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View Full Version : Rigged draw?...Probability help requested.

Graham
10-20-2003, 10:31 AM
Mathemagicians, can anyone help me with this calculation?

Suspicion is around that the (nominally random) draw for the Euro 2004 playoffs were rigged to have the higher rated sides avoid each other. (this is soccer for you lot that reckon 'football' involves running around and not touching the ball with your feet at all.... /images/graemlins/tongue.gif)

ie Russia, Spain, Turkey, Croatia and Holland have all been drawn against the 5 lesser teams (one of which, mine, Scotland, ranked 53 in world, has unfortunately drawn the biggest powerhouse of Holland, ranked 7, in the draw over the 2-leg playoff).

Basically, 10 teams are drawn against each other in pairs. What are the odds of the 5 highest ranking teams all avoiding each other in a random draw.

I can't do this type of calculation (but I'm willing to learn). It's some kind of combination problem, is it?

thanks

G

btw, The rumour is that heated balls are used in the televised draw to signify the higher ranked teams when ye olde dignitary draws numbered balls out of the bag. This could be all hokum, of course, but the same happened last time 4 years ago as well.

BruceZ
10-20-2003, 12:22 PM
If 10 teams are paired against each other in one round, the probability of the top 5 teams being paired against the bottom 5 teams is 12.7% or 6.9-1.

The total number of ways to pair 10 teams against each other is 10!/2^5/5! = 945. Imagine writing the 10 team numbers in a list. There are 10! = 10*9*8*7*6*5*4*3*2*1 ways to order the list (10 ways to pick the 1st team, 9 ways to pick the 2nd, etc). Now for each list, you can pair the 1st item of the list with the 2nd, 3rd with the 4th, etc. This will generate all the pairings, but it will count each pairing twice when the opponents are interchanged. For example, 12.... and 21... are the same. So we divide by 2^5 to remove order. This is a standard counting procedure. It is total permutations of all the members, divided by the permutations of each of the pairings. 10!/2^5 is the same as the number ways to deal hold'em hands to 5 players from a "deck" of 10 cards. The only difference is that here we don't care which "players" get which hands, the hands being the pairings. If 1 pairs with 2, we don't care which of the 5 places in the list that occurs. So we divide by 5! to remove the order of the 5 pairings. Now the number of ways to pair the top 5 against the bottom 5 is 5! = 5*4*3*2*1 = 120. That is, there are 5 teams that the 1st place team can pair with, leaving 4 that the 2nd place team can pair with, etc. So the odds of this are 120/945 = 12.7%.

The probability of it happening at least twice in 5 years, assuming there were only 5 such pairings in 5 years, is:

1 - (1-120/945)^5 - 5*(120/945)*(1-120/945)^4 = 12.4%

from the binomial distribution. This is about the same as the probability of it happening on any given draw.

Graham
10-20-2003, 01:06 PM
Thanks Bruce,

I got the same answer of 120/945 by doing the old iteration of probability of team 1 avoiding teams 2-5 (ie 5/9) multiplied by team 2 avoiding teams 3-5 (4/7) multiplied by 3 avoiding 4 &amp; 5 (3/5) multiplied by 4 avoiding 5 (2/3).

I guess your method works better with larger no's...

btw, the event only happens every 4 years, not yearly. So, I guess the chance of that happening twice in a row is 120/945 X 120/945 = 1.6%.

Dunno if that's enough to convict UEFA but it wouldn't really surprise many if that was the case - revenue money talks you know /images/graemlins/tongue.gif

G