### Beantown Blood Bowl

Basic Probability

Blood Bowl (and any dice based boardgame) relies frequently on rolling dice and understanding the basics of probability will help you understand the risk of a given action. For this discussion, we will always assume the dice to be "fair" in that an result of a single die roll is just as probable as any other result. Later we will discuss tests to determine if dice are "fair".

Let's start from the very beginning. For a given die with n sides, the probability that any given result will be rolled is

P(x) = 1/n.

Pretty basic. If I roll a 1d6 there is a probability of 1/6 that I will roll a 2. Similarly there is a 1/6 chance that I will roll a 5. In this formalism, we could write those statements as P(2) = 1/6 and P(5) = 1/6.

Compound probability

Now let's look at the probability that two independent events will occur, with probabilities P(x) and P(y) respectively. The compound probability of both occuring is the product of the two

P(xy) = P(x)P(y).

So back to our example. Let's say we roll a 1d6 and want to know the probability of first rolling a 2 and then rolling a 5. We could express that as

P(2 then 5) = P(2)P(5) = (1/6)(1/6) = 1/36.

Compound Events

There are two types of compound events. The easier of the two to grasp is the mutually exclusive case where you can have more than one result but not at the same time. For example, rolling a 1 or 2 on a 1d6. This is mutually exclusive because if you rolled a 1, clearly you did not roll a 2, and vice versa. Another way to say this is

P(x and y) = 0.

This often occurs in Blood Bowl when you need to roll a 3 or greater and the action fails on a 1 or 2. The probability of a mutually exclusive compound event is

P(x or y) = P(x) + P(y)

or in the example of rolling a 1 or a 2,

P(1 or 2) = 1/6 + 1/6 = 1/3.

Also observe the fact that the probability of all possible outcomes sums to unity, in other words for a 1d6

P( 1 or 2 or 3 or 4 or 5 or 6) = P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1.

Well what if P(x and y) <> 0? This is the case where the events are mutually inclusive, i.e. there is dependency between the probability of the two events. This actually doesn't come up too often in Blood Bowl but I'll illustrate this with a couple examples. In the case of mutually inclusive probabilities, P(x or y) is written more generally,

P(x or y) = P(x) + P(y) - P(x and y)

so if P(x and y) = 0, it reduces to the mutually exclusive case. So let's do a trivial example and then follow with an example that is conceivable, but just doesn't apply to Blood Bowl.

Example 1

What is the probability that you roll a number greater than 2 on a 1d6 or roll greater than 3 on the same 1d6? Well the results that are greater than 2 are {3,4,5 or 6} so P(2) = 4/6. The results greater than 3 are {4,5,or 6} so P(3) = 3/6. {4,5 and 6} are common to both probabilities so P(2 and 3) = 3/6 so we arrive at

P(2 or 3) = 4/6 + 3/6 - 3/6 = 4/6 or 66.7%

so this is kind of "duh" since all of the >3 rolls satisfy the >2 rolls so it makes perfect sense that you only need consider the probability of being greater than 2.