Dice probability in variable pools with receding difficulty

Question

Can't quite wrap my head around how to go about this (outside of brute forcing it with manual calculations, which isn't ideal given variable difficulties and variable numbers of dice) so I figured I'd ask here.

I'm playing in a game where I can spend limited resources at attempts to harvest and create things. This runs on a D6 system.

Difficulty for harvest, and difficulty to craft, can be 2+, 3+, 4+, 5+, 6+, or 6 and then 5+ on a following roll(counts as a "single" roll for cost purposes), average difficulty 4+. Difficulty goes down by 1, minimum 2+, each attempt.

So if I spend the resources to make 2 harvest attempts and up to 2 craft attempts:

Harvest 1: 4+ pass, 3- fail.

Harvest 2: 3+ pass, 2- fail.

Craft 1: 4+ pass, 3- fail. Irrelevant if both harvests fail.

Craft 2: 3+ pass, 2- fail. Irrelevant if both harvests don't pass.

To make things slightly less complicated, I only particularly care if at least 1 craft passes, so I'm trying to balance harvest vs craft to successfully get at least 1 pass.

TLDR: How do I go about calculating the odds of overall success when each attempt on one side regresses in difficulty on that side, down to a minimum, each attempt on the other side regresses difficulty on that side, and the cap on number of attempts on the second side is dependent on the number of successes on the first?

Answer 1

I think a certain amount of brute-forcing will be necessary if you want to calculate it exactly.

The probability of having at least one successful crafting \$p(C\ge1)\$ if we have \$n\$ harvest and \$m\$ crafting attempts is:

\$p(C\ge1) = 1 - \displaystyle\sum_i^m p(C=0|H=i,n) p(H=i|n )\$

with

\$p(C=0|H=i,n) = \displaystyle\prod_{j=0}^i (1-p_j) \$

where \$H=i, n\$ means that you had \$i\$ successful harvests from \$n\$ attempts (and thus \$i\$ crafting attempts) and \$p_j\$ is the probability of the \$j\$-th roll. Note thay \$j=0\$ refers to the "zeroth" roll and \$p_0\$ is always 0.

Now, the problem is \$ p(H=i|n)\$. If all rolls had the same probability, this would be a binomial distribution. But since they don't have the same probability, it follows a Poisson binomial distribution, and unfortunately the only way to compute this distribution (without approximating it) is to list all possible outcomes. For example, if \$n = 3\$:

\$p(H=0|n=3) = \displaystyle\prod_{j=1}^n (1-p_j) \$

\$p(H=1|n=3) = p_1 (1-p_2) (1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3\$

\$p(H=2|n=3) = p_1 p_2 (1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3\$

\$p(H=3|n=3) = \displaystyle\prod_{j=1}^n p_j \$

There is a closed form for this on Wikipedia, but it really just hides the fact that we are generating all possible combinations with set theory.

Now, if we assume you have the possibility to compute the Poisson Binomial Distribution (e.g. using a software library), then the result would be:

\$p(C\ge1) = 1- \displaystyle\sum_{i=0}^m \left(\displaystyle\prod_j^i (1-p_j)\right) PBD(i,n,p)\$

where \$PBD(i,n,p)\$ is the probability of having i successes from n rolls given the probabilities p. I'm not sure I understood the values for p correctly, but I think it's \$p_j = \min(\frac46, \frac{j}6)\$ if it's 6+ on the first roll and 2+ at the best.

Example

If \$n=2\$ and \$m=2\$ and the probabilities for rolls are \$p=(0, \frac16,\frac26)\$ (so difficulty starting at 6+ and then going down to 5+, the first zero is the probability of succeeding without rolling)

\$PDB(0,2,p) = \frac{20}{36}\$

\$PDB(1,2,p) = \frac{14}{36}\$

\$PDB(2,2,p) = \frac2{36}\$

\$ p(C\ge1) = 1 - (1-p_0)PDB(0,2,p) - (1-p_0)(1-p_1) PDB(1,2,p) - (1-p_0)(1-p_1)(1-p_2)PDB(2,2,p)\$

\$ = 1 - 1\frac{20}{36} - 1\frac56 \frac{14}{36} - 1\frac56\frac46\frac2{36}\ = 0.08950617\$

In your example you used different difficulties, though. You started from 4+ and went down to 3+. In that case the result would be around 0.277

Answer 2

If my understanding is correct ...

You need to succeed in at least one check in the first pool and at least one check in the second and each check in each pool can have a different probability of success.

Let's call the first pool \$A\$ and you have \$n\$ attempts with a probability of success of \$a_i\$. Similarly, you have \$m\$ attempts in the \$B\$ pool with a probability of success of \$b_i\$. If so, the overall probability of success is:

$$ \begin{align} p&= \left(1-\prod_{i=1}^n(1-a_i)\right)\left(1-\prod_{i=1}^m(1-b_i)\right) \end {align} $$

For your example where \$n=m=2\$ and \$a_1={3\over6}\$, \$a_2={4\over6}\$, \$b_1={3\over6}\$ and \$b_2={4\over6}\$ the answer is:

$$ \begin{align} p&= \left(1-\left(1-{3\over6}\right)\left(1-{4\over6}\right)\right)\left(1-\left(1-{3\over6}\right)\left(1-{4\over6}\right)\right)\\ &={25\over 36} \end {align} $$