0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.