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Was the cup draw fixed?

There were suspicions of a fix in the draws for both the 2007 and 2008 European Champions League quarter-finals. Here we consider the probabilities involved.


The 2007 Draw

In March 2007 three English teams made it through to the quarter-finals of the European Champions League. There was a rumour going around before the draw was made that the draw would be fixed to ensure the English clubs would avoid each other in the draw. Given the very public, and apparently random, nature of the draw this rumour was highly questionable. Nevertheless, the English teams were duly drawn apart from each other. Assuming the draw was truly random (and it really is highly unlikely not to have been) what was the probability that this would happen?

The probability is 4/7 - that's better than one in two and so not really much of a coincidence.

Here is why:

The probability that any chosen English team is drawn against a non-English team is 5/7 because 5 out of the other 7 teams are non-English.

Now, given that the first chosen English team has drawn a non-English team, there are 2 English teams left and 4 non-English. So the probability that a chosen one of the 2 English teams draws a non-English team is 4/5.

That leaves 1 English team and 3 non-English teams, so the remaining English team is certain to play a non-English team

So the probability that no English team is drawn against another is:

5/7 x 4/5 = 4/7

Here is an animation (it is a wmv file):

cup draw

An alternative way to solve this, that considers all permutations of the draw, appears at the bottom of the page.


The 2008 draw

In March 2008 FOUR English teams made it through to the quarter-finals of the European Champions League. Although this time, as we will see below, the issue of whether the draw was fixed was not about the English teams being drawn apart, it is interesting to ask what was the probability this time that none of the four teams would be drawn against eachother? The answer  is 8/35 - that's a much small chance than the previous year.

We can use the same approach to explain the answer:

The probability that any chosen English team is drawn against a non-English team is 4/7 because 4 out of the other 7 teams are non-English.

Now, given that the first chosen English team has drawn a non-English team, there are 3 English teams left and 3 non-English. So the probability that any chosen remaining English team draws a non-English team is 3/5.

That leaves 2 English teams and 2 non-English teams. The probability that a chosen one of the English teams draw a non-English team is 2/3.

That leaves 1 English team and 1 non-English who must of course be drawn against eachother.

So the probability that no English team is drawn against another is:

4/7 x 3/5 x 2/3 = 8/35

In the event two of the English teams were drawn against eachother. But the fix story was the following (this is taken from the Daily Telgraph on 14 march 2008):

A post left on a newspaper's internet forum at 10.28am this morning, 90 minutes before the draw took place in Nyon, Switzerland, correctly predicted the entire quarter-final draw.
    
The poster claimed bookmakers were no longer willing to accept bets on the draw, although it is understood that most were not running books on it anyway.

When all four ties predicted in the post came to fruition it caused fevered internet speculation that some sort of manipulation might have taken place.

But a Uefa spokesman said: "It is just a lucky guess."

Assuming it was just a lucky guess, what was the probability? The answer is 1/105 (although several newspapers got the figure wrong, with odds of 191 to 1 and 27 to 1 being wrongly claimed). It is easy to see that there are 105 different ways in which 8 teams can be drawn against each other. As in the above calculation:
So that makes 7 times 5 times 3 combinations which is 105.
So the probability of correctly guessing all four ties is 1 in 105 (which using the 'odds' definition of probability is 104 to 1 as explained here).


Alternative solution to the 2007 problem

In a quarter final there are 8 teams, call them teams A, B, C, D, E, F, G, H of whom 3 are English. We may as well assume teams A, B and C are English.

Now think of the total number of ways (i.e. permuations) in which the draw can be made. There are 8! (8 x 7 x 6...x 2 x 1) because there are 8 ways to draw the first team, then 7 to draw the second etc.

The question then reduces to asking how many of those permutations result in a match between two English teams. To answer this think about the first match drawn (i.e. positions 1 and 2 of the permutation). There are 6 ways the first match can involve two English teams, namely

A, B (meaning A drawn first B drawn second)
B, A
A, C
C, A
B, C
C, B

For EACH of those permutations there are 6! permuations of the resulting 6 positions. So there are 6 x 6! ways that two English teams could be drawn together as the FIRST match.

But, in addition to the first match (i.e. positions 1 and 2) the English teams could also be drawn against each other in 3 other matches, namely matches (3,4) (5,6), and (7,8). So we need to multiply by FOUR, i.e. there are  4 x 6 x 6! ways that two English teams can be drawn together.

Hence the probability of two English teams drawn together is:

(4 x 6 x 6!) / 8!  =  (4 x 6) / (7 x 8) = 6/14  = 3/7

and so the probability of no two English teams being drawn together is one minus that number, which is 4/7.





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Norman Fenton


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