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So the question goes-

  • Mr.Bhallu rolls $3$ standard $6$ sided fair dice together.
  • He wins if the total sum of the outcomes after rerolling once is $7$.
  • Mr.Bhallu can choose to reroll any number of dice $\left(0-3\right)$.
  • He always plays to maximize his chances of winning, calculate the probabilty of him rerolling $2$ dices.

Can someone please explain the concept(s) at use here along with probability for choosing to reroll each ( no dice ), ( $1$ dice) and ( $3$ dice ) as well $?$.

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  • $\begingroup$ You have to make a list of the possibilities, and calculate the probability that he wins. He can re-roll all three dice, or he can have a total of $1$ through $6$ and re-roll two dice, or a total of $2$ through $6$ and roll $1$ die. Once you know those probabilities, you can figure out what's best to do for any of the original rolls. $\endgroup$
    – saulspatz
    Aug 24, 2020 at 17:20
  • $\begingroup$ Devansh: You have some answers that interpret the question in a couple of different ways. Please look at them and clarify which interpretation you have in mind. $\endgroup$
    – Brian Tung
    Aug 26, 2020 at 0:44

3 Answers 3

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It's possible I don't understand the problem correctly, because I get a different answer from the one provided by G Cab. Here's how I understand it: The player rolls three ordinary fair six-sided dice. If the total is $7$, they stop (or equivalently, they re-roll zero dice). This happens with probability $5/72$.

Otherwise, they are allowed to re-roll any number of dice: one, two, or three. They are assumed to select optimally. We want to find the probability that their choice is to re-roll two dice.


With that understanding, let us consider strategy. The probability of winning when re-rolling three dice is the probability of rolling a $7$ on those three dice: again, $5/72$. That will always be an option.

Re-rolling a single die is an option when the lowest two dice total no more than $6$. Under those circumstances, the probability of winning is $1/6$, because exactly one result on the re-rolled die will work. Note that $1/6 > 5/72$, so we always prefer to re-roll one die rather than three, when one is an option.

What about re-rolling two dice? This is an option whenever the lowest die is no more than $5$. However, if the lowest die is $5$, the probability of winning when re-rolling the two other dice is the probability of getting a $2$ on two dice: $1/36$. This is lower than $5/72$. Likewise, if the lowest die is $4$, the probability of winning when re-rolling the two other dice is the probability of getting a $3$ on two dice: $1/18$. This again is lower than $5/72$. Only when the lowest die is no more than a $3$ is it better to re-roll two dice than three.

How does re-rolling two dice compare to re-rolling one? The easiest win when re-rolling two dice is when the lowest die is a $1$. Then the probability of winning is that of getting a $6$ on two dice: $5/36$. This is better than with three dice ($5/72$), but worse than with one die ($1/6$). Therefore, we always prefer to re-roll one die when that is an option.

So here's our strategy:

  1. If the lowest two dice sum up to no more than $6$, then re-roll the other die.
  2. Else, if the lowest die is no more than $3$, then re-roll the other two dice.
  3. Else, re-roll all three dice.

To find out the desired probability, therefore, enumerate those cases where the lowest two dice sum up to more than $6$, but where the lowest die is no more than $3$. There are not that many cases. Be sure to account carefully for all of the permutations. For example, there are three permutations of a single $1$ and a pair of $6$s, but there are six permutations of a $2$, a $5$ and a $6$. Count up all of the cases and divide by $6^3 = 216$, and there's your probability.

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  • $\begingroup$ the difference with my understanding, is that by re-rolling a certain die you mean to delete the contribution of that die (which one ?) in the previous sum, while I understand that it is retained. $\endgroup$
    – G Cab
    Aug 25, 2020 at 14:33
  • $\begingroup$ @GCab: Ahh, I see. Well maybe we'll have to ask OP to clarify. $\endgroup$
    – Brian Tung
    Aug 25, 2020 at 22:33
  • $\begingroup$ yes, indeed, that should be clarified $\endgroup$
    – G Cab
    Aug 25, 2020 at 23:16
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I only reroll no dice if the sum is $7$ already. The chance of getting $7$ on a roll of all three dice is $\frac 5{72}$ so I need something better to reroll $1$ or $2$ dice. If I have two dice that sum to less than $7$, I can reroll the other and get a probability of $\frac 16$. This is better than $\frac 5{72}$, so if I have two that sum to less than $7$ I don't reroll all three.

If I have a $1$ I can reroll the other two and have a chance of $\frac 5{36}$ to win. This is the best case and is still less than $\frac 16$, so if I can reroll just one it dominates over rerolling two. If I have a $2$ I can reroll the other two and have a chance of $\frac 4{36}$ and if I have a $3$ I can reroll the other two and have a chance of $\frac 3{36}$. Both of these beat rerolling all three. All of these are less than $\frac 16$, so we have a nice algorithm:

If the dice sum to $7$, stop here.
ElseIf the two lowest dice sum to less than $7$, reroll the other one.
ElseIf the lowest die is $3$ or less, keep it and reroll the other two.
Else, reroll all three.

Compute the probability of each line, and you want the chance of the second.

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  • $\begingroup$ We came to the same conclusion. :-) $\endgroup$
    – Brian Tung
    Aug 25, 2020 at 22:35
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I understand the rules of the play as follows:
a) you roll three dice a first time: if you get a sum of $7$ you win, if you get more you loose, if you get less you can proceed with step b) ;
b) you are allowed to re-roll $1$ or $2$ or $3$ dice, deciding in advance how many, to try and complement to $7$ the sum previously obtained;
c) the decision in b) is taken as to optimize the probability of getting the needed complement.

Now, the probability of getting a sum $s$ when throwing $m$ dice is given by $$ \eqalign{ & p(s,m) = {{N_b (s - m,5,m)} \over {6^{\,m} }}\quad \left| {\;0 \le {\rm integers }m \le s} \right.\quad = \cr & = {1 \over {6^{\,m} }}\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - m} \over 6}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{ m \hfill \cr k \hfill \cr} \right) \left( \matrix{ s - 1 - k6 \cr s - m - k6 \cr} \right)} \cr} $$ as explained in this post

When you tabulate this for the case of interest you get

3dice_reroll_1

and it is clear that if you do not get $7$ at the first instance, then you have better just to reroll only one dice.

In fact you have :

  • $\approx 0.07$ to center $7$ with the first three dice;
  • $\approx 0.05$ to get $6$ with the first three dice, then to get the missing $1$ the best and only way is to roll one die ;
  • $\approx 0.03$ to get $5$ with the first three dice, then to get the missing $2$ the best way is to roll one die ;
  • ....
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    $\begingroup$ Am I misunderstanding the question? If one rolls (say) $17$ with three dice, it surely cannot be optimal to re-roll only one die. The lowest one can get with a single re-roll would be $12$. $\endgroup$
    – Brian Tung
    Aug 25, 2020 at 1:23
  • $\begingroup$ @BrianTung: I added on top of the answer my understanding of the rules. $\endgroup$
    – G Cab
    Aug 25, 2020 at 14:28

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