# Throwing a cube for $20$ times and counting the amount of possibilities

We throw a regular cube with $$6$$ faces $$20$$ times. Consider $$A_i$$ to be the case where we got the number $$i$$ exactly $$i$$ times. I'm trying to calculate $$P(A_i\cap A_j)$$ where $$i\neq j$$.

The probability to get $$i$$ is $$\frac{1}{6}$$. The probability to get $$j$$ is $$\frac{1}{6}$$. The probability to get a number that is different than $$i$$ and $$j$$ is $$\frac{2}{3}$$.

Consider a vector that contains exactly $$i$$ times the number $$i$$ and $$j$$ times the number $$j$$. Ther other $$20-i-j$$ places can contain any number $$x\in\{1,...,6\}\backslash\{i,j\}$$. The number of possible vectors that contain $$i$$ times $$i$$ and $$j$$ times $$j$$ is $${20 \choose i}{20-i \choose j}$$. So we get:

$$P(A_{i}\cap A_{j})={20 \choose i}{20-i \choose j}\left(\frac{1}{6}\right)^{i}\left(\frac{1}{6}\right)^{j}\left(\frac{2}{3}\right)^{20-i-j}$$

The only thing I don't understand in this solution is why we ignore the number of possibilities to arrange the other four numbers in the other $$20-i-j$$ places. I think the solution should be: $$P(A_{i}\cap A_{j})=4^{20-i-j}{20 \choose i}{20-i \choose j}\left(\frac{1}{6}\right)^{i}\left(\frac{1}{6}\right)^{j}\left(\frac{2}{3}\right)^{20-i-j}$$

But I don't understand why my book says otherwise. Will be glad for explanation.

• If you include that extra factor, then the power of $(2/3)$ in your formula should be changed to the same power of $(1/6)$ since you'd be distinguishing the four different non-$i,j$ values. But then your formula would agree with the original one, since $4*(1/6) = 2/3$. – Ned Nov 24 '19 at 21:38

Suppose, our die has $$n$$ sides, and we rolled it $$k$$ times. Suppose, that $$i$$ is rolled $$a$$ times and $$j$$ is rolled $$b$$ times. Suppose, if we determined the exact $$a$$ rolls with $$i$$ and the exact $$b$$ rolls with $$j$$ we would have had the probability being $$(\frac{1}{n})^{a + b}(\frac{n - 2}{n})^{k - a - b}$$. Then there are $$\frac{k!}{a!b!(k - a - b)!)}$$ distinct ways of choosing those two sets. Thus the final probability will be:
$$\frac{k!}{a!b!(k - a - b)!}(\frac{1}{n})^{a + b}(\frac{n - 2}{n})^{k - a - b}$$
Now in your specific case $$n = 6$$, $$k = 20$$, $$a = i$$ and $$b = j$$. Thus your answer is:
$$\frac{20!}{i!j!(20 - a - b)!}(\frac{1}{6})^{i + j}(\frac{2}{3})^{20 - a - b}$$
And we are ignoring the number of possibilities to arrange the other four numbers, because there is no need for us to know them. If it were an unfair die with three sides: "i" with probability $$\frac{1}{n}$$, "j" with probability $$\frac{1}{n}$$ and "other" with probability $$\frac{n - 2}{n}$$, the problem would have been exactly the same.
Consider an outcome i.e. a specific sequence of twenty dice tosses, $$(x_1,\dotsc, x_{20})$$ where we got $$i$$ exactly $$i$$ times, and $$j$$ exactly $$j$$ times where $$i\neq j$$. By independence of the dice tosses the probability of observing a specific outcome is $$\left(\frac{1}{6}\right)^i\left(\frac{1}{6}\right)^j\left(\frac{4}{6}\right)^{20-i-j}\tag{0}$$ The event $$A_{i}\cap A_{j}$$ is the union of all these (disjoint) posssible outcomes and hence $$P(A_i\cap A_j)$$ is the sum of the probabilities of all these outcomes. Since each outcome has equal probabilty we only need to multiply (0) by the number of possible outcomes to determine $$P(A_i\cap A_j)$$. The number of possible outcomes equals $$\binom{20}{i}\binom{20-i}{j}\binom{20-i-j}{20-i-j}=\binom{20}{i}\binom{20-i}{j}$$ (choose $$i$$ dice tosses to correspond to rolling $$i$$, then choose $$j$$ dice tosses among the remaining $$20-i$$ dice tosses to correspond to rolling j). Hence $$P(A_i\cap A_j)=\binom{20}{i}\binom{20-i}{j}\left(\frac{1}{6}\right)^i\left(\frac{1}{6}\right)^j\left(\frac{4}{6}\right)^{20-i-j}$$