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There are several questions similar to this one but after reading those, I am still very confused.

I also did a similar problem of this one and I think I got it, but then I got stuck again.

So if four dice are rolled, the chance of getting three of a kind is: $ \binom{6}{1} \frac{1}{6}* \binom{5}{1} \frac{1}{6}*\binom{4}{1} \frac{1}{6} *\frac{1}{6}$

so if seven dice are rolled, in my understanding, the chance of getting three of a kind and four of another would be: $ \binom{7}{1} \frac{1}{6}*\binom{6}{1} \frac{1}{6}*\binom{5}{1} \frac{1}{6} *\binom{4}{1} \frac{1}{6} *\binom{3}{1} \frac{1}{6} *\binom{2}{1} \frac{1}{6} *\binom{1}{1} \frac{1}{6} $ however, the answer in the book is $ \frac{6 *5*\binom{7}{4} }{6^7} $ and I am totally lost. Please help!

additional problems The answers you guys gave kind of make sense to me but they also make me very confused. can I think of it using the way I did above? for example, another part of the questions asked about the chance of getting two fours, two fives and three sixes. I think of it as: $ \binom{7}{2} \frac{1}{6}^2*\binom{5}{2} \frac{1}{6}^2*\binom{3}{3} \frac{1}{6}^3 $ which matches the solution in the book.

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Homework ? If so, please read this. – Sasha Nov 6 '11 at 21:46
this is not a homework question. I'm just practicing problems in the book for a mid-term. – Gigi Nov 6 '11 at 21:57
I'm not sure you can. How did you obtain your expressions? In my solution below, I calculated the chance by (number of ways to obtain desired outcome / total number of outcomes). I think this is the best approach for this kind of problem. – David Mitra Nov 6 '11 at 22:25
up vote 3 down vote accepted

There are $6^7$ possible outcomes, listed as 7-tuples. Now, how many 7-tuples would consist of two distinct elements, 4 of one kind, and 3 of other. Such 7-tuples can be ordered into $[a,a,a,a,b,b,b]$ with $a \not= b$. $a$ can be chosen 6 different ways. Once $a$ is chosen, $b$ can be chosen 5 different ways. There are $\binom{7}{4}$ ways to rearrange $[a,a,a,a,b,b,b]$ tuple.

The probability is then the ratio of the number of needed outcomes, over the number of total outcomes: $$ p = \frac{1}{6^7} \binom{7}{4} 6 \times 5 $$

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why is it 7 choose 4? – Gigi Nov 6 '11 at 21:57
@Gigi: you are choosing which of the seven dice show $a$. – Ross Millikan Nov 6 '11 at 22:11

Think of filling in 7 slots; in each you have the value of a die roll. There are $6\cdot 5$ ways to choose the values for the three and, different valued, four of a kind (for example, the three of a kind is three '2's and the four of a kind is four '5's). There are $7\choose 4$ ways to select ''slots'' in which to place your 4 of a kind. The remaining three slots will then contain the three of a kind. So, there are $6\cdot5\cdot{7\choose 4}$ different ways to obtain a three of a kind and 4 of a kind. Since outcomes are equally likely here, the chance of obtaining a three of a kind and 4 of a kind is as you stated.

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