Dice combinations

I’m always a little uncertain as to how far I can rasonably talk about Project Euler problems. The site is a superb resource that challenges you to develop programs to solve mathematical problems. For myself, it has helped both to improve my Python programming skills and to broaden my appreciation of just how diverse and fascinating a field Mathematics is.

However, the site does ask you not to share solutions and I can see the reason for this, If you can just copy and past the answer to a problem, then it all becomes a little pointless.

On the other hand, I do sometimes come up with solution for which I feel (I think) justifiably proud. In the case of Problem 250, the solution I was pleased with is not the solution to the problem, so I feel reasonably justified in talking about it.

So here goes…

Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4.
Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5, 6.

Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal.

What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form 0.abcdefg

This looked to me like a simple question question of number crunching. If I know all of the combinations and frequencies that six six-sided dice can produce, and all the combinations and frequencies for nine four-sided dice, then I can simply work through every possible dice roll and see who wins.

The fun began when I decided I wanted a generic function that would give me, for any number of a dice, the frequency of every possible combination. It’s taken a while, but the final result turned out to be a lot simpler than I expected.

from itertools import product

def list_dice_combinations(number, dice):
    die = []
    for i in range(1, dice + 1):

    combinations = {}
    for roll in product(die, repeat = number):
        total = sum(roll)
        if total in combinations:
            combinations[total] += 1
            combinations[total] = 1

How you use this function is, of course, completely up to you.