ende

Football Betting Permutations: 5 Cup Finals in 5 Days


Slim chance for the favourites?

It has been a busy week in the world of football with the conclusions of many major tournaments across Europe, including the jewel in the crown, the UEFA Champions League final between Bayern Munich and Chelsea.

Having previewed many of the recent important cup games we decided to take a retrospective look at the pre-match expectations and compare them with what actually happened.

The table below summarises five of the matches we previewed and shows the probabilities we calculated using historical statistics.

Table showing the odds calculated by Soccerwidow converted into probabilities for four 2012 domestic European cup finals, and the 2012 Champions League final

Cup Final Dreams

For example, in the Swiss Cup final, we computed that FC Basel had a 52.80% chance of winning in normal time, but the game ended 1-1 after 90 minutes. In comparison, our spreadsheets showed the draw had just a 24.70% likelihood.

The only match which finished as expected was the Scottish Cup final, where Hearts defeated poor Hibernian, 5-1. In the other four matches, either the underdog won (Académica in Portugal and Napoli in Italy), or the games ended in a draw (Bayern v Chelsea and Basel v Luzern).

To compute the overall probability of a certain combination of results, simply multiply the individual probabilities together. Thus, the combined probability that all five favourites won was:

52.8% x 45.9% x 64.9% x 54.7% x 43.7% = 3.76%

Effectively speaking, the 3.76% combined chance of all five favourites winning means this would statistically happen only once in every 25 seasons. This is a fairly bleak insight for those of you who like accumulator or parlay bets!

Taking this train of thought further, if Basel, Hearts, Bayern and Sporting all won, but the Coppa Italia ended in a draw, then the probability calculation would be as follows:

52.8% x 45.9% x 64.9% x 54.7% x 34.1% = 2.93%

The probability generally that four favourites won and the fifth match ended in a draw or a victory for the underdog, was 17.78%.

Calculating this is quite simple and is purely a question of multiplying the probabilities for every permutation of this event and then adding together the results. In this case there are 10 possible permutations, summarised in the table below:

Table showing the 10 possible permutations of results involving four favourites winning and the permutation probabilities for four 2012 domestic European cup finals, and the 2012 Champions League final

10 Possible Permutations

We can now say that in five cup finals with similar probabilities, the likelihood that either all the favourites will win, or at least 4, equates to 21.54% which means this will happen on average approximately every five years:

17.78% (only 4 favourites win) plus 3.76% (all favourites win) = 21.54% (around one in five)

Again, we mention accumulator bets, which perhaps don’t seem so attractive once you know the actual statistical chances of succeeding with, in these examples, a five-fold.

As an aside, the probability in our five cup finals in general for only one favourite winning, two games ending in draws, and two victories for the underdog was 4.86%. This combination of events was therefore more statistically likely than the 3.76% chance that all the favourites would win.

Read more about this subject in further detail in our article: Combinatorics and Probability Theory in Football Betting.

We hope you enjoyed this article and also hope it has added something to your knowledge base. Please always gamble responsibly and never more than you can afford to lose.


learn to think like a bookmaker!
deciphering bookmaker mathematics

Last Update: 22 May 2012

Categories:Case Studies Odds Calculation



One Response to “Football Betting Permutations: 5 Cup Finals in 5 Days”

  1. a
    25 December 2014 at 8:16 am #

    It is just some joint probabilities of independent experiments. High school maths.

Leave a Reply

Time limit is exhausted. Please reload CAPTCHA.