Could You Have Won the Super Bowl?

The 2024 Super Bowl went to overtime, and San Francisco, having won the coin toss, chose to start on offence. Take a simplified model of football. Each time a team has the ball, it scores a touchdown ($7$ points) with probability $\tfrac13$, a field goal ($3$ points) with probability $\tfrac13$, and nothing with probability $\tfrac13$; then the other team has the ball. In overtime your team has the ball first. Whatever happens, the other team then gets one possession. If the game is still tied after that, the teams keep alternating possessions, and whichever team scores next wins at once. What is your team’s probability of winning?

The Fiddler, Zach Wissner-Gross, February 23, 2024(original post)

The official solution appeared in the post of March 1, 2024 and gives $7/12$, agreeing with the answer below. The Extra Credit half of that post is paywalled, so the value $9/16$ is my own.

Solution

Start at the end of the game. Once the scores are level after both opening possessions, the next score wins: this is sudden death. Let $s$ be the chance that the team with the ball wins sudden death. That team scores with probability $\tfrac23$ and wins on the spot. Otherwise, with probability $\tfrac13$, the other team gets the ball and is now the one with the ball in sudden death, so it wins with probability $s$ and the first team with probability $1 - s$. Hence $$s \;=\; \tfrac23 + \tfrac13 (1 - s), \qquad\text{so}\qquad s = \tfrac34 .$$ In every tied position after the opening exchange it is your team that has the ball (you had the first possession, they had the second), so a tie at that point is worth $\tfrac34$ to you.

Now condition on your opening possession, and then on the reply.

• You score a touchdown. They tie with a touchdown (probability $\tfrac13$), which sends the game to sudden death; anything else and you win. Your chance is $\tfrac23 + \tfrac13 \cdot \tfrac34 = \tfrac{11}{12}$.

• You kick a field goal. A touchdown beats you, a field goal ties, and nothing hands you the win: $\tfrac13 \cdot \tfrac34 + \tfrac13 = \tfrac{7}{12}$.

• You score nothing. Any score beats you, and only a second empty possession leads to sudden death: $\tfrac13 \cdot \tfrac34 = \tfrac14$.

The three openings are equally likely, so $$P(\text{win}) \;=\; \tfrac13\left(\tfrac{11}{12} + \tfrac{7}{12} + \tfrac{3}{12}\right) \;=\; \tfrac{21}{36} \;=\; \boxed{\ \tfrac{7}{12} \approx 58.3\% . \ }$$ Going first is worth having in this model, chiefly because the team that goes first also has the ball first in sudden death.

The computation

A Monte Carlo simulation plays overtime as described: two opening drives, then alternating drives until someone scores. Over two million games the team with the first possession wins about $58.3\%$ of the time.

import random

def overtime(rng):
    drive = lambda: rng.choice((7, 3, 0))
    lead = drive() - drive()         # your drive, then theirs
    yours = True                     # who has the ball next
    while lead == 0:
        if drive():
            lead = 1 if yours else -1
        yours = not yours
    return lead > 0

rng = random.Random(2024)
n = 2_000_000
wins = sum(overtime(rng) for _ in range(n))
print(f"{wins / n:.4f}   (7/12 = {7/12:.4f})")

It prints

0.5834   (7/12 = 0.5833)

Extra Credit

Now each team, on each possession, may choose its plan. The usual plan gives $7$, $3$ or $0$ points with probability $\tfrac13$ each. The aggressive plan gives $7$ or $0$ points with probability $\tfrac12$ each. With your team on offence first, and both teams choosing to maximise their own chance of winning, what is your team’s probability of winning?

Solution

Work backwards again, because each choice depends only on what can still happen.

Sudden death. Any score wins, so the team with the ball wants the highest chance of scoring at all: $\tfrac23$ with the usual plan against $\tfrac12$ with the aggressive one. The usual plan is better, and sudden death is worth $s = \tfrac34$ to the team with the ball, exactly as before.

Their reply to your opening. They choose the plan that is worst for you.

• You lead by $7$. A field goal is useless to them, so the usual plan gives you $1 - \tfrac13 \cdot \tfrac14 = \tfrac{11}{12}$, while the aggressive plan gives you $1 - \tfrac12 \cdot \tfrac14 = \tfrac78$. They go aggressive: $\tfrac78$.

• You lead by $3$. With the usual plan they win with a touchdown and tie with a field goal, leaving you $\tfrac13 + \tfrac13 \cdot \tfrac34 = \tfrac{7}{12}$. Aggressive, they win half the time and you win the other half: $\tfrac12$. They go aggressive: $\tfrac12$.

• Scores level at $0$. Any score wins for them. The usual plan scores more often, leaving you $\tfrac13 \cdot \tfrac34 = \tfrac14$, against $\tfrac12 \cdot \tfrac34 = \tfrac38$ for the aggressive plan. They play the usual plan: $\tfrac14$.

Your opening. With the usual plan your chance is the average over $7$, $3$ and $0$ points: $$\tfrac13\left(\tfrac78 + \tfrac12 + \tfrac14\right) = \tfrac{13}{24} .$$ With the aggressive plan the field goal disappears: $$\tfrac12\left(\tfrac78 + \tfrac14\right) = \tfrac{9}{16} .$$ Since $\tfrac{9}{16} = \tfrac{27}{48}$ beats $\tfrac{13}{24} = \tfrac{26}{48}$, you open aggressively. A field goal on the first drive is nearly worthless once the other side may answer by playing for a touchdown. $$\boxed{\ P(\text{win}) = \tfrac{9}{16} = 56.25\% . \ }$$ The extra choice costs the first team a little (down from $58.3\%$), because the second team uses it in its reply.

The computation

The program solves the game from its rules, with no use of the case analysis above. It describes each plan as a list of (points, probability) pairs, finds the value of sudden death by repeating the rule “the team with the ball picks the plan that is best for it” until the numbers settle, then lets the second team pick its worst-for-you reply to each opening and the first team pick its best opening. Run with only the usual plan, it must reproduce $7/12$.

from fractions import Fraction

USUAL = [(7, 1/3), (3, 1/3), (0, 1/3)]
BOLD = [(7, 1/2), (0, 1/2)]

def solve(plans):
    you, them = 0.5, 0.5     # your chance in sudden death
    for _ in range(200):     # with you / them on offence
        you, them = (
            max(sum(p * (1 if pts else them) for pts, p in pl)
                for pl in plans),
            min(sum(p * (0 if pts else you) for pts, p in pl)
                for pl in plans))
    def reply(lead):
        def end(b):
            return 1 if lead > b else 0 if lead < b else you
        return min(sum(p * end(b) for b, p in pl)
                   for pl in plans)
    return max(sum(p * reply(a) for a, p in pl) for pl in plans)

for plans in ([USUAL], [USUAL, BOLD]):
    v = solve(plans)
    print(f"{v:.6f}", Fraction(v).limit_denominator(100))

It prints

0.583333 7/12
0.562500 9/16