How Much Does Game 1 Matter?

You and an evenly matched opponent begin a best-of-seven series: each game is an independent coin flip, and the first to four wins takes it. How much does Game $1$ swing your chances, that is, what is your probability of winning the series after winning Game $1$ minus your probability after losing it?

The Fiddler, Zach Wissner-Gross, October 31, 2025(original post)

Solution

Game $1$ changes the outcome of the series only when the other six games split evenly, three apiece; in every other case the series is already decided the same way regardless of Game $1$. So the swing equals the probability that the remaining six games are a $3$–$3$ tie: $$\frac{\binom{6}{3}}{2^{6}} = \frac{20}{64} = \frac{5}{16} = \boxed{31.25\%}.$$

The computation

Encode the series as a recursion: $\mathrm{win}(a,b)$ is your chance of taking it when you still need $a$ wins and the opponent needs $b$, each game a coin flip. The swing is your series-win probability after winning Game $1$ minus after losing it.

from functools import lru_cache
from fractions import Fraction as F
@lru_cache(None)
def win(a, b):                     # prob you win; you need a more wins, opp needs b
    if a == 0: return F(1)
    if b == 0: return F(0)
    return (win(a-1, b) + win(a, b-1)) / 2
print(win(3, 4) - win(4, 3))       # 5/16   (after winning vs losing Game 1)

Extra Credit

For a much longer series, first to $N$ wins (a best-of-$(2N-1)$), what is the Game $1$ swing for large $N$?

Solution

The same argument holds: Game $1$ decides the series exactly when the other $2N-2$ games split $N-1$ each, so the swing is $$\frac{\binom{2N-2}{\,N-1\,}}{2^{\,2N-2}} = \frac{\binom{2N-2}{N-1}}{4^{\,N-1}}.$$ By the central binomial estimate $\binom{2m}{m}\sim 4^{m}/\sqrt{\pi m}$ with $m=N-1$, $$\text{swing} \sim \frac{1}{\sqrt{\pi (N-1)}} \approx \boxed{\frac{1}{\sqrt{\pi N}}}.$$ A single game’s influence fades only like $1/\sqrt N$: at $N=1000$ the swing is still about $1.78\%$.

The computation

The swing is exactly the probability the other $2N-2$ coin flips tie $N-1$ apiece, a central-binomial probability; compute it for growing $N$ and watch it track $1/\sqrt{\pi N}$.

from math import comb, sqrt, pi
for N in (4, 100, 1000):
    print(N, comb(2*N-2, N-1)/4**(N-1), 1/sqrt(pi*N))