Rolling two dice 10 times, what is the probability of getting all possible "doubles"?

Question

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?

Answer 1

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 $$P=\sum_{r=6}^{10} p_r\>q_r={416216045\over8463329722368}\doteq0.0000491788\ .$$

Answer 2

We can bound it as follows: One calculation is to compute the chance we get exactly six doubles, one of each type. We have to choose six of the rolls to be doubles, and order of the doubles, and the four non-double rolls, so $\frac {{10 \choose 6}6!30^4}{36^{10}}\approx 00000335$ An upper bound would be to change allow the other rolls to be anything. We will now double count the ones with one extra double. That gives $\frac {{10 \choose 6}6!36^4}{36^{10}}\approx 0.0000695$ The first subtraction is seven doubles: $6$ ways to choolse the pair of doubles, $10 \choose 2$ to locate them, ${8 \choose 5}5!$ to locate and order the other doubles, $36^3$ for the other rolls for a probability of $\frac {6{10 \choose 2}{8 \choose 5}5!36^3}{36^{10}}\approx 0.0000023$ The real result is close to the $0.0000695$ number.