What Is the Exact Formula for Calculating Dice Roll Probabilities in D&D?

[Buce]

I'm an Applied Physics major (considering switching to plain, vanilla Physics), and I've been working on this problem forever. It's not homework; only something I've been trying to work out, in bits and pieces, over the past few weeks in my spare time. It's for a game called D&D, which I'm sure some of you have heard of, this being a forum of physicists, mathematicians, and such.

The problem: You have x number of y-sided dice (denoted xdy for simplicity; though, if it alienates anyone, I'll try to avoid it), and you want to roll a number n. There are a lot of basic things anyone who's played D&D for a while can figure out; the minimum you could possibly roll is x, the maximum is xy, the average is (x+xy)/2, a plot of probability vs. n would yield a bell curve for x>2, etc.

What I'd like, though, is an exact formula to describe the number of combinations for n you could roll on xdy. Presumably, once I have that, I can calculate the probability of rolling any given number by dividing by y^x, and the probability of rolling any number n or higher by summing from n to xy. I have little education in the way of probability analysis, so I basically started listing the number of rolls for each number you could roll on 1d4, 2d4, 3d4...up to 7d4, and tried to extract a formula by intuition. So far, what I've come up with is:

f1(x, y, n) = [ $$\prod$$$$\stackrel{x-1}{i=1}$$(n-i) ]/(x-1)!
f2(x, y, n) = f1(x, y, n) - x*f1(x, y, n-y)
g2(x, y, n) = f2(x, y, x + xy - n)
g1(x, y, n) = f1(x, y, x + xy - n)

And f(x, y, n), which gives the number of cominations of dice you could have to roll n, is:

f(x, y, n) =

f1(x, y, n), x ≤ n ≤ -1+x+y​

f2(x, y, n), -1+x+y ≤ n ≤ -1+x+2y​

g2(x, y, n), 1-2y+xy ≤ n ≤ 1-y+xy​

g1(x, y, n), 1-y+xy ≤ n ≤ xy​

The middle two functions aren't needed for x<=2. When x>=6, though, the function breaks down, and it looks like I'll need to introduce two more functions; one for -1+x+2y ≤ n ≤ (x+xy)/2, and one for (x+xy)/2 ≤ n ≤ 1-2y+xy.

The thing I'm most worried about, though, is that I'll have to continue expanding the function indefinitely. As I said before, I have very little training in probability math, so is there another, easier way for me to tackle this? I've taken math courses up to Diff Eq, Calc III, and I'm currently enrolled in Discrete Math. Thanks in advance, and sorry for the poor formatting. I'm not used to forums with such an extensive typography (is there a way to make curly brackets that extend up and down several lines?).

 

[mmwave]

Hello there,

It's great to see someone using their knowledge in physics to tackle a problem in a different field! As a physicist, probability analysis is not my area of expertise, but I will do my best to offer some guidance.

Firstly, it's important to note that there is no one "exact formula" for the number of combinations for a given number on xdy dice. This is because the number of combinations will depend on the specific values of x, y, and n. For example, the number of combinations for rolling a 7 on 2d4 is different from the number of combinations for rolling a 7 on 3d6.

That being said, there are some general methods you can use to approach this problem. One approach is to use generating functions, which can help you find the number of combinations for a given number on xdy dice. A generating function is a mathematical tool that allows you to represent a sequence of numbers in a compact form. In this case, the sequence would be the number of combinations for each possible roll on xdy dice.

Another approach is to use techniques from combinatorics, specifically the concept of combinations and permutations. This involves counting the number of ways you can arrange xdy dice to get a certain number, taking into account the different values of the dice and their order.

I would also recommend looking into the concept of multinomial coefficients, which can help you calculate the number of combinations for a given number on multiple dice with different numbers of sides.

Overall, my advice would be to continue exploring different mathematical tools and techniques, and perhaps consult with a mathematician or probability expert for further guidance. I hope this helps, and best of luck with your problem!

 

[tacman]

Hello there,

As a fellow physics major, I can definitely understand your interest in finding an exact formula for the probability of dice rolling. It's always satisfying to find a precise mathematical solution to a problem.

First of all, I would like to point out that your intuition and approach are actually quite good. The formulas you have come up with are on the right track and can be used to calculate the number of combinations for a given roll. However, as you have mentioned, the problem becomes more complicated as the number of dice increases. This is because the number of combinations for each roll increases exponentially with the number of dice.

Fortunately, there is a well-known formula in probability theory that can help us with this problem. It's called the binomial distribution, and it can be used to calculate the probability of a specific outcome in a series of independent trials. In our case, rolling multiple dice can be seen as a series of independent trials, where each dice is rolled separately.

The formula for the binomial distribution is:

P(x successes in n trials) = (nCx) * p^x * (1-p)^(n-x)

Where:
n = number of trials (in our case, the number of dice rolled)
x = number of successes (in our case, the desired roll)
p = probability of success in one trial (in our case, the probability of rolling the desired number on one dice)
nCx = combination of n things taken x at a time (this is the part that takes into account the number of combinations for each roll)

Using this formula, we can calculate the probability of rolling a specific number with xdy dice. For example, let's say we want to calculate the probability of rolling a 10 with 3d6. The probability of rolling a 10 on one dice is 1/6, so p = 1/6. The number of combinations for rolling a 10 with 3d6 is 3, so nCx = 3. Plugging these values into the formula, we get:

P(10 with 3d6) = (3C3) * (1/6)^3 * (1-1/6)^(3-3)
= (1) * (1/216) * (1)
= 1/216

Similarly, you can use this formula for any number of dice and any desired roll. It may seem a bit daunting at first, but