- #1
alexmahone
- 304
- 0
What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?
My attempt:
Let us first pick the 3 different ranks. There are \(\displaystyle {13\choose 3}\) ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways \(\displaystyle ={13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}\)
Total no. of ways of selecting a five-card poker hand \(\displaystyle ={52\choose 5}\)
\(\displaystyle p=\dfrac{{13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}}{{52\choose 5}}\)
This doesn't match the answer given in the textbook. Where have I gone wrong?
My attempt:
Let us first pick the 3 different ranks. There are \(\displaystyle {13\choose 3}\) ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways \(\displaystyle ={13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}\)
Total no. of ways of selecting a five-card poker hand \(\displaystyle ={52\choose 5}\)
\(\displaystyle p=\dfrac{{13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}}{{52\choose 5}}\)
This doesn't match the answer given in the textbook. Where have I gone wrong?