Just have a small problem solving this puzzle found online.
On average, how many times do
you need to roll six dice together, to
get all six different sides on each die. it is a group roll 6 dice together
[Math] Expected number of throws of 6 dice until all 6 faces appear
probability
Related Solutions
Denote by $E_r$ the expected number of additional throws when there are $r$ numbers that we have not yet seen. The $E_r$ satisfy the following recursion: $$E_0=1,\quad E_r=1+{r\over6} E_{r-1}\ .$$ (The next throw shows with probability ${r\over6}$ a new number instead of one we have already seen.)
Probably there is a closed formula for the $E_r$. At any rate, doing the recursion manually gives $$E_6={1223\over324}\doteq 3.77\ .$$
The probability that all sides are the same is $6$ times the probability that all sides are $1$. So we don’t have to worry about the distribution of all numbers on the die, we just have to keep track of the number of $1$s. That yields a Markov chain with $7$ different states, two of which ($0$ and $6$) are absorbing. The transition matrix is
$$ \mathsf P(i\to j)=6^{-6}\binom6ji^j(6-i)^{6-j} $$
(where $0^0=1$), or in matrix form:
$$ P=6^{-6}\pmatrix{ 46656&0&0&0&0&0&0\\ 15625&18750&9375&2500&375&30&1\\ 4096&12288&15360&10240&3840&768&64\\ 729&4374&10935&14580&10935&4374&729\\ 64&768&3840&10240&15360&12288&4096\\ 1&30&375&2500&9375&18750&15625\\ 0&0&0&0&0&0&46656\\ }\;. $$
To my surprise, this matrix has a rather nice eigensystem:
$$ P=\pmatrix{6\\5&5&-5&25&-5&3725&1\\4&8&-4&-8&8&-10576&2\\3&9&0&-27&0&14337&3\\2&8&4&-8&-8&-10576&4\\1&5&5&25&5&3725&5\\&&&&&&6}\\ \times\pmatrix{6\\&38160\\&&120\\&&&2448\\&&&&120\\&&&&&648720\\&&&&&&6}^{-1} \\ \times \pmatrix{1\\&\frac56\\&&\frac59\\&&&\frac5{18}\\&&&&\frac5{54}\\&&&&&\frac5{324}\\&&&&&&1} \\\times\pmatrix{1\\-2681&981&1125&1150&1125&981&-2681\\7&-8&-5&0&5&8&-7\\-14&33&-6&-26&-6&33&-14\\1&-4&5&0&-5&4&-1\\-1&6&-15&20&-15&6&-1\\&&&&&&1}\;, $$
where the first diagonal matrix is just for normalization (so that I could write the matrices containing the left and right eigenvectors with integers) and the second diagonal matrix contains the eigenvalues.
Thus, since we start in state $1$, the probability to have reached state $6$ after $n$ rolls is
$$ -\frac{5\cdot2681}{38160}\left(\frac56\right)^n+\frac{5\cdot7}{120}\left(\frac59\right)^n-\frac{25\cdot14}{2448}\left(\frac5{18}\right)^n+\frac{5\cdot1}{120}\left(\frac5{54}\right)^n-\frac{3725\cdot1}{648720}\left(\frac5{324}\right)^n+\frac16 \\[3pt] = -\frac{2681}{7632}\left(\frac56\right)^n+\frac7{24}\left(\frac59\right)^n-\frac{175}{1224}\left(\frac5{18}\right)^n+\frac1{24}\left(\frac5{54}\right)^n-\frac{745}{129744}\left(\frac5{324}\right)^n+\frac16\;. $$
We need to multiply this by $6$ to get the probability of having reached this state for any of the $6$ numbers on the die in order to get the probablity that the number $N$ of required rolls is less than or equal to $n$:
$$ \mathsf P(N\le n)=1-\frac{2681}{1272}\left(\frac56\right)^n+\frac74\left(\frac59\right)^n-\frac{175}{204}\left(\frac5{18}\right)^n+\frac14\left(\frac5{54}\right)^n-\frac{745}{21624}\left(\frac5{324}\right)^n\;. $$
Here we can check that $\mathsf P(N\le0)=0$ and $\mathsf P(N\le1)=6^{-5}$, as they must be. The probability that we need exactly $n$ rolls is
\begin{eqnarray} \mathsf P(N=n) &=& \mathsf P(N\le n)-\mathsf P(N\le n-1) \\[3pt] &=& \frac{2681}{6360}\left(\frac56\right)^n-\frac75\left(\frac59\right)^n+\frac{455}{204}\left(\frac5{18}\right)^n-\frac{49}{20}\left(\frac5{54}\right)^n+\frac{47531}{21624}\left(\frac5{324}\right)^n \end{eqnarray}
for $n\gt0$ and $\mathsf P(N=0)=0$. Here’s a plot, in agreement with your numerical results. The expected value of $N$ is
\begin{eqnarray} \mathsf E[N] &=&\sum_{n=0}^\infty\mathsf P(N\gt n) \\[3pt] &=& \sum_{n=0}^\infty\left(\frac{2681}{1272}\left(\frac56\right)^n-\frac74\left(\frac59\right)^n+\frac{175}{204}\left(\frac5{18}\right)^n-\frac14\left(\frac5{54}\right)^n+\frac{745}{21624}\left(\frac5{324}\right)^n\right) \\[6pt] &=& \frac{2681}{1272}\cdot\frac6{6-5}-\frac74\cdot\frac9{9-5}+\frac{175}{204}\cdot\frac{18}{18-5}-\frac14\cdot\frac{54}{54-5}+\frac{745}{21624}\cdot\frac{324}{324-5} \\[6pt] &=& \frac{31394023}{3251248} \\[6pt] &\approx&9.656\;, \end{eqnarray}
also in agreement with your numerical results.
Best Answer
When we roll the $6$ dice, say consecutively, there are $6^6$ possible outcomes. There are $6!$ outcomes where the results are all different, so the probability $p$ that the outcomes are all different is given by $p=\frac{6!}{6^6}$.
The random variable $X$ that measures the number of times that we do the rolling of six until all numbers are different has geometric distribution with parameter $p$. It is a standard result that the expectation of $X$ is $\frac{1}{p}$.