Three draws, keep the most value

Question

You draw up to three numbers, each independent and uniform on $[0, 1]$ . You see each value as it appears and may keep it (ending the game) or draw again; if you reach the third draw you must keep it. Unlike the highest-card game, you now want to maximize the expected value of the number you keep.

What is the optimal policy, and what expected value does it earn?

Accepted Answer

Solve by backward induction, computing the value of the game with k draws remaining. One draw left (the third). You must keep it: expected value E1 = \mathbb{E}[U] = 0.5. Two draws left (the second). If you reject, you fall back to the third draw worth 0.5. So keep the second draw x only if x 0.5. The value is E2 = \mathbb{E}[\max(x, 0.5)] = \int0^{0.5} 0.5\,dx + \int{0.5}^1 x\,dx = 0.25 + \left(0.5 - 0.125\right) = 0.625. Three draws left (the first). Now the continuation value is E2 = 0.625, so keep the first draw x only if x 0.625: E3 = \mathbb{E}[\max(x, 0.625)] = \int0^{0.625} 0.625\,dx …

Three draws, keep the most value

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