# What is the probability of winning in a dice game

John and Lucy take turns rolling two dice:

• If John rolls a pair larger than or equal to $$10$$ he wins and the game stops.
• If Lucy rolls a pair strictly smaller than $$10$$, she wins and the game stops.
• They keep rolling in turn until one of them wins.
• Suppose John rolls first. What is the probability that Lucy wins $$?$$.

My attempt:

First, we have to figure out the probability of John winning on any particular round. That is if he rolls: (5,5), (5,6), (6,5), (4,6), (6,4), (6,6). Thus, his chances of winning are $$\frac{6}{36}=\frac{1}{6}$$. Then since Lucy wins if she rolls everything else, that means she has a $$5/6$$ chance of winning on any particular round.

Now to figure out the probability of Lucy winning if John goes first. For this to happen, John has to lose his first roll and Lucy has to win after that. We know that P(John Losing)=$$\frac{5}{6}$$ and P(Lucy winning)=$$\frac{5}{6}$$. Thus, we have $$(\frac{5}{6})(\frac{5}{6})=\frac{25}{36}$$ chance of Lucy winning the game.

However, I don't think that this is right because it does not account for playing additional rounds (i.e. she doesn't win on the second round but on some nth round afterward). Could I get some help? Thanks.

Please see the below. It is a geometric distribution.

John can win by rolling $$(6,4), (4,6), (5, 5), (5,6), (6,5)$$ or $$(6,6)$$.

So as you calculated his probability of winning in a given roll $$p = \frac{6}{36} = \frac{1}{6}$$

Lucy's probability of winning in a given roll $$q = \frac{5}{6}$$

If Lucy's chances of winning the game is $$P(W)$$, we have

$$P(W) = (1-p) \times q + (1-p) \times (1-q) \times P(W)$$

$$P(W) = \frac{5}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} \times P(W)$$

$$P(W) = \frac{25}{31}$$

Suppose John's winning probability is $$p$$. Then by the law of total probability, $$p=\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}\cdot p\implies\boxed{p=\frac{6}{31}.}$$