Expectations and variance with rolling a dice 10 times

Question

Let's say you roll a fair dice 10 times and X is the number of sides that never show up. (i.e. Roll 1 - 10 = 1424145221, X = 2 because 3 and 6 never show up)

Values of $N=0,1,2,3,4,5.\\ P(N=6) = 0$ because at least one of the numbers has to show up.

SOLUTION: Use inclusion-exclusion

Answer 1

There probably is a better way to approach this, but when worse comes to worse, one can always fall back on enumeration. Here are a couple of hints that may help.

You are looking for special sets of 10 digit numbers which are a subset of all 10 digit numbers which can be made from the numbers $[1, 2, 3, 4, 5, 6]$. How many of those are there? $6^{10} = 60,466,176$.

Let's consider $P(X = 4)$. This means the strings are made of exactly $2$ numbers. How many ways can we select these $2$ numbers from $6$? That's ${6\choose 2} = 15$. Now comes the tricky part. How many ways can we make 10 digit strings using only 2 numerals? Well, we could split them $9:1, 8:2, 7:3, 6:4, 5:5, 4:6, 3:7, 2:8,$ or $1:9$. There are $\frac{10!}{9!1!}$ ways to split the strings each $9:1$ and $1:9$. There are $\frac{10!}{8!2!}$ to split them $8:2$ and $2:8$. So, in a nutshell, the number of strings that will satisfy the numerator of $P(X = 4)$ is $$15\cdot \left(\left[2\sum_{i=1}^4\frac{10!}{i!(10 - i)!}\right] + \frac{10!}{5!5!}\right)$$

This makes $$ P(X = 4) = \frac{15\cdot \left(\left[2\sum_{i=1}^4\frac{10!}{i!(10 - i)!}\right] + \frac{10!}{5!5!}\right)}{6^{10}}$$

This enumeration can be used to calculate the other options. The tricky part is identifying the kinds of acceptable partitions for the 10-digit strings in order to calculate the multinomial. For example, when using 3 possibilities, you can have 8-1-1, 7-1-2, 7-2-1, 6-2-2, 6-3-1, etc. We know that in general, the number of ways to partition a sum $S$ into $n$ non-negative summands is ${S + n - 1}\choose{n - 1}$. In our case, I believe we have to subtract $n$ as we don't want any 0's. So there will be ${12\choose 2} - 6 = 60$ partitions for 10-length strings made out of 3 digits, 282 separate partitions for 10-length strings made of 4 digits, 996 of the 5 digit types, and 2997 unique strings using all 6 digits. This does highlight what Andre said in the comments, that this is not really a simple problem; although I'm pretty sure that there are simpler ways than brute force enumeration.

Answer 2

There has been a similar question asked a few days ago.

Using that notation,

$$a(k,m,n) = \binom{k}{m}\cdot \sum_{i=0}^{m} (-1)^i \, \binom{m}{i}\cdot (m-i)^n$$

and the required expectation is:

\begin{align*} \mathbb{E} &= \frac{1}{6^{10}}\, \sum_{j=1}^6 (6-j)\cdot a(6,j,10) \\ &= \frac{9765625}{10077696} \approx 0.969033497339074 \end{align*}

Answer 3

If you want to find the expectation (one of the tags) then this is an alternative route:

Let $X_{n}$ denote the number of sides that never show up by $n$ rolls. Here $X_{0}=6=\mathbb{E}X_0$.

Then $\mathbb{E}\left(X_{n}\mid X_{n-1}=k\right)=\left(k-1\right)\times\frac{k}{6}+k\times\left(1-\frac{k}{6}\right)=\frac{5}{6}k$.

This shows that $\mathbb{E}\left(X_{n}\mid X_{n-1}\right)=\frac{5}{6}X_{n-1}$.

Then $\mathbb{E}X_{n}=\mathbb{E}\left(\mathbb{E}\left(X_{n}\mid X_{n-1}\right)\right)=\mathbb{E}\frac{5}{6}X_{n-1}=\frac{5}{6}\mathbb{E}X_{n-1}$.

With this recursion we find easily that: $$\mathbb{E}X_{n}=6\times\left(\frac{5}{6}\right)^{n}$$