My original post included Mathematica code for calculating how close to normal the distribution of the sum of the dice is. Here I’d like to redo the code in Python to show how to do the same calculations using SymPy. [**Update**: I'll also give a solution that does not use SymPy and that scales much better.]

If you roll five dice and add up the spots, the probability of getting a sum of *k* is the coefficient of x^{k} in the expansion of

(*x* + *x*^{2} + *x*^{3} + *x*^{4} + *x*^{5} + *x*^{6})^{5} / 6^{5}.

Here’s code to find the probabilities by expanding the polynomial and taking coefficients.

from sympy import Symbol sides = 6 dice = 5 rolls = range( dice*sides + 1 ) # Tell SymPy that we want to use x as a symbol, not a number x = Symbol('x') # p(x) = (x + x^2 + ... + x^m)^n # where m = number of sides per die # and n = number of dice p = sum([x**i for i in range(1, sides + 1)])**dice # Extract the coefficients of p(x) and divide by sides**dice pmf = [sides**(-dice) * p.expand().coeff(x, i) for i in rolls]

If you’d like to compare the CDF of the dice sum to a normal CDF you could add this.

from scipy import array, sqrt from scipy.stats import norm cdf = array(pmf).cumsum() # Normal CDF for comparison mean = 0.5*(sides + 1)*dice variance = dice*(sides**2 -1)/12.0 temp = [norm.cdf(i, mean, sqrt(variance)) for i in roles] norm_cdf = array(temp) diff = abs(cdf - norm_cdf) # Print the maximum error and where it occurs print diff.max(), diff.argmax()

**Question**: Now suppose you want a better approximation to a normal distribution. Would it be better to increase the number of dice or the number of sides per dice? For example, would you be better off with 10 six-sided dice or 5 twelve-sided dice? Think about it before reading the solution.

**Update**: The SymPy code does not scale well. When I tried the code with 50 six-sided dice, it ran out of memory. Based on Andre’s comment, I rewrote the code using `polypow`

. SymPy offers much more symbolic calculation functionality than NumPy, but in this case NumPy contains all we need. It is much faster and it doesn’t run out of memory.

from numpy.polynomial.polynomial import polypow from numpy import ones sides = 6 dice = 100 # Create an array of polynomial coefficients for # x + x^2 + ... + x^sides p = ones(sides + 1) p[0] = 0 # Extract the coefficients of p(x)**dice and divide by sides**dice pmf = sides**(-dice) * polypow(p, dice) cdf = pmf.cumsum()

That solution works for up to 398 dice. What’s up with that? With 399 dice, the largest polynomial coefficient overflows. If we divide by the number of dice *before* raising the polynomial to the power `dice`

, the code becomes a little simpler and scales further.

p = ones(sides + 1) p[0] = 0 p /= sides pmf = polypow(p, dice) cdf = pmf.cumsum()

I tried this last approach on 10,000 dice with no problem.

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