Probability is so pervasive in the RPG hobby that it’s important to have a basic understanding of how it works regardless of where in the hobby you are, designer, GM, or player. Yet in my experience, far too few people regardless of their role,  have a grasp on the probabilities at work at the gaming table. I won’t name names, but I’ve seen systems where characters with what’s labeled as a “normal human level of competency” with a skill has a 0% chance to succeed at an “easy” task, GMs place opponents that their players had a 5% chance or worse to effect, and players taking actions with a 20% chance of success where failure meant certain death. You can argue that maybe these things are intentional, but I find it more likely that the individuals in question need a refresher course on how dice work.

For players, the most useful applications of probability is being able to gauge the effects of your actions in the game world given your current numbers and die rolls. GMs should have the same feel for the probabilities of certain actions, but may also occasionally want to better understand the chance their players have to overcome certain challenges or how their tweaks to standard numbers effect their game. Designers should more or less know the probabilities associated with their systems inside and out, which is beyond the scope of a single article for beginners.

Probability Values and Notation:

Probabilities are numbers from 0 (no chance) to 1 (absolutely certain). There can never be a probability less than 0 or more than 1. If you calculate one, you’ve done some math wrong somewhere. If you feel the need to convert a probability to a percent, simply multiply it by 100. We don’t use percent values in calculation because it would make extra steps in the math, but they’re easier for many people to understand, so converting your final probabilities makes sense if you’re sharing them. Probabilities are noted P(Event).

The Sample Space and Events:

In probability, we define the sample space as a set of all possible outcomes of an experiment. The sample space is denoted as S={outcomes}. The size of a sample space is denoted |S| and has a value equal to the number of outcomes in the sample space.

An event is an outcome or a group of outcomes we’re interested in. Events are commonly referred to by letters and denoted as A={outcomes}. The size of an event is denoted |A| and has a value equal to the number of outcomes in A.

If we were rolling a d6 and we wanted to find the probability of rolling higher than 3, our sample space and event would look like this:

S={1,2,3,4,5,6}
A={4,5,6}
|S|=6 |A|=3

The Simple Sample Space:

The simple sample space is a sample space in which all the possible outcomes are considered equally likely. Rolling most single dice is a simple sample space; drawing cards from some kinds of decks is a simple sample space. If an experiment uses a simple sample space, the probability of an event A is equal to |A|/|S|.

Thus, using our earlier example of rolling higher than 3 on a d6,

S={1,2,3,4,5,6}
A={4,5,6}
|S|=6
|A|=3
P(A)=|A|/|S|=3/6=50%

Let’s look at a more useful example:

Grog the barbarian is fighting a highly armored foe, and needs to roll a 16 or better to hit him. If Grog rolls a 20, he crits. What’s the chance Grog hits without critting, crits, or does either?

S={all numbers 1 through 20}
|S|=20
H (just hits)={16,17,18,19}
|H|=4
C (crits)={20}
|C|=1
E (either)={16,17,18,19,20}
|E|=5
P(H)=|H|/|S|=4/20=20%
P(C)=1/20=5%
P(E)=5/20=25%

So what should Grog do? Should he swing his axe? Run? Drink a healing potion? That depends on the situation Grog is in, but with this info you can make the right choice.

Counting Methods:

Let’s consider an example of rolling 3d6 for an ability score. We want to know the probability we’ll roll a 16 or higher.
S={all sequences of 3 numbers 1 through 6}
A={all rolls of 16 or higher}
P(A)=|A|/|S|

Figuring that out could be very daunting task. We could start counting the sample space manually: 111,112,113, etc… but luckily there are counting methods that help us find these numbers quickly and easily. The hard part of these counting methods is knowing when to use which one. Not only are they not always the same, but sometimes you need to calculate the sample space size with one of them, and the event size with a different one.

Multiplication Rule:

The multiplication rule states that if an experiment can be broken down into smaller, independent experiments,  and that if, regardless of the outcome of the individual experiments, the number of outcomes in the other experiments, though not necessarily the specific outcomes remain the same then the total number of outcomes in the complex experiment is the product of the number of outcomes of all the simple experiments.

The sample space of our ability score example can be easily figured out with this counting method. Each d6 has 6 possible outcomes, regardless of what those outcomes are. Thus, the number of outcomes in 3d6 is 6*6*6= 216

Permutations:

Permutations are a way to count the number of outcomes in a decreasing pool of options. The number of ways you can draw cards from a deck, or the number of ways you can roll unique numbers in a die pool are permutations.

Permutations are noted Pn,k, read “number of permutations of n objects taken K at a time” and the formula for determining how many there are is n!/(n-k)!

Say hello to the factorial. The factorial, represented by an exclamation mark, is a mathematical symbol used to denote the product of an integer and all preceding positive integers.  Thus  6! = 6*5*4*3*2*1 = 720

Permutations don’t come up in dice much because they usually model sampling without replacement (outcomes are removed from the pool of possible outcomes as they occur)  which doesn’t apply to dice. They do, however apply to cards a lot.

You have to role 3d6 on the GM’s home brewed critical fumble table. Doubles result in permanent injuries, and triples result in various ignominious deaths and bards writing comedic songs about your ineptitude. What’s the probability you get off easy?

S={all sequences of 3 numbers 1-6}
A={all sequences of 3 unique numbers 1-6}

We can use the multiplication rule to determine |S|=6*6*6=216. |A| however, is a permutation. You can roll any of the six numbers on the first die, any of the other 5 on the next, and any of the remaining 4 on the last. |A|=P6,3=6*5*4*3*2*1/3*2*1=120. Thus: P(A)=|A|/|S|=120/216=5/9=approx. 56%

Arrangements:

When you want to know how many ways you can arrange more than one type of otherwise indistinguishable elements (like how many ways you can get 6 successes on 9 dice, or how many ways three orcs and two ogres can be stationed at 5 guard posts, that’s an arrangement.

Arrangements are noted where n is the total number of object you’re selecting and n1,n2,n3…  are the size of each different group you’re arranging. The sizes of all the sub groups must also sum to the greater number.

The formula to calculate the number of arrangements is n!/(n1!*n2!*n3!…)

Warning! if you’re using arrangements to determine the number of ways you can split things into like-sized indistinguishable groups, you have to then divide the above formula by x! where x is the number of indistinguishable like-sized groups. For example, the number of ways the party of 6 PCs could split into 3 groups of two, is (6!/2!*2!*2!)/3! unless those groups are somehow distinguishable from one another. This situation doesn’t come up much, but when it does, it can throw your probabilities off badly!

As an example, consider the probability of rolling exactly 3 6s with a die pool of 5d6.

S={all sequences of length 5 of the numbers 1-6}
A={all of the above that contain exactly 3 6s)

Using the multiplication rule, we know that |S|=6*6*6*6*6=7776.
We can use arrangements and the multiplication rule together to determine the size of |A|. Break A into A1={rolling 3 6s} and A2={rolling 2 not sixes}. We can use the multiplication rule to figure each of these out individually, and then multiply their results together. The multiplication rule also says there are 5*5=25 ways to roll 1-5 on two dice and that there are 1*1*1=1 way to roll 6s on three dice. Now we just need to know how many ways we can order those 6s and not 6s. Since we have 5 total objects, and groups of 3 and 2, there are 5!/(3!*2!)=5*4*3*2*1/(3*2*1)*(2*1)=10 ways to arrange those two groups. Thus |A|=25*1*10=250

Thus, P(A)=|A|/|S|=250/7776=approx. 3%

Application:

To use these tools, all you have to do is decide what your sample space and event are, and count them. It’s important to always count the same type of objects. If you’re counting die pools, for example, counting your sample space in terms of numerical rolls, and your event in terms of successes and failures can result in an error. This doesn’t however, mean that you need to use the same counting methods for both the sample space and the event. As you can see in the examples above. It’s necessary to mix counting methods on occasion. Think carefully about what you’re counting and the methods you use to count them. It’s very easy to over-count complex events, so it’s worth extra time carefully considering which method to use. Finally, it’s important to make sure you remain grounded in simple sample spaces. If every event in your sample space and event aren’t equally probable, than your result won’t be valid. As with many things, practice with these techniques makes perfect, so if you’re not getting reasonable numbers, keep trying.