Generate Distribution Table

Problem

What is the probability of obtaining a given sum \( s \) of points with \( d \) dice with \( n \) faces?

Solution

The total number of possible outcomes when rolling \( d \) dice is \( n^d \). To find the number of cases where the sum of the dice equals \( s \), we need to count the number of solutions to the following equation: \[ a_1 + a_2 + \dots + a_d = s, \] where \( a_1, a_2, \ldots, a_d \) are integers representing the value of each die and each \( a_i \) takes an integer value from 1 to \( n \). To count the number of solutions to this equation, consider the function: \[ x + x^2 + x^3 + \cdots + x^n. \] Each term represents a possible dice roll, where the power is the sum and the coefficient is the number of occurrences of that sum. Multiplying this polynomial by itself \( d \) times gives the function: \[ (x + x^2 + x^3 + \cdots + x^n)^d = \left(\frac{x(1 - x^n)}{1 - x}\right)^d. \] We are interested in the coefficient of \( x^s \) in the expansion of this polynomial, as it represents the number of ways to obtain a sum \( s \) with \( d \) dice. Using the binomial theorem, we can expand the term \( (1 - x^n)^d \) as: \[ (1 - x^n)^d = \sum_{t=0}^{d} (-1)^t \binom{d}{t} x^{nt}. \] Thus: \[ x^d \cdot (1 - x^n)^d = \sum_{t=0}^{d} (-1)^t \binom{d}{t} x^{nt+d}. \] Also, the series \( \left(\frac{1}{1 - x}\right)^d \) can be expressed as: \[ (1 - x)^{-d} = \sum_{k=0}^{\infty} \binom{d + k - 1}{k} x^k. \] In order to find the coefficient of some \( x^s \), we will use the first series as a base. Notice that given the term in the summation with \( x^{nt+d} \), there is one and only one corresponding \( x^k \). We can find the value of \( k \): \[ x^{d+nt+k} = x^s \] \[ k = s - (d+nt) \] Therefore we get the term: \[ (-1)^t \binom{d}{t} x^{nt+d} \cdot \binom{d-1+s-(d+nt)}{s-(d+nt)} x^k \] \[ (-1)^t \binom{d}{t} \cdot \binom{s-nt-1}{d-1} x^s \] To consider all the combinations, notice that we need \( d+nt \leq s \) since if it is greater than \( s \), there is no corresponding \( k \) that exists that can make the term \( x^s \). \[ d+nt \leq s \] \[ \frac{s-d}{n} \geq t \] Therefore we have: \[ \sum_{t=0}^{\lfloor \frac{s-d}{n} \rfloor} (-1)^t \binom{d}{t} \cdot \binom{d-1+s-(d+nt)}{s-(d+nt)} x^s \] The number of outcomes for obtaining the sum \( s \) is given by the expression above. To find the probability \( p \), we divide this by the total number of possible outcomes \( n^d \): \[ p = \frac{1}{n^d} \sum_{t=0}^{\lfloor \frac{s-d}{n} \rfloor} (-1)^t \binom{d}{t} \binom{s - nt - 1}{d - 1}. \] This expression gives the probability of obtaining a specific sum \( s \) when rolling \( d \) dice.