How much the recent form affects the outcome of a football game?

This article is part of my new book MATH to WIN: NUMBERS can Talk. You can use it along with my first book MATH to WIN: Football to improve your betting decisions.

The recent form, usually the last six results of a team, is a highly appreciated factor by punters when evaluating a football game.

The key aspect of the idea of form is that form influences future success. For example, one might say, “Typically, I would favor Liverpool, but Everton has been in great form the last four games.”

To fully investigate the “form effect” we must answer the following question: Does a team’s performance in the last six games affect the winning chances of the next game? If the answer is yes, how much?

To answer this question, I calculate the unadjusted form factor and the adjusted form factor by giving more weight to the most recent games results.

For example, let’s see the form score for the Premier League game Huddersfield vs. Bournemouth from the season 2018-2019. The form score for Huddersfield was -0,67 and the previous six games was (from the older to newest) :

The Bournemouth score was -0,5 and the previous six results were:

I separate the games into chunks based on the form factor.  The percentages of each outcome, regardless of any factor or calculation, are: Home win 44.9%, Draw 26.5% and Away win 28.6%

The following table shows the percentages of each outcome regarding the home team’s average form factor score

As you can see, there is a very high correlation between the form score and the winning percentages. The correlation degree for the home win is 0.96 and for the away win is -0.99. The following graph shows, in graphical form, the percentages of each outcome regarding the average home team’s form factor.

From the graph above, we can extract the polynomial equations which describe the data and use them to perform our calculations.

P(1) = 0,0921x3 + 0,0694x2 + 0,1154x + 0,4396,   = 0,993

P(X) = -0,0316x3 – 0,0486x2 – 0,0115x + 0,2718, R² = 0,972

P(2) = -0,0608x3 – 0,021x2 – 0,1036x + 0,2886, R² = 0,993

 

Let’s apply these equations to our example. The un-adjusted form score for Huddersfield was -0,67, so the probabilities of each outcome calculated as follows:

 

P(1) = 0,0921*(-0,67)^3 + 0,0694*(-0,67)^2 + 0,1154*(-0,67) + 0,4396 =  0,366

P(X) = -0,0316*(-0,67)^3  – 0,0486*(-0,67)^2  – 0,0115*(-0,67) + 0,2718 = 0,267

P(2) = -0,0608*(-0,67)^3   – 0,021*(-0,67)^2  – 0,1036*(-0,67)  + 0,2886 = 0,367

 

We can convert these probabilities to decimal odds as follows :

O(1) = 1 / 0,366 = 2.73, O(X) = 1 / 0,267 = 3.75, O(2) = 1 / 0,367 = 2.73

 

The next table shows the percentages of each outcome regarding the guest team’s average form factor score.

We have a great degree of correlation between the form factor and the winning percentages. Indeed the correlation for the home win is –0.980 and for the away win is 0.949. You can verify these results if you copy the table to MS Excel and use the CORREL function.

The next graph shows, in graphical form, the percentages of each outcome regarding the average guest team’s form factor.

From the graph above, we can extract the polynomial equations which describe the data and use them to perform our calculations.

 

P(1)= -0,0669×3 – 0,0474×2 – 0,1236x + 0,4588, R² = 0,9967

P(X) = -0,002×3 – 0,0348×2 + 0,0197x + 0,2701, R² = 0,9561

P(2) = 0,0684×3 + 0,0825×2 + 0,1036x + 0,2709, R² = 0,9987

 

Let’s apply these equations to our example. The un-adjusted form score for Bournemouth was -0,5, so the probabilities of each outcome calculated as follows:

 

P(1) = -0,0669*(-0,5)^3 – 0,0474*(-0,5)^2  – 0,1236*(-0,5)  + 0,4588 = 0,517

P(X) = -0,002*(-0,5)^3  – 0,0348*(-0,5)^2  + 0,0197*(-0,5)  + 0,2701 = 0,252

P(2) = 0,0684*(-0,5)^3  + 0,0825*(-0,5)^2  + 0,1036*(-0,5) + 0,2709 = 0,231

 

We can convert these probabilities to decimal odds as follows :

O(1) = 1 / 0,517 = 1.93, O(X) = 1 / 0,252 = 3.97, O(2) = 1 / 0,231 = 4.33

If we want to use only the form factor to estimate the fair odds for this game, we must add the probabilities for each outcome and then divide them by two. Then we can convert the new probabilities to decimal odds.

P(1) = (0,366 + 0,517) / 2 = 0,441

P(X) = (0,267 + 0,252) / 2 = 0,259

P(2) = (0,367 + 0,231) / 2 = 0,300

The final odds are :

O(1) = 1 / 0,441 = 2.27, O(X) = 1 / 0,259 = 3.85, O(2) = 1 / 0,300 = 3.34

 

So, the answer to our question is obvious: Yes, the last six games’ performance affects the next game’s winning chances.

Another useful observation is that the winning percentage changes are far more significant when the form factor is bigger than zero.

The home win % changes 13.7% when the home form factor lies between -1 and 0 and 27.8% when the factor lies in positive territory. Similarly, the away win % changes 9.4% when the away form factor is in negative territory and 25.3% when the factor values are between 0 and 1.

This fact tells us that when the form is not good, other factors may affect the outcome more, but it affects the game much more when it is above average.

This article was only a small part of my new book MATH to WIN: NUMBERS can TalkYou can use it along with my first book MATH to WIN: Football to improve your betting decisions.

Leave a Reply

Your email address will not be published. Required fields are marked *