# 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:

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

• Thank you! This was a very well-written and insightful response. May 27, 2015 at 2:49
• You're very welcome. I'm happy you enjoyed, and thank you for posting this very fun problem!
– Ant
May 27, 2015 at 2:55

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 $P(B)=6/13$.

• Thank you! This was a very nice and beautiful solution. I would not have thought of it. May 27, 2015 at 2:49
• No problem, glad to help :) May 27, 2015 at 2:50
• I think both answer are making a mistake. They assume the pair of dice that can sum 12 are the first two ones, or the third and fourth, etc. But it's possible to have the following sequence: 1, 7 , 5. And Alice with with the third dice, this is case it's not in the answer. Also, Bob win if 7 appears twice in a row! not if two consecutive dice sums 7. May 27, 2015 at 4:23
• You cannot get a sum of $1$ when rolling two dice May 27, 2015 at 4:23
• It was my understanding (and also seemingly @JohnChessant as well) that they roll both dice simultaneously May 27, 2015 at 4:25

Such problems are often solved by partitioning sample space on the first throw. Let $X$ be the RV denoting the sum in the first throw and $P(B)$ be the probability $B$ wins,then: \begin{align} P(B) &= P(B,X = 7) + P(B, X\neq 7) \\ & = P(B|X = 7) P(X = 7) + P(B|X\neq7) P(X\neq 7)\tag{a}\label{eq}\end{align} Given that the first throw is 7, B wins immediately if the second throw is 7 as well. If second throw is neither 7 nor 12, then the game begins again. Hence, $$P(B|X=7) = \frac{1}{6} + \frac{29P(B)}{36}$$

For $P(B|X\neq 7)$ partition further by $X\neq 12$. \begin{align}P(B|X\neq 7) &= P(B, X\neq 12|X\neq 7) + P(B, X =12 |X\neq 7) \\ &= P(B| X\neq 12,X\neq 7)P(X\neq 12|X \neq 7) + P(B| X =12 ,X\neq 7)P(X = 12|X\neq 7) \end{align} B can't win if first roll is $12$, i.e. $P(B| X =12 ,X\neq 7) = 0$. If first roll is neither 7 nor 12, we are back to original game. Lastly, $$P(X\neq 12|X \neq 7) = \frac{P(X\neq 12,X \neq 7)}{P(X\neq 7)} = \frac{29}{30}$$ Thus, $P(B|X\neq 7) = \frac{29P(B)}{30}$

Put, all values in equation (\ref{eq}) $$P(B) = \left(\frac{1}{6}+ \frac{29P(B)}{36}\right)\frac{1}{6} + \frac{29P(B)}{30}\frac{5}{6}$$ Solves to $P(B) = \frac{6}{13}$