Probability Mass Function of infinitely re-rolled dice

Question

I play a game called Shadowrun. It is a role-playing game that uses a dice pool mechanic. A player has a dice pool of $x$ six-sided, unbiased dice. Every 5 or 6 counts as a success. The more successes, the better. Calculating this probability is easy (it's a simple binomial distribution with number of trials $n = x$ and probability of success $p = \frac{1}{3}$).

However, every player has an additional attribute called Edge, $y$. Edge can be used in one of two ways:

a) It allows a player to take all the dice that didn't roll a success in the first attempt and roll them again. Any successes achieved this way add to the result.

b) It allows a player to add their Edge to the total number of dice they roll and any dice that rolled a 6 can be rolled again and if they roll a success, that adds to the result; if any of the dice rolled a 6 again, they can be rerolled. This can be done until the dice no longer roll a six. This is called "exploding dice".

The player can only do one or the other, not both and they can use this ability a limited number of times per game.

My question is: which one is better, given a pool of $x$ and Edge $y$ - or more precisely - is there a simple way of expressing the probability mass function of method B? I have found a simple function for method A and a long, computationally intensive one for method B, but I wonder if there is a way of simplifying it. Below is my work so far.

Thanks for your help and I apologise if this question has been asked before (I looked, I could find things regarding exploding and adding the result, but not just adding 1 to the successes with infinite explosions) and sorry if I'm breaking unwritten community rules - it is my first post here.

A. The Reroll

The first method is to reroll the dice that failed on the first try. This means the methods of achieving a success are:

• Roll a 5 or 6 (probability $\frac{1}{3}$)
• Roll a 1, 2, 3, 4 (probability $\frac{2}{3}$) and then reroll to a 5 or 6 (probability $\frac{1}{3}$) Therefore the probability of succeeding on a single dice roll is: $$p = \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3} = \frac{5}{9}$$ This leads to a binomial distribution: $$Pr(X = s) = \binom{x}{s} \left(\frac{5}{9}\right)^s \left(\frac{4}{9}\right)^{x-s}$$ Where $x$ is the dice pool and $s$ is the desired number of successes.

B. The Exploding Dice

For this part I am going to simplify my notation a bit. I am going to use a subscript to indicate the number of dice rolled and I am going to ommit the $X =$ part in the probability function, e.g. $Pr_7(4)$ is the probability of achieving 4 successes on 7 dice. I will be using $s$ to denote a number of successes

1 die

I didn't have a clear idea how to approach this, so I started with a single die. In order to achieve:

• 0 successes: the roll must be 1-4
• 1 success: the roll must be 5 or a 6 followed by a 1-4
• 2 successes: the roll must be 6 followed by a ,5 or followed by a 6 followed by a 1-4 I can note that as: $$Pr_1(0) = \frac{2}{3}$$ $$Pr_1(1) = \frac{1}{6} + \frac{1}{6} \cdot \frac{2}{3} = \frac{1}{6} \cdot \frac{5}{3}$$ $$Pr_1(2) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{2}{3} = \left(\frac{1}{6}\right)^2 \cdot \frac{5}{3}$$ $$Pr_1(s) = \left(\frac{1}{6}\right)^s \cdot \frac{5}{3} \quad for \; s > 0$$

2 dice

So here my approach was similar, using combinations:

• 0 successes: the number of successes must be 0 on Dice 1 and 0 on Dice 2
• 1 success: the number of successes must be 1 on Dice 1 and 0 on Dice 2; or 0 on Dice 1 and 1 on Dice 2
• 2 successes: the number of successes must be 2 on Dice 1 and 0 on Dice 2; or 1 on Dice 1 and 1 on Dice 2; or 0 on Dice 1 and 2 on Dice 2

Similarly, I can note this using the previously calculated one-die probabilities: $$Pr_2(0) = Pr_1(0)Pr_1(0)$$ $$Pr_2(1) = Pr_1(1)Pr_1(0) + Pr_1(0)Pr_1(1)$$ $$Pr_2(2) = Pr_1(2)Pr_1(0) + Pr_1(1)Pr_1(1) + Pr_1(0)Pr_1(2)$$ $$Pr_2(s) = \sum_{n=0}^{s} Pr_1(s-n)Pr_1(n)$$

3 dice

I am going to spare you the repeat - this time I need to iterate through three dice, which I can write as: $$Pr_3(s) = \sum_{m=0}^{s-n}\sum_{n=0}^{s} Pr_1(s-n-m)Pr_1(n)Pr_1(m)$$

Any pool

So techinically, for $x + y$ dice (remembering that we add the Edge to the pool in method B) I can write an extremely cumbersome formula that basically iterates through all the possible combinations of parameters and uses single-die probabilities to calculate the total probability. The problem is that for pools even as small as 10 dice, this is thousands of possible combinations and just seems a bit brute-force-y. Does anybody have any idea how to simplify this. For completeness, I am attaching the formula:

$$Pr_{x+y}(s)= \sum_{n_{1}=0}^{s} \sum_{n_{2}=0}^{s - n_{1}} \sum_{n_{3}=0}^{s - n_{1} - n_{2}} \cdots \sum_{n_{x+y}=0}^{s - \sum_{a=1}^{x+y-1} n_a} \left[Pr_1\left(s- \sum_{b=1}^{x+y} n_b\right) \cdot \prod_{c=1}^{x+y} Pr_1(c) \right]$$

Answer 1

Scenario $B$ is equivalent to a scenario where each die has success probability $\frac23$ on the first roll and $\frac16$ on all further rolls, and is rerolled as long as it succeeds. We can use this to express the distribution as a single convolution, instead of the $(x+y)$-fold convolution that you wrote.

The probability for $k$ of $m=x+y$ dice to succeed on the first roll is

$$ \binom mk\left(\frac13\right)^k\left(\frac23\right)^{m-k}\;. $$

In the second part of the experiment, instead of rerolling $k$ dice we can imagine the same die being rerolled until it has failed $k$ times. The probability for this to happen after $s$ additional rolls, yielding $s-k$ additional successes, is

$$ \binom{s-1}{k-1}\left(\frac56\right)^k\left(\frac16\right)^{s-k}\;. $$

Thus, the probability for a total of $s$ successes is

$$ \sum_{k=0}^s\binom mk\left(\frac13\right)^k\left(\frac23\right)^{m-k}\binom{s-1}{k-1}\left(\frac56\right)^k\left(\frac16\right)^{s-k}\\ =\left(\frac23\right)^m\left(\frac16\right)^s\sum_{k=0}^s\binom mk\binom{s-1}{k-1}\left(\frac52\right)^k\;. $$

Answer 2

Having done a bunch of work on these types of distribution recently, I think I can give this a rigorous shot with the help of probability generation functions.

Why PGFs?

The question asked here is: Which option is better, given $x$ dice and $y$ edge, which I think we can reasonably assume means "which option gives a higher expectation of successes?". So, we don't actually need to find a mass function for these rolls - we just need to find a way to calculate the expectation in terms of $x$ and $y$, and once we have a generating function, finding the expectation is incredibly simple.

The Base Function Because we're dealing with a binomial distribution, our base function is incredibly simple. If we represent successes with $z$, we can use:

$$[F(z)]^x=\left(\frac{2z}{6}+\frac{4}{6}\right)^x$$

We simply replace $x$ with the number of dice we're rolling, and once we've expanded it out, the coefficient of each power of $z$ will be the probability of getting that many successes (so the coefficient of $z^1$ will be the probability of 1 success, the coefficient of $z^2$ will be the probability of 2 successes, and so forth. Even better, if we want to find it's expectation, we simply differentiate the function once with respect to $z$, and then set $z=1$. If we do that here, we get:

$$E(X)=\frac{2x}{6}$$

Option 1: Reroll Failures

Re-reading the rules for Edge use in Shadowrun, the rules state that a player can only spend one edge per roll. If we follow that rule, rerolling failures is easy to represent as a Generating Function, by simply multiplying the probability of failure by the original generating function and so we get:

$$\begin{align*} [F(z)]^x&=\left(\frac{2z}{6}+\frac{4}{6}\left(\frac{2z}{6}+\frac{4}{6}\right)\right)^x\\ &=\left ( \frac{5z}{9}+\frac{4}{9} \right )^x \end{align*}$$

And again, we can differentiate once with respect to $z$ than set $z=1$ to get:

$$E(X)=\frac{5x}{9}$$

Option 2: Extra Dice + Exploding Dice

This is where the Generating function approach benefits us greatly. Since our function effectively represents rolling a single die, we can use a recursive approach to define our exploding dice generating function. Starting with a single dice:

$$F(z)=\frac{z}{6}+\frac{4}{6}+\frac{zF(z)}{6}$$

We can then do some simple algebra to solve for $F(z):

$$\begin{align*} F(z)&=\frac{z}{6}+\frac{4}{6}+\frac{zF(z)}{6} \\ 6F(z)&=z+4+zF(z)\\ 6F(z)-zF(z)&=z+4\\ F(z)(6-z)&=z+4\\ F(z)&=\frac{z+4}{6-z}\\ [F(z)]^{x+y}&=\left (\frac{z+4}{6-z} \right )^{x+y}\\ \end{align*}$$

Since we add our Edge in dice as well, we add our dice and Edge together to set the exponent. This is all perfectly differentiable with respect to $z$, so if we do and set $z=1$, we get:

$$E(X)=\frac{2(x+y)}{5}$$

Comparing the Pair

So we have our two expectations that we can easily compare. We can see that if we set $y$ to zero and compare the two expectations, we get $\frac{5x}{9} = 0.555...$ per dice for the first, and $\frac{4x}{10} = 0.4$ per dice for the second. Since each point in Edge gives us an extra dice, each extra Edge point effectively adds $0.4$ to our expectation. As a simple table:

x	y	E(Opt1)	E(Opt2)
1	0	0.555...	0.4
1	1	0.555...	0.8
1	2	0.555...	1.2
1	3	0.555...	1.6
1	4	0.555...	2
1	5	0.555...	2.4
5	0	2.777...	2
5	1	2.777...	2.4
5	2	2.777...	2.8
5	3	2.777...	3.2
5	4	2.777...	3.6
5	5	2.777...	4
10	0	5.555...	4
10	1	5.555...	4.4
10	2	5.555...	4.8
10	3	5.555...	5.2
10	4	5.555...	5.6
10	5	5.555...	6

So, if $x$ is low, and $y$ is high, adding dice + explosions is generally the better option, while the higher $x$ is, the higher $y$ needs to be in order for adding dice + explosions to return a better result than rerolling successes.

Also, Mass Functions!

Since the question did ask for Mass Functions as well, we might as well get them done as well. The first option for probability $k$, as noted, very simple:

$$P(K=k)=\binom{x}{k}\left( \frac{20}{36} \right)^k\left( \frac{16}{36} \right)^{x-k}$$

The second option is a little more complicated, but we can adapt the mass function from this answer to come up with:

$$P(K=k)=\sum_{i=0}^{\min(k,x+y)}\binom{x+y}{i}\binom{x+y+k-i-1}{k-i}\left( \frac{1}{6} \right)^i\left( \frac{1}{6} \right)^{k-i}\left( \frac{4}{6} \right)^{x+y-i}$$

Since our exploding probability and our normal success probability are identical, we can do some simplification to end up with:

$$P(K=k)=\sum_{i=0}^{\min(k,x+y)}\binom{x+y}{i}\binom{x+y+k-i-1}{k-i}\left( \frac{1}{6} \right)^k\left( \frac{4}{6} \right)^{x+y-i}$$