finding the probability of A happens before B in die rolling I'm thinking about the following old exam question:
A die is rolled infinitely many times, what is the probability that (6,6) (i.e. two consecutive sixes) happens before (1,2)?
Since the expectations of waiting time for (6,6) and (1,2) could be calculated, I was trying to get it from there but I haven't seen the connection between them. Is it possible to get $P(\xi_1<\xi_2)$ with only $E(\xi_1)$ and $E(\xi_2)$, where $\xi_1$ and $\xi_2$ are the waiting times for $(6,6)$ and $(1,2)$ respectively? I also tried writing down all the possible events and sum them up, but the summation seems to be hard to calculate. Thanks for any help.
 A: Let $X$ be the random variable that represents the time of the first instance of $12$ and $Y$ be the random variable that represents the first time of the instance of $66$. Then $P(X=1)\neq P(Y=1)=0$ and $P(X=2)=P(Y=2)=1/36$ but $P(X=3)\neq P(Y=3)$. 
Namely $P(X=3)=\frac{6}{216}$ and $P(Y=3)=\frac{5}{216}$. That's because any sequence $*12$ satisfies $X=3$, but $*66$ only satisfies $Y=3$ if the first roll is not $6$.
So it isn't as simple as the two being equal.
One tricky way to do this is to use a Markov process. Let $s_0$ be the start state. Let $s_1$ be the state of having a string ending in $1$ and not having a prior $66$ or $12$. Let $s_6$ be the state of ending in a $6$ with no prior $12$ or $66$. And let $s_{12}$ and $s_{66}$ be the state of having reached $12$ or $66$ first.
Then you are looking for:
$$(0,0,0,P(X<Y),P(Y>X))=\lim_{n\to\infty} (1,0,0,0,0) \begin{pmatrix}
\frac 46&\frac16&\frac 16&0&0\\
\frac36&\frac16&\frac16 &\frac16&0\\
\frac46&\frac16&0&0&\frac16\\
0&0&0&1&0\\
0&0&0&0&1
\end{pmatrix}^n$$
This can be done by expressing $(1,0,0,0,0)$ in terms of the left eigenvectors of the above matrix.
Wolfram Alpha gives me $P(X<Y)\approx 0.5538$ and $P(Y<X)\approx 0.4462$.That is $12$ has a better than 55% chance of coming before $66$.
A: Consider this diagram:

Where the text inside the nodes denotes the "state," i.e. the suffix that is of interest. $*$ denotes we have nothing, $1$ denotes we have rolled a $1$, and so on.
Let $\phi(\text{state})$ denote the probability of getting to $66$ before getting to $12$ starting from $\text{state}$. It can be seen (and shown) that $\phi$ satisfies the following:
$$
\begin{align*}
\phi(12) &= 0\\
\phi(66) &= 1\\
\phi(*) &= \frac46\phi(*) + \frac16\phi(1) + \frac16\phi(6)\\
\phi(1) &= \frac36\phi(*) + \frac16\phi(1) + \frac16\phi(6) + \frac16\phi(12)\\
\phi(6) &= \frac46\phi(*) + \frac16\phi(1) + \frac16\phi(66)\text{.}
\end{align*}
$$
We are interested in $\phi(*)$, and solving, we get $\displaystyle \phi(*) = \frac6{13}$. Quick & dirty simulation in R.

An alternative solution is given by $B_{1,2}$, where
$$
B = (
I_{3 \times 3} -
\begin{pmatrix}
\frac46&\frac16&\frac16\\
\frac36&\frac16&\frac16\\
\frac46&\frac16&0
\end{pmatrix}
)^{-1}
\begin{pmatrix}
0&0\\
\frac16&0\\
0 &\frac16
\end{pmatrix}
=
\begin{pmatrix}
\frac7{13}&\frac6{13}\\
\frac8{13}&\frac5{13}\\
\frac6{13} &\frac7{13}
\end{pmatrix}\text{.}
$$
The rows of $B$ correspond to starting with nothing (first row), a $1$ (second row), or a $6$ (third row), while the columns correspond to the probability of finishing in $12$ before $66$ (first column) or in $66$ before $12$ (second column).
