# How does re-rolling $n$ of $m$ dice affect the probability distribution of the set of dice?

I'm trying to wrap my head around some dice maths with re-rolls.

Say I have $$m$$ six sided dice and roll them. I then check each die individually, and if the it is showing a face less than some threshold $$t$$ then I can re-roll up to $$n$$ of these "failed" dice. How will this change the probability distribution of each face for the set of dice?

Example:

To start with, say I just roll $$2$$ dice. The probability of each face appearing on either die is $$\frac{1}{6}$$.

Then say I can re-roll both of these dice if the result of the die is less than 2. So if I rolled $$\{1, 4\}$$ then I would re-roll the $$1$$, if I rolled a $$\{1, 1\}$$ I would re-roll both dice, and if I rolled a $$\{2, 3\}$$ then I would re-roll neither.

My understanding is that the probability of a face appearing on each face is:

$$\mathbb{P}(1) = \frac{1}{36}$$

$$\mathbb{P}(\{2\dots6\}) = \frac{7}{36}$$

However I get completely lost when I can only re-roll up to $$1$$ die, so in the example above where I rolled a $$\{2, 3\}$$ I would only be able to re-roll one of the dice.

Ideally I would be able to expand this to a formula where I can roll a larger number of dice $$m$$, and re-roll up to $$n$$ of them where the result is less than $$t$$.

• You need to specify the procedure a little more. Suppose you can re-roll a $1$ or a $2$, but only one die. If the roll is $1$-$2$, which die do you re-roll? Dec 6, 2019 at 16:13
• That's a good point. I'm trying to run stats on a board game so in reality it could be either, as it would still be considered re-rolling $1$ of $2$ failed dice. I assume if we re-rolled the minimum available it would generally produce the best result. Dec 6, 2019 at 16:17
• If you just want stats, I'd advise running simulations. This is a messy calculation, as Ross noted, so even if I did it theoretically, I'd want to run a simulation to confirm my computations. Dec 6, 2019 at 16:20
• I'm tempted to, this would end up in code anyway so may run some simulations to get a decent level of accuracy. Dec 6, 2019 at 16:21

Taking three dice where you can reroll $$2$$ or less but only two of them, we initially ignore the restriction on rerolling only two. Each die has $$\frac 1{18}$$ chance to come up $$1$$ or $$2$$ because you have to get $$1$$ or $$2$$ at the start, then the specific number for the second roll. The other numbers then have chance $$\frac 29$$ to come up.
When we consider the restriction you have $$\frac 1{216}$$ chance of rolling all $$1$$s and leaving you stuck with one of them. This would add $$\frac 1{216} \cdot \frac 56$$ chance of getting a $$1$$ where the $$\frac 56$$ is because we already counted the chance you get three $$1$$s, reroll, and this die comes up $$1$$ again. You lose $$\frac 1{216} \frac 16$$ chance of getting a $$2$$ because you can't reroll the third die and get $$2$$. The chance of each number above $$2$$ is decreased by $$\frac 1{6^4}$$ just like $$2$$.
You have $$\frac 7{216}$$ of rolling all numbers $$2$$ or below including at least one $$2$$, which sticks you with this $$2$$. This adds $$\frac 7{216}\cdot \frac 56$$ to the chance you get a $$2$$ and subtracts $$\frac 7{216}\cdot \frac 16$$ from the chance of each other number.