# Count the number of ways of four distinct numbers showing up when six dice are rolled

Suppose we roll six fair dice, how many ways can four distinct numbers show up?

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Did you at least try some combinatoric argument? I feel the inclusion-exclusion principle might come in handy, but I'm really tired right now, so it's just a wild guess. – Patrick Da Silva Dec 9 '11 at 2:34
The number of ways doesn't depend on the dice being fair. – joriki Dec 9 '11 at 2:59

I'll assume that you mean six-sided dice.

There are $\binom64$ ways of choosing the $4$ distinct numbers. They can either appear $3,1,1,1$ times or $2,2,1,1$ times. In the first case, there are $4$ choices of the number appearing thrice and $6\cdot5\cdot4$ choices for the positions. In the second case, there are $6$ choices for the two numbers appearing twice and $6\cdot5\cdot\binom42$ choices for the positions. Thus, the total is

$$\binom64\left(4\cdot6\cdot5\cdot4+6\cdot6\cdot5\cdot\binom42\right)=15\cdot6\cdot5\cdot(16+36)=23400\;.$$

Thus, the probability of this happening is $23400/6^6=23400/46656\approx50\%$.

The corresponding probabilities for the other numbers of distinct numbers are:

$$\begin{eqnarray} p(1)&=&\binom61/6^6\approx0.01\%\;,\\ p(2)&=&\binom62(2^6-2)/6^6\approx2\%\;,\\ p(3)&=&\binom63\left(3\cdot6\cdot5+3!\cdot6\cdot\binom52+\binom62\binom42\right)/6^6\approx23\%\;,\\ p(5)&=&\binom65\cdot5\cdot6\cdot5\cdot4\cdot3/6^6\approx23\%\;,\\ p(6)&=&\binom666!/6^6\approx2\%\;. \end{eqnarray}$$

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Really nice. Thank you. – geraldgreen Dec 9 '11 at 3:15