Rolling two dice, what is the probability that two consecutive $7$s happens earlier than a $12$?

Question

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!

Answer 1

Let's call the probability that Bob wins be $x$.

On the first roll of the two dice, there are three options:

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

2. 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).

3. 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:

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

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

3. 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 $1-\frac{6}{13}=\frac{7}{13}$).

Answer 2

Let $P(B)$ be the probability that Bob wins the game. We will have three cases based on the first roll of the game:

X) First roll sums to $12$

Y) First roll sums to $7$

Z) First roll sums to anything other than $12$ or $7$.

As these cases are disjoint, we can write $P(B)$ as a sum of conditional probabilities, namely $$ P(B)=P(X)P(B|X)+P(Y)P(B|Y)+P(Z)P(B|Z). $$ Clearly $P(B|X)=0$, as in this case Alice wins immediately.

I'll skip to $P(B|Z)$, as it is somewhat simpler. If neither $12$ nor $7$ is the first sum, then essentially Bob and Alice just start playing the game over again. Thus $P(B|Z)=P(B)$.

For $P(B|Y)$, either the second roll sums to $7$ and Bob wins, or the second roll sums to $12$ and Bob loses, or the second roll sums to something else and the game starts anew again. Thus $P(B|Y)=1/6\times1+1/36\times0+29/36\times P(B)$. Therefore we have $$ P(B)=1/36\times0+1/6\times(1/6+29/36 P(B))+29/36\times P(B). $$ All that is left is to solve for $P(B)$. This leads to $P(B)=6/13$.