UCL quarterfinal draw - English odds

The sun rises in the east, and all four English clubs make the Champions League quarterfinals. Now that we're here, what are the odds of none of the four facing each other in the quarters? What are the odds of all four being placed in the same half of the bracket? I'm sure I could find this on the web somewhere; I do my best thinking in the shower, however, so the numbers below are brought to you by Johnson & Johnson shampoo and Dial soap. Let's dive in...

I'm mainly going to consider how to group 8 teams into 4 pairings, without regard for home and away legs or position of the pairings in the bracket. There are 8! ways to arrange 8 objects in order, except we don't care about the 4! ways in which the 4 pairs can be arranged, and we also don't care about the 2! ways in which each of the 4 pairs can be ordered. This can be expressed as

8!
--------------------
4! * (2!2!2!2!)

which equals 105. So there are 105 possible pairings for the quarterfinals. (Note that I've ignored the position of the pairings in the bracket, and I'll continue to do so for the most part.)

Starting over, if we only consider each club as English or not, there are three cases for possible pairings:

1. EE EE xx xx
2. EE Ex Ex xx
3. Ex Ex Ex Ex

Case 1 has each English club facing another English club. Within this case, each English club may face any of 3 opponents, although once that selection is made, the other all-English matchup is determined. Likewise, there are 3 possible matchups for the 4 non-English clubs. There are 3 * 3 = 9 pairings with two all-English matchups. Please note that if Case 1 occurs, there is only a 1-in-3 chance that both all-English pairings will be placed on the same side of the bracket.

Case 2 has exactly one all-English matchup. Selecting this matchup requires choosing 2 objects from 4, and there are 6 possibilities. There is also exactly one non-English matchup, for which there are also 6 possibilities. We then know the clubs for the remaining two games, although there are 2 possibilities for selecting which English club will play which non-English club. There are 6 * 6 * 2 = 72 pairings with exactly one all-English matchup. Please note that if Case 2 occurs, there is a 2-in-3 chance that the all-English matchup and the non-English matchup will be on opposite sides of the bracket.

Case 3 has no all-English matchups, so that each pairing includes one English club and one non-English club. For each English club, we need to select the non-English club it will play, which is simply arranging 4 objects in order. There are 4! = 24 pairings with no all-English matchups.

Note that 9 + 72 + 24 = 105; two different methods, same result.

A couple of calculations to sum up:

• Chance of at least one all-English matchup in quarterfinals:
  81/105 = 77.1%
• Chance of all English clubs being placed in the same half of the bracket:
  9/105 * 1/3 = 2.9%
• Chance of Liverpool knocking out Chelsea:
  (no math required)