Spin a dial with two re-spins, price the game

Question

You spin a dial that lands uniformly on the interval $[0, 1]$ and get paid that value in dollars. After the first spin you may re-spin; after the second you may re-spin again. You must keep the third spin. Each re-spin discards the previous result.

Find the optimal strategy and the fair value of the game.

Accepted Answer

Work backward, treating each stage's continuation value as the bar to beat. Last spin (no options left): worth E[U] = 0.5. One re-spin remaining: keep the current spin x if it beats the continuation value 0.5. You keep when x 0.5 (probability 0.5, average value 0.75) and re-spin otherwise: V1 = 0.5 \cdot 0.75 + 0.5 \cdot 0.5 = 0.375 + 0.25 = 0.625. Two re-spins remaining: now the bar rises to V1 = 0.625. Keep when x 0.625 (probability 0.375, average value 	frac{0.625 + 1}{2} = 0.8125), else re-spin: V2 = 0.375 \cdot 0.8125 + 0.625 \cdot 0.625 = 0.3047 + 0.3906 = 0.6953. Optimal strategy: on …

Spin a dial with two re-spins, price the game

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