What's the probability to see all 6 numbers after odd number of rolls? Let X be the random variable which is the number of rolls required to see all 6 faces of a die. What's P(X is odd)?
I found a very indirect way to get an analytic answer, a fraction, to this question.
Let $E_2$ be average times a pair of 6-sided dice must be rolled until all sides appear at least once. Let $E_1$ be be average times one 6-sided die must be rolled until all sides appear at least once. Then, $2E_2 = E_1 + P(\text{X is odd})$, where $E_2$ can be obtained by Markov chain, which is quite messy but doable.
I'm wondering if there is an elementary way to get this fraction. I think, which could be wrong,
$$
P(X=n) = \frac{6*5^{n-6}{n-6 \choose 5} *5!}{6^n}
$$.
So, in other words, I'm asking how you simplify the infinite sum of this expression over all odd n. In fact, if there is an easy way to solve this question. I should be able to find $E_2$ without markov chain.
 A: Here is an approach using exponential generating functions.  Readers unfamiliar with generating functions can find many resources in the answers to the question How can I learn about generating functions?
To simplify the discussion, let's assume the final roll is a $6$.  Then the rolls up to that point must include at least one of each of $1$ through $5$ and no $6$. The exponential probability generating function for $n$ rolls with at least one of $1$ through $5$ and no $6$ is $(e^{x/6}-1)^5$.  If we want the number of rolls to be even, then the EGF is
$$f(x) = \frac{1}{2} [(e^{x/6}-1)^5 + (e^{-x/6}-1)^5] $$
I.e.,
$$f(x) = \sum_{n=0}^{\infty} p_n \frac{1}{n!} x^n$$
where $p_n$ is the probability of rolling a die $n$ times with a result of at least one each of $1$ through $5$ and no $6$, with $n$ even.  Then $(1/6) f(x)$ is the EGF for the probability that we get a complete set of $1$ through $6$ with the final $6$ occurring on roll $n+1$, with $n+1$ odd.  So the total probability of ending with a $6$ on an odd number of rolls is
$$\begin{align}
\frac{1}{6}\sum_{n=0}^{\infty} p_n &= \frac{1}{6} \int_0^{\infty} e^{-x} \; f(x) \; dx \tag{*} \\
&= \frac{1}{12} \int_0^{\infty} e^{-x} \cdot [(e^{x/6}-1)^5 + (e^{-x/6}-1)^5] \; dx \\
&= \frac{461}{5544}
\end{align}$$
But of course the final roll may be any number from $1$ to $6$, not just $6$, so the answer to the problem is
$$6\cdot \frac{461}{5544} = \boxed{\frac{461}{924}}$$
At $(*)$ we have taken advantage of the identity
$$n! = \int_0^{\infty} x^n e^{-x} \; dx$$
A: $X$ is the sum of six independent geometric random variables $X_k$, the number of rolls needed to see the next distinct value when $k$ distinct values have been rolled. It is easy to work out that $Q_k=\Pr(2\nmid X_k)=\frac6{6+k}$; since $X_0$ is always $1$ and always odd,
$$\Pr(2\nmid X)=\prod_{k=1}^5(1-Q_k)+\sum_{s\in[5],|s|=2}\prod_{k\in s}Q_k\prod_{k\not\in s}(1-Q_k)+\sum_{s\in[5],|s|=4}\prod_{k\in s}Q_k\prod_{k\not\in s}(1-Q_k)$$
which still takes some effort, but should be less work than the Markov chain approach.
A: To do it with a Markov chain, I used states of the form $(n,k)$ where $0\leq n\leq6$ and $0\leq k\leq1$.  $n$ indicates that we have seen $n$ numbers so far, $k=0$ means we have made an even number of rolls, and $k=1$ means we have made an odd number of rolls.  The chain starts in state $(0,0).$
For writing a computer program, I encoded $(n,k)$ as state $2n+k$.  To get exact answers, I used sympy.  Initially, I encoded the transition probabilities as $6$ times the actual value, and then divided the matrix by $6$.  I used the formulas and notation from the Wikipedia page on absorbing Markov chains.
from sympy import zeros, eye

P = zeros(14)  
for n in range(6):
    for k in range(2):
        this = 2*n + k
        that = 2*n + (1-k)
        other = 2*(n+1)+(1-k)
        P[this,that] = n
        P[this,other] = 6-n
P[12,12] = 6
P[13,13] = 6
P = P/6

Q = P[:12,:12]
R =  P[:12,12:]
N=(eye(12)-Q).inv()
B = N@R

The matrix $B$ is
Matrix([
[463/924, 461/924],
[461/924, 463/924],
[461/924, 463/924],
[463/924, 461/924],
[331/660, 329/660],
[329/660, 331/660],
[ 82/165,  83/165],
[ 83/165,  82/165],
[  28/55,   27/55],
[  27/55,   28/55],
[   5/11,    6/11],
[   6/11,    5/11]]) 

We want the probability absorption in state $13$ if the chain starts in state $0$.  This is given by the second element of the first row of $B$.
