First to roll a five or six, taking turns

Question

Two players alternate rolling a fair die, with the first player rolling first. A roll of 5 or 6 wins the game (probability $\tfrac13$ each roll); anything else passes the turn.

What is the probability the first player wins?

Accepted Answer

Let P be the win probability of the player about to roll, with p = 	frac13. They win now with probability 	frac13; otherwise (probability 	frac23) the opponent rolls next and wins with probability P, so the current player wins with probability 1 - P: P = 	frac13 + 	frac23(1 - P) = 1 - 	frac23 P \;\Longrightarrow\; 	frac53 P = 1 \;\Longrightarrow\; P = \frac35 = 0.6. Takeaway. With p = 	frac13, the mover's advantage is \frac{1}{2 - p} = \frac35; the more likely the winning event, the more going first pays.

First to roll a five or six, taking turns

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