# Finding the probability of soccer game's results

It's a little basic question. I'm so much into soccer and I love how statistics and football can be integrated. But I have little knowledge about statistics, yet I do have information about the probability subject.

So let's say there is a team A and team B. We know that A's winning, losing, drawing probability for the next game aside from its opponent. The same thing applies B. So, in our hand we have:

A - Win : 0.7 A - Draw : 0.2 A - Lose : 0.1

B - Win : 0.5 B - Draw : 0.3 B - Lose : 0.2

So, how do i predict probabilities of results of the game according to these.

• To the close voters: Probability is unquestionably an appropriate topic here, is it just because the question is motivated by soccer?
– Jim
Mar 16 '13 at 16:52
• It will be difficult to take into account the for-profit arrangements that are made by professional gamblers. Mar 16 '13 at 17:02
• I am not a gambler by the way, I don't bet. I just interested in predicting the unpredictable. Mar 16 '13 at 17:03

If you want to predict the probability of $A$ or $B$ winning and intend to use historical probability, I highly suggest you use the historical results of all matches played between $A$ and $B$.

In other words, consider the probability model you have again: The probability of $A$ winning is $0.7$; drawing is $0.2$ and losing is $0.1$. This actually is the proportion of all games that $A$ has played. Right now, only pick the games that $A$ played with $B$ and recalculate this probability.

Same thing goes for $B$. Only select the games that $B$ played with $A$ and calculate the same probability.

• Thanks for the answer. There is something that came to my mind. Can I use joint probability of these values. For example: we know that if A win happens, B lose happens too. So if I take the joint probability of these, would it be logical? Mar 16 '13 at 17:02
• You can technically use joint probability, but you soon will realize that probability A winning = probability B losing, probability A drawing = probability B drawing and probability A losing = probability B winning. You could, however use joint probability to find out goal differences ie P(A wins given goal difference = 3) etc. That will be more meaningful. Mar 16 '13 at 17:14

add wins of a to losses ofb. draws of a to draws of b. losses of a to wins of b. divide all totals by2. a wins=45%,a draws=25%, a losses=30%

There are so many methods ... The one I prefer is to divide the teams into classes depending on theit table positions. Then work out the probability of class A versus class B, class B versus class C and so on. There are quite a few refinements.Such as neutral venues and non neutral venues, early season-late season a.s.o. This gives me the basic strength of the teams but there may be of course quite a few qualifying factors at play, such as player injuries, team motivation to win.

Now if two teams of different countries are facing each other, this method more or less fails and you have to look for other methods.

My method is old but as I said there are many methods. There are so many sites in the internet providing us with answers. A more important question is to establish a common criterion to measure the effectiveness. Signal information theory provides us with such criteria, but they 're not being used. Football probability calculation is not an exact science. Nobody has a definite answer and nobody can say "this match is the exact equivalent of N1 red balls in a jar, N2 white and N3 red". Everybody makes his own estimate -never mind that those estimates are quite often similar. But information theory can supply us with universal benchmarks. As it is what we have now is various chaps who bolster "we are the greatest" but we don't really know.