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I saw the question What is the average of rolling two dice and only taking the value of the higher dice roll?.

What about the case that instead of rolling two dice simultaneously, rolling of the dice in the game is this:

  1. The player rolls the dice.

  2. The player is asked whether he/she want to roll the dice again.

  3. If the player want to roll the dice again, he/she will roll the dice again and the points obtained will be the final score; if the player does not want to roll the dice again, he/ she will have the dice roll points in 1. as the final score

The purpose of the game is to get the as high score as possible.

Am I correct that the expected value of the game score is also 4.47 (as in the question mentioned at the beginning of this question)? Am I correct that the optimal strategy is that if a player get a dice roll point smaller than 5, he/ she should roll it again?

If yes, does that mean this game is equivalent mathematically to the game mentioned at the beginning of this question?

Is such result extensible to more than 2 dice (e.g. 3 dice and rolling a die for 3 times)?

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    $\begingroup$ Note that there is a difference in strategy between "maximize the expected value of the score", "maximize the probability of having a score of at least 5", "maximizing the probability of beating an opponent", ... I assume we here investigate only the first problem $\endgroup$ Commented Sep 27, 2019 at 14:26
  • $\begingroup$ The title says that two dice are being rolled, but the description sounds like it's only one. Otherwise, how can you expect a score of $4.47$? Please clarify. $\endgroup$
    – saulspatz
    Commented Sep 27, 2019 at 14:29
  • $\begingroup$ @saulspatz I changed the title. 4.47 was from another question linked at the beginning of this question. $\endgroup$
    – Aqqqq
    Commented Sep 27, 2019 at 14:31
  • $\begingroup$ Assuming we merely want to maximize the expected value of the score, we should reroll any time we rolled $3$ or less. This would give an expected result of $\frac{1}{2}(5+3.5)=4.25$. It makes sense that this is smaller than the result in the linked question because there would be times where we rolled a 4 or 5 and chose to keep the value instead of trying for a higher value and doing worse as a result while in the linked question there would be no reason not to try. $\endgroup$
    – JMoravitz
    Commented Sep 27, 2019 at 14:34
  • $\begingroup$ When you are rolling only one, it is a die, not dice. Dice is plural. $\endgroup$ Commented Sep 27, 2019 at 14:38

2 Answers 2

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The expected score of a single die role is $3\frac 12$. Hence if I roll a die and decide to retry whenever the first roll was $n$ or lower, the expected value of the score is $$\begin{align} \underbrace{\frac 16\cdot 3\frac12+\ldots \frac 16\cdot 3\frac12}_n+\frac 16\cdot (n+1)+\frac 16\cdot (n+2)+\ldots+\frac16\cdot 6&=\frac{7n}{12}+\frac{42-n(n+1)}{12}\\&=\frac{42+6n-n^2}{12}\\&=\frac{51-(n-3)^2}{12}\\&\le \frac{17}{4}=4\frac 14\end{align}$$ with equality iff $n=3$.

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  • $\begingroup$ Where does the sum of $n$ $\frac 16\cdot 3\frac12$ come from? $\endgroup$
    – Aqqqq
    Commented Sep 27, 2019 at 14:50
  • $\begingroup$ @Aqqqq That can be thought of as the $n/6$ chance that you choose to reroll the die times the expected value of the reroll (3 1/2). $\endgroup$ Commented Sep 27, 2019 at 14:53
  • $\begingroup$ Still does not explain the role of $n$ here, which is the dice point. $\endgroup$
    – Aqqqq
    Commented Sep 27, 2019 at 15:00
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The two games are not equivalent. Roll two and take the highest will give you $6$ from a roll of $5,6$. Roll one and decide will stop at $5$ and give you $5$.

In fact, roll one and decide should stop at $4$ as well, because the average roll is $3.5$ and you are above that. The average roll for roll one and decide is $\frac 12 \cdot 5 + \frac 12\cdot 3.5=4.25$ because the average roll you keep is $5$, which you do with chance $\frac 12$ and the rest of the time you average $3.5$.

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  • $\begingroup$ I give a more general approach to the question of more die rolls with a d20 here $\endgroup$ Commented Sep 27, 2019 at 14:37
  • $\begingroup$ What do you mean by "rest of the time"? When I did not get 4-5 in the first-time roll? $\endgroup$
    – Aqqqq
    Commented Sep 27, 2019 at 14:57
  • $\begingroup$ The time you did not keep the first roll because you got $4,5,$ or $6$. $\endgroup$ Commented Sep 27, 2019 at 15:06
  • $\begingroup$ Sorry, why would I not keep the first roll because I got 4, 5 or 6? $\endgroup$
    – Aqqqq
    Commented Sep 27, 2019 at 15:25
  • $\begingroup$ Because if you roll again you have to keep what you get. The average of that is $3.5$, so you are better off keeping anything $4$ or better. $\endgroup$ Commented Sep 27, 2019 at 15:40

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