## A brief introduction to dice rolling probabilities

Disclaimer: what follows assumes some basic familiarity with probability theory and combinatorics. This is also no substitute for a proper textbook, as we will skip over a lot of details.
Dice rolling is common in many games, but especially in Role Playing Games (RPGs) it can get quite complicated, with several dice of different shapes rolled together. In the following we will take a look at how to compute the probability to get a given (desired?) result in a number of common situations. And yes, this is largely an excuse to do a bit of combinatorics.

### A single die

A proper axiomatic definition of probability is beyond the scope of this tutorial, but since we are only going to deal with dice rolling, we can get away with a very simple frequentist definition: the probability \(P(A)\) that \(A\) happens, is the ratio between the number of ways \(A\) can happen and the total number of ways anything can happen.This might look like a weird definition if it is the very first time you encounter it, but it makes sense: the probability to roll a 5 on a 6-face die is the number of ways I can get 5 (just one), divided by the number of ways anything can happen (there are 6 possible outcomes, so 6), i.e. \(P(5)= 1/6\), and since there is only one way to roll a 4, also \(P(4)= 1/6\), and so on. In general, the probability to get any specific roll from a 6-sided die is \(1/6\).

What if we have dice with a different number of faces? Nothing changes. The probability to roll a 5 on a 20-sided die is \(1/20\), the probability to roll a 5 on a 8-sided die is \(1/8\) etc. Notice that everything we said so far is independent on whether a given die can be Physically realized, we just assume that the die is fair.

#### Probability to roll higher/lower than....

What is the probability to roll higher than 3 with a 6-sided die? Using the same approach as before, and noticing that 4, 5, and 6 are the possible rolls higher than 3, we can write: \[ P(\#>3)=\frac{\text{number of ways we can roll higher than 3}}{\text{number of ways we can roll}} = \frac{3}{6}=0.5=50\% \] Similarly, the probability to roll higher than 3 with a 12-sided die is: \[ P(\#>3)=\frac{\text{number of ways we can roll higher than 3}}{\text{number of ways we can roll}} = \frac{9}{12}=0.75=75\% \] A couple of things to notice:*cumulative probability*.

Also notice that, since the probability that anything happen is always 1, we can do calculations like (e.g. for a 6-sided die) \[ P(\#\le3)=1-P(\#>3)=1-0.5=0.5=5\% \] and \[ P(\#< 3)=P(\#\le3)-P(\#=3)=0.5-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{2}{6} \simeq 0.333 = 33.3\% \]

##### A few common cases

Rolling a single 6-sided die# | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

P(#) | 16.6% | 16.6% | 16.6% | 16.6% | 16.6% | 16.6% |

P(≥#) | 100% | 83.3% | 66.6% |
50% | 33.3% | 16.6% |

P(≤#) | 16.6% | 33.3% | 50% | 66.6% | 83.3% | 100% |

Rolling a single 20-sided die

# | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% | 5% |

P(≥#) | 100% | 95% | 90% | 85% | 80% | 75% | 70% | 65% | 60% | 55% | 50% | 45% | 40% | 35% | 30% | 25% | 20% | 15% | 10% | 5% |

P(≤#) | 5% | 10% | 15% | 20% | 25% | 30% | 35% | 40% | 45% | 50% | 55% | 60% | 65% | 70% | 75% | 80% | 85% | 90% | 95% | 100% |

### Multiple (identical) dice

When we roll multiple dice we open ourself to a lot more different possible questions about what kind of result we are interested in. We also need some form of notation to express in a compact form. Thankfully there is a very standard notation we can borrow: \(nDm\) means we roll \(n\) \(m\)-sided dice.#### Probability to get a given roll exactly once

Let's say we roll \(3D6\) (three 6-sided dice), what is the probability that exactly one is a 6? The number of ways anything can happen is the number of ways we can roll the first die (6), times the number of ways we can roll the second die (also 6), times the number of ways we can roll the third die (6 once again), for a total of \(6\times6\times6=216\). Counting the number of ways we can get exactly one 6 out of three rolls is slightly more complicated: there are 3 ways to choose which die will be the one rolling a 6, one way to choose what this die will roll (i.e. only a 6), and 5 ways each of the other two dice can roll (since neither can roll a 6), resulting in \(3\times1\times5\times5=75\). So we have \[P(\text{one 6 rolling 3D6}) = \frac{75}{216} \simeq 0.347 = 34.7\%\] We can use the same strategy to find the probability to get a given result exactly once on a \(nDm\) roll (i.e. rolling n m-sided dice): the total number of possible rolls is \(m^n\), the number of ways we can choose which die will roll the result we want is \(n\), the number of ways we can choose the roll of this specific die is just 1, the number of ways one of the other die can roll is \(m-1\), and there are \(n-1\) of those. So we finally get \[ P(\text{given result exactly once on a }nDm\text{ roll}) = \frac{n\times1\times(m-1)^{n-1}}{m^n} \] (notice that, since all rolls are equiprobable on a fair die, which specific result you are looking for is irrelevant for the calculation of the probability.)##### A few common cases

Roll \(3D6\)# | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 0.5% | 1.4% | 2.8% | 4.6% | 6.9% | 9.7% | 11.6% | 12.5% | 12.5% | 11.6% | 9.7% | 6.9% | 4.6% | 2.8% | 1.4% | 0.5% |

P(≥#) | 100% | 99.5% | 98.1% | 95.4% | 90.7% | 83.8% | 74.1% | 62.5% | 50.% | 37.5% | 25.9% | 16.2% | 9.3% | 4.6% | 1.9% | 0.5% |

P(≤#) | 0.5% | 1.9% | 4.6% | 9.3% | 16.2% | 25.9% | 37.5% | 50.% | 62.5% | 74.1% | 83.8% | 90.7% | 95.4% | 98.1% | 99.5% | 100% |

#### Probability to get a given roll at least once

This problem looks very similar to the previous one, but it is instructive because there is a very common trick to make the calculation much simpler, based on the idea that the probability the something happen plus the probability of its negation must add to exactly 1. So, since we want the probability to get a given roll*at least*once, we can obtain ite as 1 minus the probability we never got any: \[ P(\text{at least once})=1-P(\text{never}) = 1 -\left(\frac{m-1}{m}\right)^n \; . \]

#### Probability to get a given roll exactly *k* times

In some games the degree of success is decided by how many times you get a given result when rolling multiple dice. To find the probability we get a given result exactly \(k\) times when rolling \(nDm\), we can use a similar approach to the one we used to calculate the probability to get the result exactly once, but this time we need the number of ways we can choose which \(k\) dice among the \(n\) we rolled got the result we want, and to do that we need to start talking about combinatorics.
##### Combinatorics basics

*factorial*\(n!\) (we are going to use this notation a lot below).

*permutations*of \(k\) items among \(n\) possibilities is

*combinations*(i.e. the number of permutations where we do not care about the order) is \(\frac{n!}{k!(n-k)!}\). Notice that this can be interpreted as the number of ways we can split a group of \(n\) items into a group of \(k\) items and a group of \((n-k)\) items (when we do not care about the exact arrangement, but only about which item is where).

##### The binomial distribution

We are now ready to find the probability to get a given roll exactly \(k\) times when rolling \(nDm\). Let's call the probability that a roll has exactly the result we want \(p\) (which will usually equal to \(1/m\) for a fair dice): the probability to get the roll we want \(k\) times is \(p^k\), the probability that the other \(n-k\) rolls are NOT the one we want is \((1-p)^{n-k}\), and the number of ways we can get \(k\) items (rolls) out of \(n\) is \(\frac{n!}{k!(n-k)!}\), resulting in a probability to get exactly \(k\) times the roll we want out of \(n\) dice equal to \[P_n(k)=\frac{n!}{k!(n-k)!} (p)^k (1-p)^{n-k} .\] This is known as the*binomial distribution*and it appears everywhere when dealing with the probability of discrete events (like dice rolls).

#### Probability that \(n\) dice sum to exactly \(s\)

This is a more complex, but very common problem. E.g., if you ever played Dungeons&Dragons, you had to roll \(3D6\) and sum them to obtain the value of your statistics (Strength, Intelligence etc.) You probably intuitively know that getting a 3 or a 18 when rolling \(3D6\) is a lot less likely than getting an average value like 10 or 11, but quantifying this exactly requires a bit of work, and the introduction of a new concept.##### The probability-generating function

Assume there is a random event you know the probability distribution of, e.g. you know that the probability to get a given result on a fair die is the same for all results and is \(p_i=1/m\). The*probability-generating function*is defined to be \[ G(x) = \sum_i p_i \, x^i \; , \] so (e.g.) the probability-generating function for the rolling of a 6-sided die is \(G(x) = \frac{1}{m}(x+x^2+x^3+x^4+x^5+x^6)\).

So, why did we bother to define such a strange mathematical object? The point is that the properties of exponentials makes it very easy to deal with sums of random events. To see how consider two random events, one where each outcome \(i\) has a probability \(p_i\), and one where the outcome \(j\) has probability \(p_j\). Calculating directly the probability of \(i+j\) is possible but complicated, while we can use the fact that \(x^{i+j}= x^i x^j\) to write \(G_{i+j}(x)=G_i(x)\,G_j(x)\). So the probability-generating function that \(3D6\) sum exactly to \(S\) is \[ G_S(x)=G_{1D6}(x) \times G_{1D6}(x) \times G_{1D6}(x) = \frac{(x+x^2+x^3+x^4+x^5+x^6)^3}{6^3} \; , \] and the probability to get exactly the sum \(S\) is the coefficient of the \(x^S\) term in \(G_S(x)\).

#### Probability to get a sum \(S\) when rolling \(n\) \(m\)-sided dice and keeping the \(k\) highest ones

This is the hardest problem we will deal with here, and it is one that is very common in many Role-Playing Games. E.g. in Dungeons&Dragons when you roll with*advantage*you roll twice and keep the highest die, or a common method to roll the character statistics is to roll \(4D6\) and keep the 3 highest (which, on average, lead to higher scores than simply roll \(3D6\)).

Let's call \(l\) the smallest value among the \(k\) highest dice, \(i\) the number of dice that rolled less than \(l\), and \(j\) the number of dice that rolled exactly \(l\) (which means that \(n-i-j\) dice rolled more than \(l\)).

If \(i+j=n\), there are no dice that rolled higher than \(l\), but since \(l\) is the lowest roll of the \(k\) highest, the only option is that all those \(k\) dice rolled exactly \(l\), and they sum to exactly \(l\times k\).

If \(i+j< n\) (\(i+j\) can't be larger than \(n\)), then \(k-(n-i-j)\) dice rolled exactly \(l\), and sum to \(l\times(k-(n-i-j))\), and the rest of the \(k\) highest dice rolled higher than \(l\). So the number of ways we can get the highest \(k\) dice to sum to \(S\) is the number of ways \(n-i-j\) dice can sum to exactly \(S-l\times (k-(n-i-j))\), keeping in mind that these dice can only have results between \(l+1\) and \(m\). We call this \(N\), and using the probability-generating function discussed above we can see that it is the coefficient of the \(x^{S-l*(k-(n-i-j))}\) term of \((x^{l+1}+x^{l+2}+\ldots+x^m)^{n-i-j}\).

The number of ways we can separate \(n\) dice into a group of \(i\) elements, a group of \(j\) elements, and a group of \(n-i-j\) elements is \(\frac{n!}{i! j! (n-i-j)!}\), and there are \((l-1)^i\) ways to choose the values of the \(i\) dice that are smaller than \(l\). So, we can put all the piece together to find the probability to get a sum \(S\) when rolling \(n\) \(m\)-sided dice and keeping the \(k\) highest ones: \[ P(S) = \frac{1}{m^n}\sum_{i,j,l} \frac{n!}{i! j! (n-i-j)!} (l-1)^i \; N \; . \] There are still a few steps before we are done. First of all, if \(l=1\), then there is no way a die can roll lower than \(l\), and thus \(i=0\). In this case the number ways to choose the values of the \(i\) dice that are smaller than \(l\) is

**not**\((l-1)^i\) (which would give us an indeterminate \(0^0\)), but just 1. The second missing step is that we haven't specified yet over what range \(i\), \(j\), and \(l\) should be summed over:

So we finally get: \[ P(S) = \frac{1}{m^n}\sum_{l=1}^{m} \sum_{i=0}^{n-k} \sum_{j=n - k - i + 1}^{n-i} \frac{n!}{i! j! (n-i-j)!} (l-1)^i \; N \; , \] (keeping in mind our definition of \(N\), and the fact that the \(i+j=n\) and \(l=1\) cases need to be dealt with separately).

##### A few common cases

Roll \(4D6\) and keep the 3 highest ones# | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 0.1% | 0.3% | 0.8% | 1.6% | 2.9% | 4.8% | 7.% | 9.4% | 11.4% | 12.9% | 13.3% | 12.3% | 10.1% | 7.3% | 4.2% | 1.6% |

P(>=#) | 100.% | 99.9% | 99.6% | 98.8% | 97.2% | 94.3% | 89.5% | 82.5% | 73.1% | 61.7% | 48.8% | 35.5% | 23.1% | 13.% | 5.8% | 1.6% |

P(<=#) | 0.1% | 0.4% | 1.2% | 2.8% | 5.7% | 10.5% | 17.5% | 26.9% | 38.3% | 51.2% | 64.5% | 76.9% | 87.% | 94.2% | 98.4% | 100.% |

Roll \(2D20\) and keep the highest one

# | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 0.2% | 0.8% | 1.2% | 1.8% | 2.2% | 2.8% | 3.2% | 3.8% | 4.2% | 4.8% | 5.2% | 5.8% | 6.2% | 6.8% | 7.2% | 7.8% | 8.2% | 8.8% | 9.2% | 9.8% |

P(>=#) | 100.% | 99.8% | 99.% | 97.8% | 96.% | 93.8% | 91.% | 87.8% | 84.% | 79.8% | 75.% | 69.8% | 64.% | 57.8% | 51.% | 43.8% | 36.% | 27.8% | 19.% | 9.8% |

P(<=#) | 0.2% | 1.% | 2.2% | 4.% | 6.2% | 9.% | 12.2% | 16.% | 20.2% | 25.% | 30.2% | 36.% | 42.2% | 49.% | 56.2% | 64.% | 72.2% | 81.% | 90.2% | 100.% |

(notice how both distributions are skewed toward high values).

#### Probability to get a sum \(S\) when rolling \(n\) \(m\)-sided dice and keeping the \(k\) lowest ones

This is the twin of the previous problem, and not surprisingly we can solve it by repeating the very same reasoning process.Let's call \(l\) the highest roll among the \(k\) lowest rolls, \(i\) is the number of dice that rolled more than \(l\), and \(j\) is the number of dice that rolled exactly \(l\) (which means that \(n-j-i\) dice rolled less than \(l\)).

If \(i+j=n\), there are no dice that rolled lower than \(l\), but since \(l\) is the highest roll of the \(k\) lowest, the only option is that all those \(k\) dice rolled exactly \(l\), and they sum to exactly \(l\times k\).

If \(i+j< n\), then \(k-(n-i-j)\) dice rolled exactly \(l\), and sum to \(l\times(k-(n-i-j))\), and the rest of the \(k\) lowest dice rolled less than \(l\). So the number of ways we can get the highest \(k\) dice to sum to \(S\) is the number of ways \(n-i-j\) dice can sum to exactly \(S-l\times (k-(n-i-j))\), keeping in mind that these dice can only have results between \(1\) and \(l-1\). We call this \(M\), and using the probability-generating function discussed above we can see that it is the coefficient of the \(x^{S-l*(k-(n-i-j))}\) term of \((x^1+x^{2}+\ldots+x^{l-1})^{n-i-j}\).

The number of ways we can separate \(n\) dice into a group of \(i\) elements, a group of \(j\) elements, and a group of \(n-i-j\) elements is \(\frac{n!}{i! j! (n-i-j)!}\), and there are \((m-l)^i\) ways to choose the values of the \(i\) dice that are higher than \(l\). So, we can put all the piece together to find the probability to get a sum \(S\) when rolling \(n\) \(m\)-sided dice and keeping the \(k\) highest ones: \[ P(S) = \frac{1}{m^n}\sum_{i,j,l} \frac{n!}{i! j! (n-i-j)!} (m-l)^i \; M \; . \] Similarly to before, if \(l=m\), then there is no way a die can roll higher than \(l\), and thus \(i=0\). In this case the number ways to choose the values of the \(i\) dice that are higher than \(l\) is

**not**\((m-l)^i\) (which would give us an indeterminate \(0^0\)), but just 1. The arguments for the maximum/minimum values \(l\), \(i\), and \(j\) can take are exactly the same as before, so we finally get: \[ P(S) = \frac{1}{m^n}\sum_{l=1}^{m} \sum_{i=0}^{n-k} \sum_{j=n - k - i + 1}^{n-i} \frac{n!}{i! j! (n-i-j)!} (m-l)^i \; M \; , \] which is almost the same formula as before.

##### A few common cases

Roll \(4D6\) and keep the 3 lowest ones# | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 1.6% | 4.2% | 7.3% | 10.1% | 12.3% | 13.3% | 12.9% | 11.4% | 9.4% | 7.% | 4.8% | 2.9% | 1.6% | 0.8% | 0.3% | 0.1% |

P(>=#) | 100.% | 98.4% | 94.2% | 87.% | 76.9% | 64.5% | 51.2% | 38.3% | 26.9% | 17.5% | 10.5% | 5.7% | 2.8% | 1.2% | 0.4% | 0.1% |

P(<=#) | 1.6% | 5.8% | 13.% | 23.1% | 35.5% | 48.8% | 61.7% | 73.1% | 82.5% | 89.5% | 94.3% | 97.2% | 98.8% | 99.6% | 99.9% | 100.% |

Roll \(2D20\) and keep the lowest result

# | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

P(#) | 9.8% | 9.2% | 8.8% | 8.2% | 7.8% | 7.2% | 6.8% | 6.2% | 5.8% | 5.2% | 4.8% | 4.2% | 3.8% | 3.2% | 2.8% | 2.2% | 1.8% | 1.2% | 0.8% | 0.2% |

P(>=#) | 100.% | 90.2% | 81.% | 72.2% | 64.% | 56.2% | 49.% | 42.2% | 36.% | 30.2% | 25.% | 20.2% | 16.% | 12.2% | 9.% | 6.2% | 4.% | 2.2% | 1.% | 0.2% |

P(<=#) | 9.8% | 19.% | 27.8% | 36.% | 43.8% | 51.% | 57.8% | 64.% | 69.8% | 75.% | 79.8% | 84.% | 87.8% | 91.% | 93.8% | 96.% | 97.8% | 99.% | 99.8% | 100.% |

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