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Consider a dice game in which you try to obtain the largest number (e.g. you win a prize based on the final dice roll).

  1. You roll an 8-sided die, with numbers 1–8 on the sides.
  2. You may either keep the value you rolled, or choose to roll a 12-sided die, with numbers 1–12 on the sides.

What's the best strategy for choosing what to do in step #2?

I know the 8-sided die has expected payoff of 4.5, and the 12-sided die has expected payoff of 6.5. So I think relying on the 12-sided die is better — but how do I show the probability of this?

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What does "Stop if you want" mean? –  Henry Sep 14 '12 at 14:12
    
Sorry it just means you do not have to go on to roll 1:12. You can stop at 1:8 if happy with that payoff/number. –  number8 Sep 14 '12 at 14:15
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I have suggested a revision to clarify the question, based on my understanding of the question from Henry's answer below. –  Niel de Beaudrap Sep 14 '12 at 15:23
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Since you know the result from the roll of the 8 sided die wouldn't a 7 or an 8 beat the expected payoff of 6.5 for the 12 sided die. I would say roll the 12 sided die if you get below 7 with the 8 sided and stop if you get a 7 or 8. This assumes expected gain as the the criterion to maximize. –  Michael Chernick Sep 14 '12 at 17:13
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3 Answers

up vote 19 down vote accepted

If you are trying to maximise the expected score, then since the expected value of the 12-sided die is $6.5$, it makes sense to stop when the 8-sided die shows greater than $6.5$, i.e. when it shows $7$ or $8$, each with probability $\frac18$. So with probability $\frac34$ you throw the 12-sided die.

The expected score is then $$7 \times \frac18 + 8 \times \frac18 + 6.5 \times \frac34 = 6.75.$$

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+1 Henry. You showed how much that strategy would increase the overall expected gain. My comment was a simple conditioning argument to motivate the strategy and to indicate why it maximizes the expected gain. –  Michael Chernick Sep 14 '12 at 17:16
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There are 256 possible strategies. To know which one to choose we would need to know the criteria for deciding how good the result is.

On the assumption that the utility of the result is nondecreasing (either a 2 is just as good as a 1, or better, etc.) there are only eight sensible strategies: keep a 2 or higher but reroll 1s, keep a 3 or higher but reroll 1s and 2s, ..., keep 8 but reroll 1-7, or reroll always.

Of those strategies "keep 7 or higher, reroll 6 or lower" is the one which maximizes the average, as explained by Henry. But others could make sense. If you're playing a Russian Roulette variant where 1 means bullet and 2 or higher means blank, you want to reroll a 1 and keep everything else. If you win $\$n^2$ on a roll of $n$ (and are risk-neutral), you should keep 8 but reroll everything lower.

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I do not see more than nine strategies: Strategy 1: Reroll when the initial roll is 1 or higher (aka always); Strategy 2: Reroll when the initial roll is 2 or higher (aka almost always); ...; Strategy 8: Reroll when the initial roll is 8 or higher (aka almost never); Strategy 9: Reroll when the initial roll is 9 or higher (aka never). Where do you get 256 possible strategies? –  emory Sep 15 '12 at 3:13
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Excellent point. The OP described the payoff function as nondecreasing, but this is not enough to answer the question. The Russian Roulette variant is an excellent example. –  emory Sep 15 '12 at 3:16
    
@emory: Other strategies include "reroll 1, 4, 6, and 8, keep others" which makes sense if the payoff is 1 for primes and 0 otherwise. The nondecreasing assumption reduces this to 8 or 9 strategies. I did not include the 9th "never reroll" because it is never better than "reroll 1, keep others" and only as good if you're indifferent to the results. –  Charles Sep 15 '12 at 4:13
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This is easy enough to break down with game theory.

As the objective is to get the highest number, you only want to roll again when you're more likely to roll a number that's greater than your first score.

N | P(>)
=========
1 | 11/12
2 | 10/12
3 |  9/12
4 |  8/12
5 |  7/12
6 |  6/12
7 |  5/12
8 |  4/12

As you can see it's a linear regression toward 0. The important point is that when the first roll is 6 the odds are 50/50 that you'll improve your score.

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When the first role is $6$, the probability of improving your score by rolling the second is indeed $6/12$, but the chance of worsening your score is only $5/12$ –  Henry Sep 14 '12 at 18:53
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