# How to calculate cumulative Poisson probabilities without adding each one if no. of outcomes is large and hence, adding (without excel) is difficult?

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).

• The double-Poisson method would only work if the scores were independent, which they probably aren't. Jan 15, 2013 at 10:56
• Goals mostly are independent. Although, you are right some "features" can be added like "scoring rate" (which increases as the match progresses), but for a basic model, poisson is used by most bookmakers with those extra features and adjustments i mentioned, which are a topic for some other time.
– O P
Jan 15, 2013 at 10:59
• A model of a competition in which the scores are independent of each other might work for some games of pure chance, but for a contest like a football match it appears less than fruitful: misleading or useless might be more accurate. This does not make the mathematical question any less interesting, but suggests that if you are actually interested in estimating football scores, you should change your model.
– whuber
Jan 15, 2013 at 13:23
• I think my research so far have been all wrong as after reading several sources and the comments here, i thoroughly started searching again and found that poisson indeed underestimates 0:0 and 1:1 draws and overestimates 1:0 results, since goals are not independent in a football game. Can someone suggest a better model than poisson. My research points me to bivariate poisson but i am unfamiliar with it and am currently researching it
– O P
Jan 15, 2013 at 13:27
• A bivariate Poisson model would allow for dependence. A negative binomial model would allow for over-dispersion. It's probably best to research the subject area literature thoroughly & then come back with any questions. Jan 15, 2013 at 14:32

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.)

• very helpful indeed. Checking it out now.
– O P
Jan 15, 2013 at 12:46
• The spreadsheet shows that under the assumptions of the question, Team1 (with a mean of $1.05$ goals per game) has a $9.14$% chance of beating Team2 (with a mean of $2.5$ goals per game) by one goal. The chance of a win is $13.98$%, a tie $17.07$%, and a loss $68.95$%, all found by summing appropriate ranges of chances in the spreadsheet (corresponding to the red, gray, and blue bars in the plot, respectively).
– whuber
Jan 15, 2013 at 18:12