Rolling two dice 10 times, what is the probability of getting all possibble "doubles": (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) among our rolls?
For instance, (1,1),(2,1),(3,4),(2,2),(3,3),(6,6),(5,5),(4,4),(5,1),(2,1) is a good roll, because all possible doubles occur.
My idea is to use the inclusion–exclusion principle:
$$
X=36^{10} – 6*35^{10} + {{6}\choose{2}}*34^{10} – {{6}\choose{3}}*33^{10} +… =\sum_{i=0}^{6}{(-1)^{i}*{{6}\choose{i}}*(36-i)^{10}}
$$
Therefore the solution is:
$$
X/36^{10}
$$
Could you check if my solution is proper or show me the simpler one?
Best Answer
We have a drawing with replacement of $10$ balls out of an urn containing $6$ red balls, numbered from $1$ to $6$, and 30 white balls.
The number $R$ of red balls drawn is binomially distributed; i.e., the probability $p_r$ that we draw exactly $r$ red balls is given by $$p_r={10\choose r}\left({1\over6}\right)^r\left({5\over 6}\right)^{10-r}\qquad(0\leq r\leq 10)\ .$$
Assuming that we draw exactly $r$ red balls, the numbers on these $r$ balls define a map $f:\ [r]\to[6]$, and all $6^r$ such maps are equally likely. Out of these maps $6!\left\{\matrix{r \cr 6\cr}\right\}$ are surjective, where $\left\{\matrix{r \cr 6\cr}\right\}$ (called a Stirling number of the second kind) denotes the number of ways to partition $[r]$ into $6$ nonempty blocks. Therefore the probability $q_r$ that on the $r$ red balls drawn all $6$ numbers are present, is given by $$q_r={6!\left\{\matrix{r \cr 6\cr}\right\}\over 6^r}\qquad(6\leq r\leq10)\ .$$ It follows that the overall probability $P$ of the event described in the question is given by $$P=\sum_{r=6}^{10} p_r\>q_r={416216045\over8463329722368}\doteq0.0000491788\ .$$