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First you are dealt two cards, they are two tens. And then three new cards are dealt. What is the probability of getting another ten, and another two equal valued cards. I think that's called a full house.

$\frac{?}{50 \choose 3}$

There is $50 \choose 3$ possible ways. I have been thinking about this problem for a few hours. One solution i tried is this.

$\frac{ {2 \choose 1} \times 12 \times {4 \choose 2} }{{50 \choose 3}}$
There are two ways to get one of the two possible tens. That's the first factor. The second factor is 12, because there are 13 cards with equal value. Substract 1 for the tens. And for every "section" of 4 cards with identical number, there is $4 \choose 2$ or 6, possible ways. I think I'm close. This gives $0.73$% and the answer is suppose to be $0.98$%.

Edit: I assumed wrong. I assumed that a full house only would be possible getting another ten, and then two more equal cards. But as pointed out below, three cards with the same number would of course be a full house too. I don't know poker.

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    $\begingroup$ Are you supposed to find the probability of a full house, or the probability of a full house involving three 10's? $\endgroup$
    – Austin Mohr
    Nov 20, 2012 at 15:44
  • $\begingroup$ Yes, the probability of a full house involving three 10's. Given that you already have two tens. $\endgroup$
    – Algific
    Nov 20, 2012 at 15:46
  • $\begingroup$ @Algific: That's strange, since you ask a different question in the title and $0.98\%$ happens to be the correct answer to that other question whereas your answer $0.73\%$ happens to be the correct answer to this question. $\endgroup$
    – joriki
    Nov 20, 2012 at 15:47
  • $\begingroup$ Sorry! I meant the possibility of a full house involving the two tens already dealt. So you guys are right. It would be a full house if the three cards have the same value. It didn't occur to me as a valid possibility. $\endgroup$
    – Algific
    Nov 20, 2012 at 15:51

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Alternatively, you could be dealt three of a single value of card, which also completes a full house. There are $12\times\binom{4}{3}$ ways of doing this. That should be enough to get you to 0.98%.

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  • $\begingroup$ Nice catch! Written out: $\frac{{2 \choose 1} \times 12 \times {4 \choose 2}+ 12 \times {4 \choose 3}}{50 \choose 3}$ $\endgroup$
    – Algific
    Nov 20, 2012 at 16:01

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