Can you work out this for me? Champions league

[shyguy]

It has been in the news. There are four English Teams in the 8 teams for the Champions League draw. Out of the four pairs picked without replacement, what is the probability that one pair are both English? How do you work it out?

 

[shubhra]

probability

I think the probability would be

[4/8]*[3/7]=3/14

Tell me if i am wrong

 

[Muppet06]

I think the probability would be

[4/8]*[3/7]=3/14

Tell me if i am wrong

Isnt that just the probability that the first pair drawn are English?

 

[4EverStudent]

Combinatorial Approach

Yes, I agree with you Muppet06.

Time to brush off the old combinatorics skills .....!

The number of distinct draws is (assuming that 2 draws are the same if the order of pairings and order within a pairing are unimportant):

8!/(4!*(2^4)) = 105

The number of distinct draws where no 2 English teams meet is:

4! = 24

The number of distinct draws where all 4 English teams meet is:

3*3 = 9

Thus, the number of distinct draws where exactly 2 English teams meet is:

105 - 24 - 9 = 72

So, the probability of a given draw containing one, and one only, English team pairing is:

72/105

Thus, given any draw the probability of a particular pairing being an all English tie is:

(24/105)*0 + (9/105)*(1/2) + (72/105)*(1/4)

(I think! I am more than happy to be corrected)

 

[MarkC]

I did something slightly different, and got a different answer...

I think you can find this probability by considering the equivalent (and far less interesting!) problem of filling in the eight spaces

< _ _ | _ _ | _ _ | _ _ >​

with four E's and four X's. An E represents an English team, and an X represents a non-English team. The vertical lines are just there to make it obvious how the eight teams are paired up after the draw.

How many distinct draws are there? 8C4 = 8! ÷ (4! x 4!) = 70, because we're essentially choosing which four spaces the E's should go in, and with this done the four X's then fall into place.

In how many of these draws is there precisely one {EE} pair? 4 x 6 x 4 ÷ 2 = 48, because:

• We've got four pairs of spaces to fill in. There are 4 ways of choosing which of them should be the {EE};
• With this done, there are then 6 possible positions for the third E (any of the spaces in the three remaining empty pairs);
• With this done, there are then only 4 possible positions for the fourth E (any of the spaces in the two remaining empty pairs);
• But in the last two steps we have double-counted all the possibilities (because we have not distinguished between, say, putting E's in space #2 followed by space #7, and putting E's in space #7 followed by space #2). So we need to divide by 2.

Since each possible draw is (assumed to be) equally likely, the probability of there being precisely one {EE} pair is 48/70 = 68.6% (ish).

 

[shyguy]

Thanks all

Hello,
Thank you for all your attempts to explain this.
My question was driven by the statement in the press (or was it SKYSPORTS NEWS?) that the chances of at least two English teams meeting in the quarter finals of the Champion League was over half. I liked the use of numbers of permtuatons of identical objects (if my A level maths terminology is correct - it was so long ago).

As explained previously there are 2^4 = 16 ways of 4 pairings of E v X for the last 8 and 4*3*2*2 = 48 for one EE, one XX and 2 EX pairings and finally 4!/(2!2!) = 6 for the parings EE, EE, XX, XX. 16+48+6=70 which gives me confidence in the last approach.

Thank you all and hope you got the exams you wanted.