Find the number of full houses in a poker hand, that is, the number of poker hands with three of a kind and two of a kind.?

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Find the number of full houses in a poker hand, that is, the number of poker hands with three of a kind and two of a kind.?

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## Answers

You've got at least two ways to approach this math problem. I'll put them into 2 categories:

Category #1 - The number of 3-of-a-kinds is 13. The number of pairs that can go with any one of those 3-of-a-kinds is 12. (You can't have Aces full of Aces.) So that's pretty simple math to figure out.

Category #2 - If the suits matter, then it's more complicated. In other words, "Aces Full of Sevens" would be just 1 hand in the first category. But if suits mattered, then you could have "Ace of Hearts, Ace of Spades, Ace of Clubs, Seven of Hearts, and the Seven of DIAMONDS", OR you could have almost the same hand, (but NOT the same hand), with "Ace of Hearts, Ace of Spades, Ace of Clubs, Seven of Hearts, and the Seven of CLUBS". If you count those 2 hands as being different just because one of the suits is different, then you've got to figure out the number of permutations of suits with a 3-of-a-kind and a pair, then multiply that times the answer for the Category #1 answer.

Try to figure it out. Post your work and either I or someone will tell you if you're on the right track. You've got to at least try.