Optimal Stopping for a Three-Roll Die Game
Summary
The document solves a finite-horizon stopping problem: roll a fair die up to three times, stop when desired, and maximize the expected value of the final retained roll. It uses backward induction. With one opportunity remaining after a roll, compare the observed result with the expected value of rerolling; this yields a threshold that determines whether to keep the current face or continue. That continuation value then sets the threshold for the first roll in the three-roll game.
The answers give the resulting policy: keep the highest two faces on the first roll, while later decisions use the two-roll threshold. The calculation illustrates how a stop-or-continue decision depends on the value of future opportunities, rather than on a fixed preference for rerolling low outcomes alone. The result assumes a fair six-sided die, independent throws, and a payoff equal to the final face. It does not consider altered dice, costs, or other payoff objectives, and one reply in the discussion incorrectly applies the two-roll rule at every stage.
Key ideas
- Use backward induction to value the remaining opportunities before deciding whether to stop.
- With one throw left after the current roll, reroll low faces and retain faces at or above the continuation value.
- For three total rolls, the first-roll threshold is higher than the later-roll threshold.
- The policy assumes independent throws of a fair six-sided die and a payoff equal to the final face.
Tags
Full text
# Оptimal strategy when throwing dice
# Оptimal strategy when throwing dice
Given a dice, you can throw it no more than three times, and you can stop at any time. How should you act so that on average you get as many as possible in the last throw?
## Answer by dm63 (score 2)
https://quant.stackexchange.com/a/77016
Using backward induction: suppose there are only 2 throws instead of 3. You roll the first dice, and the decision should be to keep a 4,5,6 and to roll again on a 1,2,3 (in which case the expected value on the second throw will be 3.5). But then the expected value of the 2 throw game is 1/6 * 6 + 1/6 * 5 + 1/6 * 4 + 1/2 * 3.5 = 4.25.
Hence, if you have a 3-throw game, on the first throw you will keep a 5 or a 6, and you will throw again a 1,2,3 or 4 because the latter produce a result less than the 2 throw game is worth.
## Answer by test (score 2)
https://quant.stackexchange.com/a/77019
Your decision is based on the expected value of the following throws. If it is higher than what you have right now, you throw again. In a two throw game, you would use the second throw if your first throw is 1, 2 or 3 since $E[T_2] = 3.5$.
For a three throw game you can use make the same strategy but the decision rule after 1st throw is slightly more difficult to come up with. You should use backward induction. You would use the 3rd throw if the 2nd is 1, 2 or 3 (see above). So you keep the 2nd throw in case of 4, 5 or 6. The expected value of the 2nd and 3rd throw combined is $E[T_2\text{ or }T_3]=\frac{1}{6}6 + \frac{1}{6} 5 + \frac{1}{6} 4 + 0.5 E[T_3] = \frac{1}{6}6 + \frac{1}{6} 5 + \frac{1}{6} 4 + \frac{1}{2}3.5 = 4.25$. So for the 1st throw you only keep 5 or 6.
## Answer by Rylan (score 1)
https://quant.stackexchange.com/a/77018
This is a standard interview question in quantitative finance. This article provides a clear description, as does Mark Joshi's "Quant Interview Questions and Answers".
## Answer by KaiSqDist (score -2)
https://quant.stackexchange.com/a/77015
Hi and welcome to the forum. Given that the expected value of a single throw is SUM(1:6)/6 = 3.5, I would always reroll (up to 3 times) as long as I get a 3 and below. I would stop so long as I get a 4 and above.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.