# Probability of goals in a football match?

##### New Member
I am trying to work out the probability of teams scoring a goal in a football match. Obviously there are many factors to consider, and the probability is impossible to predict correctly as there are many factors that will change in a football match. However i want to figure out the probability using only the following information. A lot of work has been done on predicting football matches using poisson distribution, but in this case, i only want it to be based on the stats that i have.

So for this example it is Team A vs Team B.

Team A Scores in 70% of their home matches (a) Team B Concedes in 50% of their away matches (b)

Team B Scores in 10% of their away matches (c) Team A Concedes in 30% of their home matches (d)

I have managed to calculate the probability of a goal by using the following formula:

(a+c)-(a*c)

I believe this calculation to be correct but it only takes into account the stats for team A or team B SCORING, it does not take into account the stat of the opposite team conceding.

So basically i am after a formula that takes into consideration team A's ability to score against team B's ability to concede, and the same for team B to score and team A to concede.

#### Prometheus

##### Member
**** you for coming up with an interesting question when i should be studying something else.

Had a quick think about it, but couldn't come up with anything sensible. I'm thinking a Bayesian perspective would be suited to this problem - i'll have a go when time allows.

Didn't realise there was automatic censoring of swear words on the forum. Bizarrely, in this case, i think it makes the swear word sound far worse than it actually was. It was supposed to be a tongue in cheek 'dam*' as opposed to the aggressive 'f**k' that comes across when i reread it. I would promise not to swear again on this forum - but there is lots of swearing involved whenever i try stats.

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#### Prometheus

##### Member
To concede a goal is to be scored against. So if team A and B are playing, team A scoring necessarily means team B conceding.

#### hlsmith

##### Omega Contributor
This problem just seems like it has some type of fallacy or you have to write out a bunch of assumptions, they had the same prior schedule, they only score one goal, each game is not independent, etc. I rewrote the information since your prior presentation was a little confusing.

a) Team X Scores in 70% of their home matches
d) Team X Concedes in 30% of their home matches

b) Team Y Concedes in 50% of their non-home matches
c) Team Y Scores in 10% of their non-home matches

What is the probability that X Scores given they are playing Y at home?
What is the probability that Y Scores given they are playing X not at home?

#### BGM

##### TS Contributor
You have to first understand what do those given probabilities mean.

I believe that the probability that Team A score (at least) one goal in their home match is 70% means: When Team A playing at home and randomly selecting their opponents, in a long run / on average there are 70% of these matches Team A will score at least 1 goal. And this should be similarly applied to other probabilities.

Now you are giving that Team A is playing against Team B at home. Obviously, as said above, Team A scoring at least 1 goal is equivalent to Team B concede at least 1 goal. When you are given that they are playing against each other, the conditional probability of A scoring will change and not necessary to be 70% anymore; but it will equal to the conditional probability of B conceding (as the two events are equivalent) which is not necessary to be 50% too.

The probability you want to calculate is A team and B team scoring at least 1 goal in the same match (not a goaless draw). It is the union of two events, but it need be conditional on A vs B. You can still apply the inclusion-exclusion principle, but they are no longer independent and thus you cannot decompose the probability of intersection of events to be the product of individual.