One can calculate the probability of a correct score of a football match by Poisson(Number of Team1 goals, Mean average Team1 goals) x Poisson(Number of Team2 goals, Mean average Team2 goals)
So for 0:0 scoreline, Poisson(0, 1.05) x Poisson(0, 2.5)
if mean Team1 is 1.05 and mean Team2 is 2.5.
But how does one calculate the probability/odds of "home team to win by exactly one goal"? Adding double Poissons for each 2:1, 1:0, 3:2, etc scorelines might be one of the methods but it will be a tedious task and I am trying to avoid Excel for the time being to find the simplest method to calculate this. How do you involve the other team in your calculations?
Was hoping an easier, simpler way was there (something like calculating P=0 and then 1-P(0) to find odds of P>=1).
Best Answer
The Skellam distribution is the distribution of the difference in counts of two independent Poisson variates:
http://en.wikipedia.org/wiki/Skellam_distribution
But you will still need a computer to calculate the modified Bessel function. Here is an Excel example.
This image (which can be magnified by zooming in the browser) shows the formulas at the left and the values for the same sheet at the right.
The spreadsheet accepts the (positive) means $\mu_1$ and $\mu_2$ as input (blue text), computes three intermediate quantities involved in the Skellam probabilities ($\exp(-\mu_1-\mu_2)$, $\sqrt{\mu_1 \mu_2}$, and $\sqrt{\mu_1/\mu_2}$, shown in green), and then uses those to compute a table of probabilities for the score difference $k$ using the hyperbolic Bessel function $I_{|k|}$. This table (shown as computed for $-12, -11, \ldots, 12$, but computable for any range of integers) is plotted at the right as a bar chart, with positive differences to the right (in red) and a zero difference (in gray) exactly in the middle. As a check, the sum of the probabilities is computed and displayed: if it is any less than $1.0000$, the table may be missing some important probabilities and should be extended in one or both directions. As usual, cumulative probabilities can be found with little extra effort by running a cumulative sum down a parallel column. (This distribution has no convenient, general, closed form formula for its CDF.)