# Probability question, not sure if I'm doing this right… at least vs exactly.

So I'm trying to figure a few things out with probability/counting. These a probability questions, but my understanding of the counting behind them is a little fuzzy still.

For example, The probability of getting 3 of a kind out of a 5 card poker hand is:

COMBIN(13, 1) to choose 1 kind of out the 4 kinds

COMBIN(4, 3) to choose 3 cards out of the 1 kind

COMBIN(12, 2) to choose the 2 kinds for the last 2 cards (the first card took 1 kind already, so it's 12 instead of 13)

(COMBIN(4,1))^2 to choose the 2 cards out the the last kind(s)

How would this change if I were to look for AT LEAST 3 cards of the same kind instead of EXACTLY 3 cards of the same kind?

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The probability of at least $3$ is the probability of exactly $3$ plus the probability of exactly $4$. (I am assuming that you do not want to count the full house hands in your "at least $3$ of a kind. If you do, the answer changes. And if you mean a hand that is $3$ of a kind or better, in the poker sense, then the answer changes again.)

You did the probability of exactly $3$ correctly. Counting the $4$ of a kind hands is even easier. By the reasoning that you used, it is $\binom{13}{1}\binom{48}{1}$. We need to choose the kind. Ater we have done that, we have $\binom{48}{1}$ ways to choose the (almost) useless card.

Remark: You did not ask for an explanation of the "$3$ of a kind" calculation, so I assume it is understood.

The subtlety there is that we must avoid a full house, which is a much better hand. So after the kind and the $3$ cards are chosen, we must choose two different kinds for our remaining cards.

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Yep, I've got the 3 of a kind idea down. So, lets say it was two of a kind (or 1 pair), which is really similar to 3 of a kind, but you wanted AT LEAST 1 pair instead of EXACTLY one pair... would it be similar, only you would have to choose the remaining 3 cards to be all of different kinds? –  user56763 Mar 12 '13 at 4:56
Hard to know what "at least one pair" means. In poker, it probably means a hand that is one pair or better. But if by at least one pair you mean $1$ pair or $2$ pairs, then yes, you find the probability of $2$ pairs, add the probability of (exactly) $1$ pair. For the latter, you do indeed have to choose $3$ different kinds. The $2$ pairs probability is actually the tricky one, it is quite easy to overcount by a factor of $2$. –  André Nicolas Mar 12 '13 at 5:02
Thanks, makes sense. Very helpful! –  user56763 Mar 12 '13 at 5:36

There are 3 kinds of poker hands that have "at least" 3 of the same rank: 3 of a kind, 4 of a kind, full house.

One way to count how to get exactly 3 cards of the same rank: 13 * (4 C 3) * (48 C 2) and add this to the ways to get exactly 4 cards of the same rank: 13 * 48.

Note also that getting exactly 3 of the same rank is 3 of a kind AND full house, thus there are more ways (higher probability) to do this than getting what the poker world calls "3 of a kind."

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