Brainsteem Mathematics Challenges: Quarter Finals Quirk
Friday saw the draw for the quarter-finals of the UEFA Champions League. There was a lot of press chatter in England as the draw appeared to be "unfair" to the English Premier League teams. Few pundits seemed to know how to actually calculate the odds of such a draw happening! Let's see if you can do better.
As an aside, there were also two Italian teams in the draw, but they avoided playing each other, so no conspiracy there!
Question
Eight teams are drawn randomly into pairs to play in the quarter-finals (or round of 8). Each team comes from a national league. Assuming two English teams and three Spanish teams (and the other three teams from different countries), what is the probability that the two English teams will be drawn to play each other and the three Spanish teams avoid being drawn against each other?
The first correct answer and further interesting comments will be rewarded with an upvote.
Enjoy!
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Both answers are correct, but I did it a different way using combinatorics.
It just dawned on me that there is only one way that draw could happen:
EE, Sx, Sy, Sz, x, y and z are the other 3 teams, giving 6 combinations.
Then it was a matter of finding out how many ways any quarter-final draw can be made. That gives 8!/4!2!2!2!2! = 105 possible random draws.
The way I see it, we can work it out like this:
Suppose an English team (Man U) is drawn for spot "A"... then the possibility of them facing another English team (Arsenal) is (1/7). Arsenal's only chance to be paired with Man U is naturally in spot B.
Suppose next, an Spanish team (Sevilla) is drawn for spot "C". The probability that they won't be paired with an Spanish team (Madrid) is 3/5.
For the remaining Spanish team the proability that they dont' get put against a Spanish teams is 2/3.
Multiplying these numbers gives: 6/105 = 2/35 or about 5.7%, not an impossibility, but a more-likely-than-not scenario!
Note: Alternatively, we could calculate the probability by similarly placing Sevilla in slot C, then calculating that Madrid is placed in a favorable position: 4/5. Similarly, the remaining team has a 1/2 chance of being placed in a favorable position. This gives 2/35, the same result as above.
Your calculations seem to be right. I got the same answer
Freely place the first british team: propbability 1 free spaces 3x2+1
British teams together: probability 1/7 free spaces 3x2
Freely place Spanish team: probability 1 free spaces 2x2+1
Second Spanish team not with the first one: probability 4/5 free spaces 1x2+1+1
Third Spanish team with neither one of the Spanish: probability 2/4=1/2
So a total of 1x1/7x1x4/5x1/2=4/70=2/35 which indeed is approximately 5,7%
I just saw it now since I was in school before. Well done!
Haha! I guess it depends which nation's press one reads. Yes, there was also the astonishment that the 3 Spanish teams avoided each other and the 2 Premier League teams were paired.
It seems a bit odd, but I suppose if you run the numbers it's not impossible. By the way, I've posted what I hope is an interesting question on my own blog, and if you're interested, I want to invite you to take a look.
I adore Mathematicians ... I am still figuring out my basic linear equation. lolx.. Thumbs up @rycharde
Wow, welldone sir. I'm finding interesting the calculative predictions of these probability deductions..