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I have to apologise if this question has been asked and answered in the past; if there is one field of Mathematics that I struggle to understand, it's definitely statistics.

I'm to calculate the probability to acquire a pair and triplet of any sort. I'm only allowed to draw five cards from the stack. I am uncertain how to approach the problem - in what avenue to begin.

Any help would be greatly appreciated, and please, the more you treat me as a dummy, all the better! :)


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How familiar you are with combination? Think the deck is composed of 13 numbers each repeated 4 times. Try to find all the possible cases. For the success outcomes find the way to select the number for the triplets and for the pair from the 13 numbers and, once you choose the numbers, the way of select 2 and 3 cards from the 4. – Kolmo Mar 21 '12 at 18:26
up vote 4 down vote accepted

You are asking for the probability of what in poker is called a full house.

There are $\binom{52}{5}$ ways to choose $5$ cards from the standard deck of $52$. All these ways are equally likely.

Now we count the number of hands that have a triple and a pair.

The kind of card you have three of can be chosen in $\binom{13}{1}$ ways, or, more simply, in $13$ ways. (By kind here we mean Ace, or King, or Queen, and so on.) For every choice of kind, there are $\binom{4}{3}$ ways of choosing the actual cards of that kind. The number $\binom{4}{3}$ is simply $4$, but I want to concentrate on the structure, so that you can apply similar reasoning to other problems.

For every way of doing the above two tasks, there are $\binom{12}{1}$ ways of choosing the kind of card you have two of. And for every such choice, there are $\binom{4}{2}$ ways of choosing the actual two cards.

Thus the total number of "full house" hands is $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}.$$ Divide by $\binom{52}{5}$ to find the probability.

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Thank you so much! :) I assume that you reduce 1 from 13 as you can't have triplets, if you've already got a pair in your hand? – Zanathel Mar 21 '12 at 18:34
@Zanathel: Our pair cannot be of the same kind as what we have three of. One could have, however, counted another way, and said there are $13$ ways to choose the kind we have two of, and for each of these ways, there are $12$ ways to choose what we have three of. Now that we have settled the kinds, there are $\binom{4}{2}$ ways to choose the actual pair, and for each of these, there are $\binom{4}{3}$ ways to choose the actual triple, for a total of $(13)(12)\binom{4}{2}\binom{4}{3}$, naturally same answer. And that's not the only way to organize the count. – André Nicolas Mar 21 '12 at 18:43

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