Lets say I have a probability of the number of goals in a game, given the league that the game is played in.

A = #Goals B = league

Using Bayes theorem: P(A|B) = P(B|A)P(A)/P(B)

Lets say B = 'Premier League'

How do you compute P(B|A) if B is a category? In my head P(B|A) means "Compute the probability of seeing B='Premier League' over every possible value of A".

Lets say A=2, I could look at the probability that A=2 in each league I have data for. Lets say that in the premier league it is 0.3, in Serie A it is 0.2, and in La Liga it is 0.4. I could then say P(B|A) = 0.3/(0.4+0.2+0.3) = 1/3.

Does that make sense? I have been trying to think how to do this for a few days and I cant get my head around B being categorical. If it was continuous P(B|A) would just be "What is the probability of seeing your observed value of B, for each possible value of A" I think.. Which makes more sense.


1 Answer 1


The way that I would think about it is the following:

First I would try to define the sample space of that problem, i.e the space that will contain all the possible results. In order to define that I would consider what values the events $A$ and $B$ can take.

$A$ which is the number of goals can take the following value $\left \{ 0,1,2,... \right \}$, while $B$ the league has three possible categories $\left \{ B_{1},B_{2},B_{3} \right \}$.

Then I would define the sample space $\Omega$ as the collection of all possible combinations/results of values of $A$ with the values of $B$. So, $\Omega$ it will contain the event that you used $(A=2)\cap B_{1}$ and $(A=2)\cap B_{2}$ and $(A=2)\cap B_{3}$ and all such events that make sense for your problem.

Then for the conditional probability of $B$ given $A$ you will have the probability rule

$$\mathbb{P}(B|A) = \frac{\mathbb{P}(B\cap A)}{\mathbb{P}(A)}$$

however, in order to make inference you have to specify the events $B\cap A$ and $A$.

So you could write all the events that you are interested in $B=B_{1}$ with $A=2$ or $B=B_{2}$ with $A=2$ etc.

And this will be transalated in probability as

$\mathbb{P}(B=B_{1}|A=2)=\frac{\mathbb{P}((B=B_{1})\cap (A=2))}{\mathbb{P}(A=2)}$,

where the events $(B=B_{1})\cap (A=2)$ and $(A=2)$ belong into your sample space $\Omega$ so you can define probabilities on them.

So, overall if you approach the conditional probability as the probability of a particular event happening (i.e. a combination of $A$ and $B$ happening) then it would be easier to understand it.

  • $\begingroup$ Thanks! Is P(A=2) in the denominator the probability of A being equal to 2 across all games (i.e. taken from a probability distribution of just A)? $\endgroup$ Commented May 10, 2021 at 8:03
  • $\begingroup$ @ChristopherCollins Exactly! Which can be expressed as $\mathbb{P}(A=2) = \mathbb{P}(A=2|B=B_{1})\mathbb{P}(B=B_{1})+\mathbb{P}(A=2|B=B_{2})\mathbb{P}(B=B_{2})+\mathbb{P}(A=2|B=B_{3})\mathbb{P}(B=B_{3})$ $\endgroup$
    – Fiodor1234
    Commented May 10, 2021 at 8:21
  • 1
    $\begingroup$ Amazing, thanks! This has really helped my understanding $\endgroup$ Commented May 10, 2021 at 16:03
  • $\begingroup$ @ChristopherCollins Glad to hear that! $\endgroup$
    – Fiodor1234
    Commented May 10, 2021 at 16:08

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