Need help with his advance dice probability scenario(concept)

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 $$?$$.

• 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. Aug 24, 2020 at 17:20
• 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. Aug 26, 2020 at 0:44

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.

• 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. Aug 25, 2020 at 14:33
• @GCab: Ahh, I see. Well maybe we'll have to ask OP to clarify. Aug 25, 2020 at 22:33
• yes, indeed, that should be clarified Aug 25, 2020 at 23:16

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.

• We came to the same conclusion. :-) Aug 25, 2020 at 22:35

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

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 ;
• ....
• 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 $12$. Aug 25, 2020 at 1:23
• @BrianTung: I added on top of the answer my understanding of the rules. Aug 25, 2020 at 14:28