Probability of winning when rolling a standard fair die with two players

Question

We have two players, lets call them A and B, a fair die is rolled and if either player gets a 5 or a 6 they win, but if they get a 2,3 or 4 the game continues (until someone gets a 5 or a 6). If two 1's come up in a row the game ends in a draw, if A roll's first find the probability that A wins ?

I have considered the case where there are no draws ie getting a 1,2,3 or a 4 continues the game and this is simply an infinite geometric series. s.t $P(A)=\frac{2}{6} + (\frac{4}{6})^2\times \frac{2}{6} +(\frac{4}{6})^4\times \frac{2}{6} +... $ but I am not sure how to do this when we consider draws?

Answer 1

To begin with, there's a simpler way to solve the no-draws case. Let $p$ be the probability that A wins. Then we have $p = \frac13 + \frac23(1-p)$, because:

• $\frac13$ of the time, A wins immediately.
• the remaining $\frac23$ of the time, A cedes the turn, which is equivalent to starting the game over, with B taking the first turn instead of A (and A, as the second player, having a $1-p$ chance of winning).

We can solve the second game in the same way, but we need more variables. While the game is still going, it either in a "no-1" state (normally) or a "1" state (if the last roll was a 1). So we define:

• $p_0$ to be the probability that, in a "no-1" state, the player to roll the die eventually wins.
• $q_0$ to be the probability that, in a "no-1" state, the other player eventually wins.
• $p_1$ to be the probability that, in a "1" state, the player to roll the die eventually wins.
• $q_1$ to be the probability that, in a "1" state, the other player eventually wins.

For these, also, we can write down some equations: \begin{align} p_0 &= \frac13 + \frac12 q_0 + \frac16 q_1 \\ q_0 &= \frac12 p_0 + \frac16 p_1 \\ p_1 &= \frac13 + \frac12 q_0 \\ q_1 &= \frac12 p_0 \end{align}

(For example, the first equation comes from the fact that the player to roll the die in a "no-1" state wins immediately $\frac13$ of the time, becomes the second player of a "no-1" state $\frac12$ of the time, and becomes the second player of a "1" state $\frac16$ of the time.)

Then we just solve a system of linear equations.