How to find the probability distributions for multiple dice rolls for dice with a differing number of sides

convolutiondiceprobabilityprobability theoryrandom variables

I have been learning how to play Dungeons and Dragons recently, and have bought my first set of dice.

The standard DnD dice are:

1d4
1d6
1d8
1d10
1d10 × 10 [10, 20, 30… 90, 100]
1d12
1d20

I was thinking about inventive ways to determine successes or failures as a DM, and I came up with the Super Roll! A super roll is rolling all seven of the standard dice, and being asked to roll above a particular number. I think this limit should be constant when asked to make a super roll, but I need to determine what this limit should be. I therefore want to calculate the probability distribution so I can pick a limit with an appropriate chance of succeeding at a super roll.

I could simulate this with Python and get an approximation of the distribution, but where's the fun in that? I want to do this in a proper way and learn some maths along the way. But I've hit a wall.

Researching this, there is a lot of information about rolling multiple dice. But in every case they always roll dice with the same number of sides.

I watched 3Blue1Brown's video about convolutions where Grant begins by explaining how convolutions can be used to add two random variables. In his example, he selects two 6-sided dice.

He goes on to say how this becomes a useful tool if, for example, the weightings of these probabilities for a given side isn't uniform.

And he gives the formula for a discrete convolution:

$$(a * b)_{n} = \sum_{\substack{i,j \\ i+j=n}} a_{i}\cdot b_{j}$$

But I don't want this. My dice aren't biased, instead it is only the number of side that is changing.

I have worked through an example by hand using the sliding windows method to convolve two dice rolls, 1d4 with 1d6.

It worked, and I methodically calculated the probabilities P(X=x).

However in the video Grant then uses python to calculate a convolution. He convolves $(1,2,3) * (4,5,6) = (4,13,28,27,18)$.

For my example, 1d4 and 1d6. The convolution is $(1,2,3,4) * (1,2,3,4,5,6) = (1,4,10,20,30,40,43,38,24)$, or if I do it in the order I did for the sliding windows, $(1,2,3,4) * (6,5,4,3,2,1) = (6,17,32,50,40,30,20,11,4)$

And I don't see how this is helpful to my problem. What do I need to do here? How can I use convolutions to calculate all values of P(X=x).

Any help would be greatly appreciated.

Best Answer

The entries are the frequencies of each face. For fair dice, each face will have the same frequency, which you'll typically want to set to 1. So d4 + d6 would be

$$ \left(1, 1, 1, 1\right) * \left(1, 1, 1, 1, 1, 1\right) = \left(1, 2, 3, 4, 4, 4, 3, 2, 1\right) $$

The result is effectively an unfair die, which you can then convolve with the next die in turn.

Related Solutions

[Math] Probability of hitting a target number on x number of dice with y number of sides

One way is to use generating functions.

If you have $n$ dice, with sides $a_1,\dots,a_n$, then the coefficient on $x^k$ of the polynomial $$ \prod_{i=1}^n \left( \frac{1}{a_i} \sum_{j=1}^{a_i} x^i \right) $$ yields the probability of a sum of $k$ with these dice.

Summing the appropriate values will give you the probability of a sum above a given value.

For example, if you have a three-sided, a four-sided, and a six-sided die, then we want to look at $$ \left(\frac{1}{3}\sum_{i=1}^3 x^i\right)\left(\frac{1}{4}\sum_{i=1}^4 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{72} x^{13} + \frac{1}{24} x^{12} + \frac{1}{12} x^{11} + \frac{1}{8} x^{10} + \frac{11}{72} x^9 + \frac{1}{6} x^8 + \frac{11}{72} x^7 + \frac{1}{8} x^6 + \frac{1}{12} x^5 + \frac{1}{24} x^4 + \frac{1}{72} x^3. $$ Then, for example, the probability that the sum is greater than $10$ is $\frac{1}{12}+\frac{1}{24}+\frac{1}{72}=\frac{5}{36}$.

Another example. For the probability of throwing 9 or greater with two six-sided dice, we look at $$ \left(\frac{1}{6}\sum_{i=1}^6 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{36} x^{12} + \frac{1}{18} x^{11} + \frac{1}{12} x^{10} + \frac{1}{9} x^9 + \frac{5}{36} x^8 + \frac{1}{6} x^7 + \frac{5}{36} x^6 + \frac{1}{9} x^5 + \frac{1}{12} x^4 + \frac{1}{18} x^3 + \frac{1}{36} x^2 $$ Summing the coefficients on $x^9$, $x^{10}$, $x^{11}$, and $x^{12}$, we find the probability of throwing $9$ or higher is $\frac{5}{18}$.

Probability – Calculating Probability with Custom Dice

The programmer's answer to this question is different from the mathematician's answer. The mathematician's answer is:

Without knowing the actual distribution of pips on the faces of the dice, the question is impossible to answer.

I think this is why nobody has responded to your question.

But you said you want to write an app that knows what the dice are and does the calculation, and this is easy. Let's suppose, just for concreteness, that $R$ and $G$ are six-sided, and that $G$ has the numbers 0 through 5 while $R$ has the standard numbers 1 through 6. Then imagine a table like this:

$$\begin{array}{c|cccccc} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & R & - & G & G & G & G \\ 2 & R & R & - & G & G & G \\ 3 & R & R & R & - & G & G \\ 4 & R & R & R & R & - & G \\ 5 & R & R & R & R & R & - \\ 6 & R & R & R & R & R & R \\ \end{array} $$

The row shows the number rolled on the red die, the column shows the number rolled on the green die, and the table shows the winner: $R$ for red, $G$ for green, and $-$ for a tie. We count up $21 R$ and $10 G$, so that $R$ has a $\frac{21}{10+21} \approx 67.7\%$ chance to win, assuming that on a tie you throw it away and start over. (Or, if ties aren't do-overs, $R$ has a $\frac{21}{36} \approx 58.3\%$ chance to win and a $\frac5{36}\approx13.9\%$ chance to tie.)

It is quite easy for the computer to do the counting, without even constructing the table. I am going to write some pseudocode here. Let's say $R$ and $G$ are as above, and represent them in the program as follows:

    R = [0, 1, 1, 1, 1, 1, 1]
    G = [1, 1, 1, 1, 1, 1]

Here R[3] is the number of faces of the $R$ die that show the number 3. R[0] is 0 because $R$ has no faces that show zero, but G[0] is 1 because $G$ does have a zero. If we had a die with the faces $1,2,2,2,7,7$, we would represent it as [0, 1, 3, 0, 0, 0, 0, 2]. If one player is rolling two standard dice, which is $R+R$, we can instead pretend they are rolling one 36-sided die that we represent as

    [0, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1].

I hope this is clear. The computer can easily calculate the list for $R+R$ from the list for $R$ itself.

Now we can compare two dice $d_1$ and $d_2$ like this:

    total = d1_wins = d2_wins = ties = 0

    for i from 0 to d1.max
      for j from 0 to d2.max
        prob = d1[i] * d2[j]
        total = total + prob
        if i > j then
          d1_wins = d1_wins + prob
        else if i < j then
          d2_wins = d2_wins + prob
        else
          ties = ties + prob
        end
      end
    end

    print "Probability of D1 winning: ", d1/total
    print "Probability of D2 winning: ", d2/total

Or, if ties are do-overs, just change total = total + prob to

    if i ≠ j then
      total = total + prob
    end

and leave everything else the same. If you want to convert the probabilities to percentages, you simply multiply by 100%.

I hope this is helpful and answers your question.

Best Answer

Related Solutions

[Math] Probability of hitting a target number on x number of dice with y number of sides

Probability – Calculating Probability with Custom Dice

Related Question