First to spin a winner, alternating turns

Question

Two players alternate spinning a wheel, with the first player spinning first. Each spin is a winner with probability $\tfrac14$ ; otherwise the turn passes. The first winning spin takes the game.

What is the probability the first player wins?

Accepted Answer

Let P be the win probability of the player about to spin, with p = 	frac14. They win now with probability 	frac14; otherwise (probability 	frac34) the opponent spins next and wins with probability P, so the current player wins with probability 1 - P: P = 	frac14 + 	frac34(1 - P) = 1 - 	frac34 P \;\Longrightarrow\; 	frac74 P = 1 \;\Longrightarrow\; P = \frac47 \approx 0.571. Takeaway. The first-mover win probability \frac{1}{2 - p} equals \frac47 when p = 	frac14, sitting between the coin's 	frac23 and the rare-event limit of 	frac12.

First to spin a winner, alternating turns

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