Rolling two dice, what is the probability that two consecutive $7$s happens earlier than a $12$? Alice and Bob are playing a game involving two dice. If a sum of 12 appears, Alice wins and they stop playing. If a 7 appears twice in a row, Bob wins and they stop playing. What is the probability that Bob wins this game?
My thought was to draw a tree diagram, which I did, but I can't seem to wrap my head around the recursion that is in the problem. How do I put this into the tree? Thanks!

Let's call the probability that Bob wins be $x$.
On the first roll of the two dice, there are three options:


*

*The dice show a sum of $12$ (probability $\frac{1}{36}$) and Alice would win immediately.

*The dice show a sum of $7$ (probability $\frac{1}{6}$) and this is a little more complicated (see next tree diagram, where let the probability that Bob would eventually win in this case be $a$ for the time being).

*The dice show a sum of some other number (probability $\frac{29}{36}$) and this would be a recursion back on itself. The key here is that if the number rolled wasn't a 7 or a 12, then the situation is exactly the same as it would be had the number not been rolled at all.
The crux is that the probability of Bob winning ($x$) is equal to:
$$x = \frac{1}{36}(0) + \frac{1}{6}(a) + \frac{29}{36}(x)=\frac{a}{6} + \frac{29x}{36} = \frac{6a+29x}{36}$$
Do you see where I got this from?
Now for the other tree diagram. This is assuming that the dice showed a sum of 7 on the first roll. Again, there are three possibilities:


*

*The dice show a sum of $12$ (probability $\frac{1}{36}$) and Alice would win immediately.

*The dice show a sum of $7$ (probability $\frac{1}{6}$) and Bob would win immediately.

*The dice show a sum of some other number (probability $\frac{29}{36}$) and this would be a recursion back to the first tree (and the probability of Bob eventually winning would be $x$ again).
So, $a$, the probability that Bob wins given that they reached this tree, is:
$$a = \frac{1}{36}(0) + \frac{1}{6}(1) + \frac{29}{36}(x) = \frac{1}{6} + \frac{29x}{36} = \frac{6+29x}{36}$$
Substitute this into the other equation and solve for $x$:
$$x = \frac{6a+29x}{36} = \frac{6\left(\frac{6+29x}{36}\right)+29x}{36} = \frac{\frac{6+29x}{6}+29x}{36}=\frac{6+203x}{216}$$
$$216x = 6+203x$$
$$13x = 6$$
$$x = \boxed{\frac{6}{13}}$$
The probability that Bob wins is $\frac{6}{13}$ (and the probability that Alice wins is $1-\frac{6}{13}=\frac{7}{13}$).
