NCAA quota bracket

Question: It feels like there’s more parity in college basketball’s March Madness than ever, with lower-seeded teams advancing further in the tournaments at the expense of the favorites. This year’s Sweet $16$ on the men’s side consists of two $1$ seeds, two $2$ seeds, two $3$ seeds, two $4$ seeds, a $5$ seed, a $6$ seed, a $7$ seed, an $8$ seed, a $9$ seed and a $15$ seed. This got Jeremy wondering about the likelihood that the Sweet 16 consists of exactly one of each seed: one $1$ seed, one $2$ seed, etc., up to one $16$ seed.

Suppose each team is equally likely to win any given game. What are the chances that the Sweet $16$ does indeed consist of one of each seed?

Extra credit: Looking at historical data on the men’s side, Jeremy estimates that the probability that seed A will defeat seed B is $\frac12 + 0.033 (B−A).$ Using these probabilities, what are the chances that the Sweet $16$ consists of one of each seed?

(FiveThirtyEight)

Solution

In the NCAA tournament, there are $4$ equivalent subtournaments (crudely, four geographic regions of the US), and each will send $4$ teams to the “sweet sixteen”.

If a team makes it to the sweet sixteen, it means that they won their sub-bracket of $4$ teams.

For example, a $1$-seed’s sub-bracket has them play a $16$-seed, followed by the winner of the $8$ and $9$-seed.

So, if we call $P(i,j)$ the probability of team $i$ beating team $j,$ the probability of any given $1$-seed making the sweet sixteen is

Likewise, for a general sub-bracket

the probability of team $i$ making the sweet sixteen is

We’re interested in the case where one of each kind of seed makes the sweet sixteen (referred to from here on as a “quota bracket”), so the probability of any given realization is just

Since we have to pick one of each kind of seed, there are $4!$ ways to choose seeds $\{1,16,8,9\}$ and likewise for the other sub-brackets.

This makes $4!^4$ possible ways to select seeds $1$ throught $16$ and, so, the overall probability is

When the teams are evenly matched, this is just $4!^4/4^{16},$ or $\approx 0.0077\%$ of the time.

When they’re not evenly matched, we need to set up some accounting to handle all the substitutions. Now the probability that seed $i$ beats seed $j$ is

First, we need to represent the structure of the sub-brackets:

tournament = {
               {{1,16}, {9, 8}}
             , {{13, 4}, {12, 5}}
             , {{10, 7}, {15, 2}}
             , {{14, 3}, {11, 6}}
             }

Next, we need a function that takes a sub-bracket in the form $\{\{i,j\},\{m,n\}\}$ and calculates the probability of team $i$ winning:

p[{a_, b_}, f_] := 1/2 + f (b - a);

firstWinsBracket[b_, f_] :=
   p[b[[1]], f] *
    (p[{b[[1]][[1]], b[[2]][[1]]}, f] * p[b[[2]], f]
      + p[{b[[1]][[1]], b[[2]][[2]]}, f] * p[Reverse@b[[2]], f])

Next, we need a function that takes a sub-bracket and returns the four variants $\{\{i,j\},\{m,n\}\}, \{\{j,i\},\{m,n\}\}, \{\{m,n\},\{i,j\}\}$ and $\{\{n,m\},\{i,j\}\},$ corresponding to each team winning the sub-bracket:

makeWinners[b_] := 
      { 
        b
      , {Reverse@b[[1]], b[[2]]}
      , {b[[2]], b[[1]]}
      , {Reverse@b[[2]], b[[1]]}
      }

Finally, we need to form all $16$ sub-brackets, calculate the win probabilities, multiply them together, and scale by the number of ways to distribute them across the $4$ geographic sub-tournaments:

brackets = Flatten[makeWinners[#] & /@ tournament, 1];

pQuotaBracket[f_] := (
  winProbs = firstWinsBracket[#, f] & /@ brackets;
  pOverall = (4!)^4 Fold[Times, winProbs];
  Return[pOverall]
  )

As the advantage to the better team grows (to a maximum of $f = 1/30$), the likelihood of a quota bracket drops to zero:

At the empirical seed advantage, $f = 0.033,$ the probability of a quota bracket stands at a mere $8.53\times10^{-10}.$