Getting a straight- dice style!

So let us generalize this beauty. Consider a dice with $m$ sides and we throw it $n$ times. We are first interested in the number of outcomes that do NOT appear among the $n$ throws. For $i\in \{1,...,m\}$ , we let $A_i$ be the event that the outcome $i$ does not appear. For $K\subseteq \{1,...,m\}$ ,We count the number of outcomes such that none of the elements of $K$ appear. So we have $\# \bigcap_{i\in K} A_i = (m-\#K)^n$ Using the inclusion exclusion rule, we get $\#\bigcup_{i=1}^m A_i = \sum_{k=1}^n (-1)^{k-1}\binom{m}{k}(m-k)^n$ Now by using De Morgan's law, we get $\#\bigcap_{i=1}^m A_i^c = \sum_{k=0}^n (-1)^{k}\binom{m}{k}(m-k)^n$ Now we find all the samples which exclude exactly $j$ of the $m$ values. To generate those, we can first choose the $j$ values to be excluded, which can happen in $\binom{m}{j}$ ways. Then we find the number of samples such that the remaining are not excluded. From our previous calculation, we get $\sum_{k=0}^n (-1)^{k}\binom{m-j}{k}(m-j-k)^n$ Thus the total number of samples that excludes exactly $j$ values is $\binom{m}{j}\sum_{k=0}^n (-1)^{k}\binom{m-j}{k}(m-j-k)^n$ The probability that $j$ values are excluded is then just the above number divided by all possible samples, which is $m^n$ , we get $\sum_{k=0}^n (-1)^{k}\binom{m-j}{k}(1-\frac{j+k}{m})^n$ The probability that $0$ values are excluded is then just setting $j = 0$ : $\sum_{k=0}^n (-1)^{k}\binom{m}{k}(1-\frac{k}{m})^n$ So specializing to $m=6$ and $n=9$ , we get $\sum_{k=0}^9 (-1)^{k}\binom{6}{k}(1-\frac{k}{6})^9$ This gives $\frac{245}{1296} \sim 0.189043209876543209876543209876543209876543209876543209876543...$

Divide this problem into three cases.

CASE ONE: 123456ABC

• ABC are distinct: $\binom63$
• Rearranging: $\times\dfrac{9!}{2!2!2!}$
• $=907200$

CASE TWO: 123456AAB

• A and B are distinct: $2\binom62$ (multiply by two because there are two A's but only one B
• Rearranging: $\times\dfrac{9!}{3!2!}$
• $=907200$

CASE THREE: 123456AAA

• A: $\binom61$
• Rearranging: $\times\dfrac{9!}{4!}$
• $=90720$

THEREFORE

• Number of cases matching criteria = $907200$ + $907200$ + $90720$ = $1905120$ .
• Total number of cases = $6^9$ = $10077696$
• Required probability = $\dfrac{1905120}{10077696}$ = $\dfrac{245}{1296}$ = $0.1890\overline{432098765}$