Dice rolling: When rolling n>3 die, probability for getting any combination from a fixed set of random 3-element-combinations.

combinationscombinatoricsdicepermutationsprobability

Good afternoon,

fyi: I am currently working to balance a dice-based combat system for tabletop rpg, which heavily relies on selecting certain combinations of die from a larger pool to outbid your opponent.

Using Excel, I've worked my way through determining the number of possible combinations for rolling n 6 sided die, as well as the probability of getting a certain combination from a larger dice pool. However, I am stuck now. Simply put, I want to know the probability for getting the (say) 6 highest ranking 3d6-combinations out of a nd6 dicepool.

In particular: Say I have a given set S of several (not all) 3-element-combinations (order doesn't matter) from rolling 3 6-sided-die like

S = {{1,1,6}, {6,6,6}, {1,1,5}, {5,5,5}, {1,1,4}, {4,4,4}, {1,1,3}, {3,3,3}}

Given S, I want to count all 4-element-permutations (now from rolling 4 6-sided-die) of wich at least one element of S is a subset. So basically, I want to know how likely I am to get a combination from S when rolling 4 die. Preferably, I would like to generalize the method for any number >1 of die.

Is there a "mathmatical" way to do this without using brute force (that would be counting done by Excel)?

Best Answer

There is a mathematical way to do it. First, you need to figure out the probability of hitting any single combination, like $\{1,1,6\}$. This can be done recursively; for each possible result of the first die, you add up the probability that the remaining $n-1$ die hit the combination with one instance of the first roll removed. There are ways to speed this up by avoiding recalculating repeated cases.

Then, given a set $S$ of combinations, you need to find the probability of hitting at least one. This is can be done with the principle of inclusion/exclusion. That is, add up the probabilities of hitting each combination in $S$, then subtract the probability of hitting the combinations in each pair of $S$, add in the triplets, etc.

I made a program that does this algorithm at the link below. It works for any number of dice, and any subset $S$ of combinations to hit at least one of, and you can check it against a simulated probability. The program will be too slow if $S$ has $20$ or more combinations.

https://repl.it/@mearnest/Tabletop-game-probability

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[Math] Probability of getting specific (one) number when 5 dice rolling at the same time

When doing these kinds of computations in probability, it's often easier to consider the opposite event of what you're actually interested in. Here, the event you care about is "at least one die is a 6" (for example), so the opposite event is "no die is a 6." This means that die 1 is not a 6, and die 2 is not a 6, and dice 3, 4, 5 are all not a 6. The key in this computation is the word "and," because the independence of the dice means we can multiply those individual probabilities. The chance that no die is a 6 is therefore $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 = \frac{3125}{7776}$$ and the probability of the event you want is thus $ 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.5981$.

To answer your question a bit more comprehensively, you asked about when to use the addition rule and when to use the multiplication rule. At a basic level, multiplication corresponds to the word "and," and addition corresponds to the word "or." However, there are caveats; multiplication corresponds to "and" when events are independent, and addition corresponds to "or" when events are disjoint. Recognizing when these things happen is one of the main challenges of learning probability theory and comes only with a fair bit of experience in the subject, as far as I can tell.

In this case, the reason I knew to consider the complementary event was that the main event you wanted to consider had to be expressed in terms of several "or" events that were relatively complicated (i.e. "exactly 1 die is a 6," "exactly 2 dice are a 6," etc.) but the opposite event was not complicated in that way.

How to calculate the probability of rolling a (six-sided) dice and having b dice be x or lower

The probability of rolling any given side of a six sided die is equal, so the probability of rolling any of certain collection of numbers, e.g. $\{1,2,3,4\}$ is just the size of the collection over 6, here $4/6 = 2/3$. In your case it seems like you want to feed in a finite sequence of $N$ numbers between 1 and 6 and want to know the probability that if you roll $N$ die that the $j$the roll is $\leq$ the $j$th element of your sequence precisely $b$ times. Since the probability of rolling any given sequence is equal, really we just want to count how many ways this can happen and they will divide by the total number of possible dice roll outcome, which is just $6^N$.

Some notation: call the sequence of upper bounds $X$ and the sequence of dice rolls $D$, so these are both sequences of length $N$ and we are looking to count the number of $D \in \{1,...,6\}^N$ s.t. $D_j \leq X_j$ precisely $b$ times.

This is perhaps hard to immediately count, but it we fix any size $b$ subset $S$ of $\{1,...,N\}$, we can count the number of ways that $D_j \leq X_j \iff j \in S$. Then we just sum over all these $S$.

For some fixed such $S$, we see that $D_j \leq X_j \iff j \in S$ iff for $j \in S$, $D_j \in \{1,...,X_j\}$ and for $j \not \in S$, $D_j \in \{X_j+1,...,6\}$. Thus there are $\left(\prod_{j\in S} X_j\right) \left(\prod_{j \in S^C} (6-X_j)\right)$ different $D$ such that $D_j \leq X_j \iff j \in S$.

Therefore the total number of ways to satisfy your property is $\sum_{S \subseteq \{1,...,N\}, |S| = b} \left(\prod_{j\in S} X_j\right) \left(\prod_{j \in S^C} (6-X_j)\right).$ The probability is then that quantity over $6^N$. This can be easily computed with code.

Edit: Here's Python code.

import itertools
#number of die rolled
N=5
#your number b
b=2
#your vector of upper bounds
X=[1,1,1,1,1]
number_of_ways = 0
#make a list of integers 0,...,N-1
base_set =[]
for i in range(N):
    base_set.append(i)
#some error checking
if len(X) != N:
    print("Make your X vector the right length")
if not (0 <= b <= N):
    print("b not in correct range")
#iterate throough the subsets of {0,...,N-1} of size b and add up the number of ways as I describe
for S in list(itertools.combinations(base_set,b)):
    temp = 1
    for i in range(N):
        if i in S:
            temp = temp*X[i]
        else:
            temp = temp*(6-X[i])
    number_of_ways+= temp
#normalized by the total number of ways
probability = number_of_ways/6**N
print(probability)

Best Answer

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