Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This morning's CL draw fell along lines that are likely to generate some suspicious of it being rigged - i.e. the strongest teams all avoided each other. I was trying to figure out the approximate probability of this draw. I imagine it's a relatively simple question and I want to check my thoughts. Basically I would divide the 8 teams into 2 pots: Pot 1-Stronger and Pot 2-Weaker (Dortmund and Juventus are probably the only debatable choices in my lists)

Pot 1: Barcelona Real Madrid Bayern Munich Dortmund

Pot 2: Juventus PSG Gala Malaga

My question is: what is the probability of all 4 matches containing exactly one Pot 1 team and one Pot 2 team? For any given draw the probability is 4/7 that you get a team from the opposite pool but what happens then? How do I sum the probabilities for the remaining draws (presumably the drawn teams are taken out of the pool for the next round)?

Thanks!

share|improve this question

4 Answers 4

up vote 0 down vote accepted

You are correct that the first matchup has $\frac 47$ chance of coming one from each pool. Now delete those two teams and draw again. The same logic says you have $\frac 35$ chance the second match comes out this way. The third is $\frac 23$ and the fourth is sure. So $$\frac {4 \cdot 3 \cdot 2}{7 \cdot 5 \cdot 3}=\frac {24}{105}=\frac 8{35}\approx 0.23$$

share|improve this answer
    
Ok, great, that's close to my 1st guess (my 1st guess I had removed one team at a time for some reason - I knew something was a little off). So, it's not so suspicious after all! :-) Thanks! –  Joe Brown Mar 15 '13 at 15:42

Another route:

Arrange the eight teams in any one of $8!$ ways. pairing first and second, third and fourth, etc to give all the possible game pairings.

Except: reversing any two teams in a pairing does not make a change. Moving any pairing from its position to any other position does not make a change. Therefore, there are $$\frac{8!}{2! 2! 2! 2! 4!}=105 $$possible four game schedules.

Now, take the four Pot 1 teams in any one particular order, say alphabetical, or date of founding, or whatever. The four Pot 2 teams can line up beside them to pair off in $4!$ or $24$ ways. So the probability of this "suspicious" match up is $\frac{24}{105}$

share|improve this answer

I thought about the probability that no German teams are playing against each other and no Spanish teams are playing against each other:

We have 105 cases.

In $\frac{6!}{2!2!2!3!}=15$ cases German teams play against each other.

Since we have three Spanish teams, they play in $3*15=45$ cases against each other.

There is only one case where German teams meet AND Spanish teams do not meet.

So the probabily that Germany and Spanish teams don't meet is:

$\frac{105-15-45+1}{105}=0.438$

Is that correct?

share|improve this answer
    
Anybody interested in calculating the probability that no Spanish and no German teams meet and no teams from pot 1 meet? –  user1141785 Mar 15 '13 at 17:29

I was wondering the exact same question as user1141785 posted, what would be the probability that no teams from the same country would meet each other in the quarter finals? And what about the europa league, in which there were 3 english clubs and the remaining clubs are from 5 different countries. What are the chances of no match between clubs from the same country in both CL and EL combined, as happened this week?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.