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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
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The number of ways doesn't depend on the dice being fair. –  joriki Dec 9 '11 at 2:59

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up vote 4 down vote accepted

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

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