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.