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You throw a dice 5 times, what is the probablity of making a full house? (ex 12112)

I'm pretty sure I'm right but would like some confirmation.

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There are 6*5 full houses and (6+5-1) Choose (6-1) hands (combos w/ rep). So P = 30 / ( 10 choose 5).
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No. There are 6*5*C(5,2) full houses if we count each ordering of the dice, since the pair can come in C(5,2) positions. The total number of ways to throw the dice is 6^5, so the probability of a full house is

6*5*C(5,2) / 6^5 =~ 3.86%.

I understand that you are ignoring order, but your expression evaluates to about 11.9%. If you explain your C(10,5) term, I will explain why it isn't right.

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Well, I was trying to count the total number of hands s.t that order didn't matter (ex 12345 = 54321). So C(10,5) is the number of combinations of size 5 from a set of size 6 w/ repetition. And then I did the same for the top for example treating 11122 and 22111 as the same.