Calculating odds distribution with 6-sided dice

Question

I'm trying to calculate the odds distribution of a changing number of 6-sided die rolls. For example, 3d6 ranges from 3 to 18 as follows:

3:1, 4:3, 5:6, 6:10, 7:15, 8:21, 9:25, 10:27, 11:27, 12:25, 13:21, 14:15, 15:10, 16:6, 17:3, 18:1

I wrote this php program to calculate it:

function distributionCalc($numberDice,$sides=6) {
for ( $i=0; $i<pow($sides,$numberDice); $i++)
    {
    $sum=0;
    for  ($j=0; $j<$numberDice; $j++)
        { $sum+=(1+(floor($i/pow($sides,$j))) % $sides); }
    $distribution[$sum]++;
    }
return $distribution;
}

The inner $j for-loop uses the magic of the floor and modulus functions to create a base-6 counting sequence with the number of digits being the number of dice, so 3d6 would count as:

111,112,113,114,115,116,121,122,123,124,125,126,131,etc.

The function takes the sum of each, so it would read as: 3,4,5,6,7,8,4,5,6,7,8,9,5,etc. It plows through all 6^3 possible results and adds 1 to the corresponding slot in the $distribution array between 3 and 18. Pretty straightforward. However, it only works until about 8d6, afterward i get server time-outs because it's now doing billions of calculations.

But I don't think it's necessary because die probability follows a sweet bell-curve distribution. I'm wondering if there's a way to skip the number crunching and go straight to the curve itself. Is there a way to do this with, for example, 80d6 (range: 80-480)? Can the distribution be projected without doing 6^80 calculations?

I'm not a professional coder and probability is still new to me, so thanks for all the help!

Stephen

Answer 1

You are looking for a Binomial Distribution

Answer 2

Ok, so let's start with rolling just one die. We know that the mean is 3.5. We can also calculate the variance,

sum(p(x) * (x - M)^2), where M is the mean, x is a dice result, and p is the probability of that dice result.

Using this formula, the variance of a single dice roll is 35/12 = 1/6*((-2.5)^2 + (-1.5)^2 + (-0.5)^2 + 0.5^2 + 1.5^2 + 2.5^2)

It's also a fact that for multiple independent samples from the same distribution, their variances add. So, if you roll N dice, you should get a new distribution with mean 3.5*N and variance 35*N/12.

So, if you generate a normal distribution with mean 3.5*N and variance 35*N/12, it will be a pretty good fit, assuming you're rolling a decent number of dice.