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.