The one-in-four wipeout wheel

Question

You play a wheel game. Each spin, with probability $\tfrac14$ the wheel lands on BUST and your whole accumulated pile is wiped to zero. Otherwise (probability $\tfrac34$ ) it adds a cash amount to your pile that averages $10. You may stop and bank your pile at any time before spinning.

At what pile size should you stop, to maximize the expected banked amount?

Accepted Answer

Look one spin ahead at pile size t: - With probability 	frac14 you bust and lose t, a change of -t. - With probability 	frac34 you gain \10 on average, a change of +10. Expected change of one more spin: \frac{1}{4}(-t) + \frac{3}{4}(10) = -\frac{t}{4} + 7.5 = \frac{30 - t}{4}. This is positive while t < 30, so you keep spinning below \30 and stop once your pile reaches \30 or more. The intuition: each safe spin is worth \10, but the bust risk costs you a quarter of your current pile in expectation. Those balance exactly when the pile is \30, since 	frac14 	imes 30 = 7.5 equals the 	frac3…

The one-in-four wipeout wheel

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