You roll $1$ die repeatedly until you get a $1$ immediately followed by a $2$ or a $2$ immediately followed by a $2$.
What is the probability that a $1$ then a $2$ is rolled before a $2$ then a $2$?
Progress
At first I thought they were equally likely. I thought "if you roll a 1 then you have a 1/6 chance of rolling a 2... and same thing with you roll a 2 you have a 1/6 chance of rolling another 2.." but then i realized that rolling a 1 means that if you roll a 2 that 2 doesn't count to rolling another 2. I.e. the 1/2 outcome cuts into the 2/2 outcome so its some dependence going on and I don't know how to account for that.
Let $a$ be the required probability. Call it the probability that we (ultimately) win. It is reasonably clear that $a$ exists.
Let $b$ be the probability we (ultimately) win given that a $1$ has just been rolled. Let $c$ be the probability we (ultimately) win given a $2$ has just been rolled (and we have not just lost).
We condition on the result of the first toss. Suppose it is something other than $1$ or $2$. Then the probability we win is $a$.
If the first result is $1$, then by definition the probability we win is $b$.
If the result is $2$, then the probability is $c$. Thus $$a=\frac{4}{6}a+\frac{1}{6}b+\frac{1}{6}c.\tag{1}$$
Now suppose a $1$ has just been rolled. If we next roll anything but $1$ or $2$, the probability we win is $a$. If we roll a $1$, the probability we win is $a$, and if we roll a $2$ then we have won, the probability is $1$. Thus $$b=\frac{4}{6}a+\frac{1}{6}b+\frac{1}{6}.\tag{2}$$
Suppose finally that we have just rolled a $2$ (and not lost). An argument similar to the previous two shows that $$c=\frac{4}{6}a+\frac{1}{6}b.\tag{3}$$
Solve the system of three linear equations in three unknowns. We get $a=\dfrac{7}{12}$.
Remark: The problem is really a Markov chain problem, with three states, "neutral," "just rolled a $1$," and "just rolled a non-losing $2$." There is a natural transition matrix, closely related to our three equations.