a)).
2 of the 4 cards of that rank. This can be done in ways.
Again, note that in each of the steps in our constructive count, we don't care about the order in which the cards are chosen.
So there are full houses. Thus, the probability is
Final answer
\frac{6}{4165}