# Successive dice pool probabilities

So, I cannot wrap my head around how to calculate this. Here is my problem: In Warhammer you roll a pool of attacks, the attacks hit (for example) on a result of 4+ on a six sided dice. Then all the hits form a new dice pool and get rolled again transforming into wounds on a 3 or more. How do I calculate my probability to get an arbitrary number of wounds? Calculating the average is fairly straightforward but it seems to me that the fact that size of the second pool of dice is dependent on the result of the first should have some kind of bearing on the formula that I am not seeing.

• So, if I may: say you first roll 5 die. Two come up with 4+, you then roll those two. Say 1 comes up with 3+. Then one wound has been dealt. Is that correct? May 31 at 5:03
• Yes, this is correct. May 31 at 13:38

Imagine that instead of reducing the dice pool after the first throw, we simply roll all the dice again and count any one die as a success if it rolls a hit on the first throw and a wound on the second roll, so the die succeeds with probability $$p_h p_w$$, where $$p_h$$ is the probability of rolling a hit and $$p_w$$ is the probability of rolling a wound. In the given example, $$p_h = 3/6$$ and $$p_w = 4/6$$.

If there are $$n$$ dice in all then the total number of successes has a Binomial distribution with parameters $$n$$ and $$p= p_h p_w$$, i.e. the probabiity of $$i$$ successes is $$\binom{n}{i} (p_h p_w)^i (1- p_h p_w)^{n-i}$$ for $$0 \le i \le n$$, and the expected number of success is $$n p_h p_w$$.

• Thank you so much. The explanation of rolling all dice again and counting those who came up correct twice is really good. Jun 1 at 4:00

Let the probability of one die coming up $$\ge a$$ be $$p=\frac{7-a}{6}$$, then the probability of $$m$$ out of $$n$$ six sided dice coming up $$\ge a$$ is

$$p^m(1-p)^{n-m}\binom{n}{m}$$

So then the probability of getting $$w$$ wounds would be the same as sum over the probabilities of getting $$0\le m\le n$$ rolls out of $$n$$ and getting $$w$$ rolls out of $$m$$, and considering that $$p_1=\frac{1}{2}$$ for the first round of rolls and $$p_2=\frac{2}{3}$$ for the second, we get the sum:

$$f(n,w,p_1,p_2)=\sum_{m=0}^n{p_1^m(1-p_1)^{n-m}\binom{n}{m}p_2^w(1-p_2)^{m-w}\binom{m}{w}}=\sum_{m=w}^n{2^{w-n}3^{-m}\binom{n}{m}\binom{m}{w}}=2^{n-w}3^{-n}\binom{n}{w}$$

The first equality follows from simplification and recognizing that the sum is $$0$$ for $$m, and the second equality was found by plugging into Wolfram Alpha.

Additionally, the average number of wounds for a given $$n$$ is just the sum $$\sum_{w=0}^n{w*f(n,w,p_1,p_2)}=\frac{n}{3}$$, again found using Wolfram Alpha.

The general sum solves to:

$$f(n,w,p_1,p_2)=(p_1p_2)^w(1-p_1p_2)^{n-w}\binom{n}{w}$$

And its average: $$np_1p_2$$

Let's denote $$X =$$ number of 4+ dice on first throw and $$Y =$$ number of 3+ dice on second throw.

Use the law of total probability and condition on $$X$$:

$$\mathbb{P}(Y=y) = \sum_{x=0}^6 \mathbb{P}(Y=y|X=x)\mathbb{P}(X=x) \\ = \sum_{x=0}^6 {x\choose y}\left(\frac{4}{6}\right)^y\left(\frac{2}{6}\right)^{x-y} \cdot {6\choose x}\left(\frac{3}{6}\right)^x\left(\frac{3}{6}\right)^{6-x} \\ = \color{red} { \frac{2^{y-6}}{729}\left(729{1 \choose y} + 1458{1 \choose y} + 1215{2 \choose y} + 540{3 \choose y} + 135{4 \choose y} + 18{5 \choose y} + {6 \choose y}\right) }$$

Here's a Desmos plot and numerical values calculated in a table.

RED seems wrong!!

• I must've made some error copying to/from Wolfram Alpha in that last line of formula because it seems to give different values. May 31 at 7:11