Spin a ten-wheel, two dollars to respin

Question

A wheel is equally likely to land on any whole dollar amount from $1 to $10, and it pays you that amount. After a spin you may pay $2 to respin (discarding the current result), as often as you like, keeping whatever you finally stop on.

Find the optimal stopping rule and the fair value of the game.

Accepted Answer

Let V be the value of a fresh spin. Respinning costs \2 and returns you to a fresh spin worth V, so it is worth V - 2. You keep a result x when x \ge V - 2. Guess you keep 5 through 10 and respin 1, 2, 3, 4, then check. The kept values sum to 5+6+7+8+9+10 = 45, and four values trigger a respin: V = \frac{1}{10}\Big[45 + 4(V - 2)\Big] = \frac{1}{10}\big[45 + 4V - 8\big] = \frac{4V + 37}{10}. Solving 10V = 4V + 37 gives 6V = 37, so V = \frac{37}{6} \approx 6.17. The threshold is V - 2 = 	frac{37}{6} - 2 = 	frac{25}{6} \approx 4.17, so you keep 5 or more and respin 1, 2, 3, 4, exactly the gues…

Spin a ten-wheel, two dollars to respin

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