I asked:

> In the third round of the English FA Cup (football) there are 64 teams and
> therefore 32 ties. 20 of the 64 teams are from the Premiership. This year
> was an apparently rare event in that no Premiership teams met each other.
> What is the actual probability of this occurring?

I received a range of answers with a clear consensus of 0.012, i.e. about
once every 83 years. Correct respondents were Michael Meyners, Peter Lynn,
David Goda, John Wood, Timothy Mark, Andrew Rose, Giles Thomas, Claus Mayer,
Graeme MacLennan, Ken Rice, Allan White and Christopher Davison. The latter
wins a car.

The method of working varied between respondents from the practical:

The probability of the first premiership team (Arsenal, say) getting
non-premier opposition is 44/63. Given this draw, there are 61 teams that
Aston Villa could play, of which 43 are non-premier. Given these 2 draws,
there are 59 teams that Birmingham could play, of which 42 are non-premier,
etc Hence the probability is (44 x 43 x 42 x ... ...25)/(63 x 61 x 59 x
......25) = 0.0120675

or

We can think of each Prem team being drawn and ask if their opponent is
non-prem. For the first prem team, the prob of
non-prem opposition is 44/63. At this stage, 19 prem and 43 non-prem teams
remain. The prob that the second prem team gets non-prem opposition is
43/62. This continues until 1 prem and 25 non-prem teams remain. So the
required prob is (44/63)(43/62)...(25/25) which I reckon is about 0.0121 or
1/82.9.

to the more theoretical:

Total number of different ways of pairing teams = 64!/(32!x2^32) = a
Total number of pairings in which no 2 premiership teams meet =
44!/(12!x2^12) = b
Probability = b/a which is approx 0.012.

or

I worked out there are n!/(2^(n/2)*(n/2)!) number of ways to choose the n/2
ties from n teams. And I figure there are 44P20 ways of choosing 20
premiership teams to meet 20 of the 44 non-premiership teams. So altogether
the probability should be:

44P20*24!(2^12*12!)/(64!/2^32*32!) = 0.012

or

The probability will be given by N/D where D is the total possible number of
ways in which the 64 teams can be paired and N is the number of ways in
which they can be paired such that no premier team is paired with any other
premier team.
D is fairly straightforward and is simply 64!/(32!*2^32).
N is more complicated. My reasoning runs as follows. The number of ways of
constructing pairs 20 pairs of teams such that each pair contains exactly
one premier team and exactly one non-premier team is 44!/24!. This
"utilises" 20 non-premier teams. However, twelve pairs must be constructed
from the remaining 24 non-premier teams. By the same argument as was used
for D, this can be done in 24!/(12!*2^12) ways. N is obtained as the product
of these two numbers, i.e. 44!/(12!*2^12).
Therefore N/D is (44!32!*2^20)/(64!12!). This comes out as a probability of
approximately 0.012

Thank you for all your responses,

Oliver