One free restart
A fair ten-sided die has faces 0 to 9. Each roll of 1 to 9 adds the face to your running total, and after any roll you may stop and take the total in dollars.
The first time a 0 comes up, your total is reset to 0, but you may keep playing. A second 0 ends the game and you get nothing.
While the reset is still unused, what is the smallest total at which stopping is optimal? You may use the fact that the game without any reset (a 0 ends it with nothing), played optimally from a total of 0, is worth about .
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
First solve the game without a reset: compare stopping at with rolling once more and then stopping.
With the reset unused, a 0 does not end the game: it sends you to the start of the no-reset game, worth about 17.68.
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Compare stopping with one more roll. With the reset unused, a 0 is not a disaster: it moves you to a fresh no-reset game. That safety net raises the stopping level by exactly the value of that game.
Step 1: without a reset. At total , one more roll gives 0 with probability and for otherwise, so on average . This beats when , so you stop at 45 or more. Played this way from 0, that game is worth .
Step 2: with the reset. Now a 0 sends you to the no-reset game at total 0, worth . One more roll is worth on average which beats stopping at when .
Step 3: why the one-step rule is right. Once , a roll either raises (you stay in the stopping region) or uses the reset (and is far below ). So stopping is optimal from 63 on, and below 63 one more roll beats stopping. The answer is 63.
The idea. An option to recover from a bad outcome makes you take more risk. Here the reset is worth more than 12 dollars from the start: the game is worth about 29.9 with it, against 17.7 without.