# Alternate moves, first one to roll a 6 wins: what's the probability of winning?

Question: Players A and B play a sequence of independent games. Player A starts and throws a fair die and wins on a "six". If she fails then player B throws the same die and wins on a "six" or fails otherwise. If she fails then player A gets the die and throws again, and so on. The first player to throw a "six" wins. Find the probability that player A wins.

Here is what I have done:

A wins right away: $\dfrac{1}{6}$

A fails B wins: $\dfrac{5}{6}\times \dfrac{1}{6}$

A fails B fails A wins: $\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}$

Unlike another similar question which the player B can win on "five" and "six" after player A fails on the first throw, in this case, the games do not seem to have an endpoint. How should I calculate the probability that player A wins then?

Player A can win straightaway with probability $\frac{1}{6}$ and with probability $\frac{5}{6}$ we are in exactly the same situation from player B's perspective. Let's name $p$ the probability of winning (in any number of moves) for the player who is about to roll the die. Then:

$$p = \frac{1}{6} \cdot 1 + \frac{5}{6}(1-p) = \frac{6-5p}{6} \iff 6p = 6-5p \iff p= \frac{6}{11}$$

Another way: $A$ wins the game if either $A$ wins the turn, or both $A$ and $B$ loose the turn and $A$ wins the game from there.

$$p_A= \tfrac 16 + (\tfrac 56)^2p_A$$

Thus $p_A=6/11$, $p_B= 5/11$

in this case, the games dont seem to have a end point (don't go default).

It is almost certain that that the game will eventually end in a victory for one player or the other, rather than continue indefinitely.   The only question is "when?" will the game finish, not "if?"

In any round of two, $B$ can win iff $A$ doesn't win on her turn,

thus odds in favor of $A = 1:\frac56= 6:5,\; P(A\;wins) = \frac6{11}$