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 $P(\xi_1&lt;\xi_2)$ 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.
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:

$$(0,0,0,P(X&lt;Y),P(Y&gt;X))=\lim_{n\to\infty} (1,0,0,0,0) \begin{pmatrix}
\frac 46&amp;\frac16&amp;\frac 16&amp;0&amp;0\\
\frac36&amp;\frac16&amp;\frac16 &amp;\frac16&amp;0\\
\frac46&amp;\frac16&amp;0&amp;0&amp;\frac16\\
0&amp;0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;0&amp;1
\end{pmatrix}^n$$

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 $P(X&lt;Y)\approx 0.5538$ and $P(Y&lt;X)\approx 0.4462$.That is $12$ has a better than 55% chance of coming before $66$.
Assuming (1,2) means 1 has to come before 2:
   Assuming the question means that both (6,6) and (1,2) need to be consecutive: The answer is 1/2. No fancy math here, just logic: Each roll of the dice doesn't depend on the previous roll. The probability of hitting 6 is 1/6. The probability of consecutive (6,6) is 1/36. The probability of 1 is 1/6; the probability of 2 is 1/6. Hence, the probability of consecutive (1,2) is 1/36. Therefore, both (6,6) and (1,2) have equal chance of happening.
   If only (6,6) needs to be consecutive: The answer is Sigma of (#rolls-1)/((#rolls-1)!+(#rolls-1)) where #rolls goes from 2 to infinity. First roll both 6 and 1 have equal probability of 1/6. Second roll has 5/6 probability of not being 6 and needing to restart 6&amp;6 sequence over again; while not hitting 2 on the second roll does not necessitate restarting the 1&amp;2. Therefore probability of getting non-consecutive (1,2) in 2 rolls=1/36, 3 rolls=1/36+1/36+1/36=3/36(first roll probability of 1/6 multiplied by second roll probability of 1/6 or first multiplied by third or second multiplied by third) , 4 rolls=6/36(1*2,1*3,1*4,2*3,2*4,3*4), 5 rolls=10/36, 6rolls=15/36, 7rolls=21/36, 8rolls=28/36, 9rolls=36/36; formula (#rolls-1)!/36. While the probability of consecutive (6,6) is always 1/36 independent of the number of rolls; probability of rolling consecutive (6,6) in 2 rolls = 1/36, 3 rolls = 2/36 (1+2 or 2+3), 4 rolls = 3/36 (1+2,2+3,3+4), 5 rolls = 4/36, 6 rolls = 5/36, 7 rolls = 6/36, 8 rolls = 7/36; formula (#rolls-1)/36. Therefore the probability of getting consecutive (6,6) before non-consecutive (1,2) is: 1/2 in 2 rolls, 2/5 in 3 rolls, 3/9 in 4 rolls, 4/14 in 5, 5/20 in 6, 6/27 in 7, 7/35 in 8, 8/44 in 9 rolls; formula (#rolls-1)/((#rolls-1)!+(#rolls-1)).