Can You Play the Favorite?

In a four-team region seeded $1$ through $4$, when an $M$-seed plays an $N$-seed the $M$-seed wins with probability $\tfrac{N}{M+N}$. The bracket is $1$ vs $4$ and $2$ vs $3$, and the winners meet. What is the probability the $1$-seed wins the region?

The Fiddler, Zach Wissner-Gross, March 21, 2025(original post)

Solution

The $1$-seed beats the $4$-seed with probability $\tfrac45$, then meets the $2$-seed (who arrives with probability $\tfrac35$) or the $3$-seed (probability $\tfrac25$): $$P=\frac45\left(\frac35\cdot\frac23+\frac25\cdot\frac34\right) =\frac45\cdot\frac{7}{10}=\frac{14}{25}\approx\boxed{56\%}.$$

The computation

Encode the bracket: the $1$-seed must beat its first opponent, then beat whichever of the $2$- and $3$-seeds survives, each path weighted by the model’s win probabilities.

from fractions import Fraction as F
beat = lambda m, n: F(n, m + n)             # P(m-seed beats n-seed)
print(beat(1, 4) * (beat(2, 3)*beat(1, 2) + beat(3, 2)*beat(1, 3)))   # 14/25

Extra Credit

For a region of $2^k$ teams under standard seeding and the same model, what is the $1$-seed’s probability of winning?

Solution

Folding the bracket up recursively (each subregion produces a distribution over its possible winners, then the two halves meet) gives no tidy closed form; the probability drifts slowly toward $\tfrac12$: $$\begin{gathered} k=1:\ \tfrac23,\qquad k=2:\ \tfrac{14}{25},\\[2pt] k=3:\ \tfrac{225148}{426465}\approx0.528,\qquad k=4:\ \approx\boxed{0.519}. \end{gathered}$$ (The general value is paywalled; these are my own.)

The computation

Build each subregion’s full distribution over who emerges, then collide the two halves under the win model; standard seeding orders the bracket so favourites meet late.

from fractions import Fraction as F
def region(seeds):
    if len(seeds) == 1: return {seeds[0]: F(1)}
    h = len(seeds) // 2; L, R = region(seeds[:h]), region(seeds[h:]); out = {}
    for a, pa in L.items():
        for b, pb in R.items():
            out[a] = out.get(a, F(0)) + pa*pb*F(b, a + b)
            out[b] = out.get(b, F(0)) + pa*pb*F(a, a + b)
    return out
def order(k):
    o = [1, 2]
    for r in range(1, k):
        m = 2**(r + 1) + 1; o = [x for pr in zip(o, [m - x for x in o]) for x in pr]
    return o
for k in (3, 4): print(k, float(region(order(k))[1]))   # 0.5279, 0.5192