I am particularly bad at counting ….
During a game of poker, you receive five-card hand.
What is the probability that you have exactly one pair?
The denominator is ${52 \choose 5}$.
For the numerator, I put down $13{4 \choose 2} * 12{4 \choose 1} * 11{4 \choose 1} * 10{4\choose 1}$,
which turns out to be bigger than the denominator.
The numerator, according to the answer is $13{4 \choose 2} * 64{12 \choose 3}$.
Why was my answer wrong?
Suppose that you pick out any pair, that's $13{4 \choose 2}$.
Now, you have 12 denominations left, each of which has 4 cards.
If you pick one, that's $12{4 \choose 1}$.
With 11 denominations left, when you pick one, that's $11{4 \choose 1}$.
With 10 denominations left, when you pick one, that's $10{4 \choose 1}$.
Clearing I am not thinking about this correctly. Thanks!
Best Answer
Let us consider the case where we get a pair of kings. This happens in $4 \choose 2$ ways. Now, we need to choose 3 single cards. As Peter said, the order of selecting these does not matter. So we do this in $4 \choose 1$*$4 \choose 1$*$4 \choose 1$ ways. So, the total number of ways of choosing two kings is $4 \choose 2$*$4 \choose 1$*$4 \choose 1$$4 \choose 1$.
Now, there are 13 other possible pairs so the total number for each of these possible pairs is the same so we have 13*$4 \choose 2$*$4 \choose 1$*$4 \choose 1$$4 \choose 1$.
And, to calculate the total probability, we divide this by $52 \choose 5$ as you said.
The percentage seems really small but I think that it is correct?