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).