Probability of a team scoring a goal?

by mattadams84   Last Updated November 17, 2017 17:19 PM

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.



Answers 1


This is a problem where mathematics can at best supply a "toy" answer, as there is so much more to sporting outcomes than simple statistics.

But this looks fun, so let's crack on with it anyway. Take everything that follows with a pinch of salt.

We have two estimates, $a$ and $b$ for team A scoring. Let's go with the average, $(a+b)/2=0.6$, and say that's equal to $\mathbb{P}(S_A)\ge 1$, the probability that the number of goals scored by team A, $S_A$, is at least 1. Let's model $S_A$ as a homogeneous Poisson process, which therefore has rate parameter $\lambda_{AB}=-\log\left(1-(a+b)/2\right)\approx 0.916$. This means we expect team A to score $k$ goals against team B with probability $e^{-\lambda_{AB}}\lambda_{AB}^k/k!$, like this:

goals | probability
------+------------
  0   |   40.0%
  1   |   36.7%
  2   |   16.8%
  3   |    5.1%
  4   |    1.2%
  5   |    0.2%

Similarly, the number of goals that team B scores against team A could be modelled as Poisson with rate $\lambda_{BA}=-\log\left(1-(c+d)/2\right)\approx 0.223$, with the probabilities working out like:

goals | probability
------+------------
  0   |   80.0%
  1   |   17.9%
  2   |    2.0%
  3   |    0.1%

Finally, we treat those independently (multiply the probabilities) to find the most likely overall scores:

score A : B | probability
------------+------------
      0 : 0 |   32.0%
      1 : 0 |   29.3%
      2 : 0 |   13.4%
      0 : 1 |    7.1%
      1 : 1 |    6.5%
      3 : 0 |    4.1%

So there we are. But please don't put any money on it!

Creosote
Creosote
September 05, 2015 20:10 PM

Related Questions



Probability problem [hard]

Updated December 18, 2016 08:08 AM

Complete and Sufficient Statistics from Two Samples

Updated April 08, 2018 02:19 AM