You are dealt a hand of five cards from a standard deck of playing cards. Find the probability of being dealt a hand consisting of four-of-a-kind.
If possible, please provide a hint first before the answer.
One of the first things that came to me was $\dfrac{52}{52} \cdot \dfrac{51}{51} \cdot \dfrac{3}{50} \cdot \dfrac{2}{49} \cdot \dfrac{1}{48}$ but this was of course wrong.
Then, I realized that the third card could be the same as the first OR second card, so I tried $\dfrac{52}{52} \cdot \dfrac{51}{51} \cdot \dfrac{7}{50} \cdot \dfrac{2}{49} \cdot \dfrac{1}{48}$, which is also wrong.
Then I realized that we probably need to add up the probabilities of situations where the first card is the one that doesn't match, or the second is the one that doesn't match, etc. I think this method leads to the solution, but I don't think we're intended to solve it this way. I think the solution looks something like $\dfrac{x}{ {{52}\choose{5}}}$, but I'm not sure what should be in the numerator.
Best Answer
First, because there are 13 sets of four-of-a-kind, number of ways of selecting one of the 13 ranks is $\begin{pmatrix}13\\1\end{pmatrix}$
Then, once you have chosen the rank, the number of ways to draw the other three cards of that rank is $\begin{pmatrix}4\\4\end{pmatrix}$
Finally, the number of ways to draw the last card. which is to choose from 48 other cards: $\begin{pmatrix}48\\1\end{pmatrix}$
The number of possible 5 card hands dealt from a 52 card deck is $\begin{pmatrix}52\\5\end{pmatrix}$
So the final answer is:$$\frac{\begin{pmatrix}13\\1\end{pmatrix}\begin{pmatrix}4\\4\end{pmatrix}\begin{pmatrix}48\\1\end{pmatrix}}{\begin{pmatrix}52\\5\end{pmatrix}}$$