Football League coincidence

Question

In the English Championship division, there are 24 teams, 8 of which have names starting with the letter B (e.g.Brighton) Tonight, all 24 teams in this division are playing each other. By a coincidence, the 8 teams starting with B are playing each other, i.e. 4 of the games involve these 8 teams (there are 2 teams per game!) What are the odds on this happening ?

Answer 1

We will assume that the pairings were done with all pairings equally likely. Line up the "B" teams in full alphabetical order. The probability that the first B team in the list got assigned a B team is $\frac{7}{23}$. Given that this happened, the probability that the first unassigned B team got assigned a B team is $\frac{5}{21}$. Continue in this way. The required probability is $\frac{7}{23}\cdot\frac{5}{21}\cdot\frac{3}{19}\cdot \frac{1}{17}$.

Translation of the answer to the language of odds, if desired, is mechanical.

Remark: One can also count in various ways. For example, imagine we list the teams at random, and pair $1$ and $2$, $3$ and $4$, and so on. There are $24!$ equally likely listings.

Now we count the "favourables," the listings that have all the B's paired. So we must choose $4$ of the odd numbers from the $12$ available to place one of the B's into. There are $binom{12}{4}$ ways to do this. Once this is done, the locations occupied by B's are determined, also the locations occupied by non-B's. The B's can be scrambled among B positions in $8!$ ways, and for each of these the rest can be scrambled in $16!$ ways. That gives probability $$\frac{\binom{12}{4}(8!)(16!)}{24!}.$$ After some cancellation, this turns out to be the same number as the one reached in a more straightforward way in the main answer.