How many $3$-of-a-kinds are possible if one ace is missing?

Question

So I know how to do the simpler 'How many $3$-of-a-kinds are possible' question - it has been asked on here multiple times too. But what if say an Ace is missing. How would you go about starting the process of that?

Answer 1

This needs to be divided into two cases:

• Three aces
• Three of some other kind

If you want three aces, then there is only one way to pick the aces. Other than that, you need to pick two more cards that are not a pair. That can be done in $\binom{48}{2} - 12\cdot 6 = 1056$ ways (the number of two-card combinations minus the number of pairs).

If the three-of-a-kind doesn't consist of aces, then there are $48$ different three-of-a-kinds. For each of those, we need to pick two more cards that are not a pair, and not of the same kind as the three-of-a-kind cards. There are $\binom{47}{2}$ different two-card combinations not of the three-of-a-kind value, and $11\cdot 6 + 3$ of them are pairs. Therefore, the number of three-of-a-kind-that-is-not-aces hands is $48\left(\binom{47}{2} -69\right) = 48\,576$.

In total that means $1056 + 48\,576 = 49\,632$ different three-of-a-kind hands.