The probability is 8/14 - that's better than one in two and so not really much of a coincidence.

Here is why:

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

and so the probability of no two English teams being drawn together is one minus that number, which is 8/14.

Norman Fenton

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