How Much Does Game 1 Matter?

Fiddler

You and your opponent are beginning a best-of-seven series, meaning the first team to win four games wins the series. Both teams are evenly matched, meaning each team has a 50 percent chance of winning each game, independent of the outcomes of previous games.

As the team manager, you are trying to motivate your team as to the criticality of the first game in the series (i.e., “Game 1”). You’d specifically like to educate them regarding the “probability swing” coming out of Game 1—that is, the probability of winning the series if they win Game 1 minus the probability of winning the series if they lose Game 1. (For example, the probability swing for a winner-take-all Game 7 is 100 percent.)

What is the probability swing for Game 1?

Solution

$N$ wins are necessary for a best-of-$(2N-1)$ series. Examining the two groups,

1. Let $W_w(N)$ represent winning the series given Game 1 is a win:
   • The last game is also a win.
   • $N-2$ additional wins are necessary between the first and last game.
   • $0 \le L \le N-1$ losses are possible between the first and last game.
   • The total number of games is $N+L$.
   • The probability is $2^{N+L-1}$.


2. Let $L_w(N)$ represent winning the series given Game 1 is a loss:
   • The last game is a win.
   • $N-1$ additional wins are necessary between the first and last game.
   • $0 \le L \le N-2$ additional losses are possible between the first and last game.
   • The total number of games is $N+1+L$.
   • The probability is $2^{N+L}$.

Let $P_s(N)$ represent probability swing for a best-of-$(2N-1)$.

\begin{align*} P_s(N) &= W_w(N) - L_w(N) \\ &= \boxed{\sum\limits_{L=0}^{N-1} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{N-2} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}}} \\ P_s(4) &= \sum\limits_{L=0}^{3} \frac{(2+L)!}{2!L!} \cdot \frac{1}{2^{3+L}} - \sum\limits_{L=0}^{2} \frac{(3+L)!}{3!L!} \cdot \frac{1}{2^{4+L}} \\ &= \left(\frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64}\right) - \left(\frac{1}{16} + \frac{4}{32} + \frac{10}{64} \right) \end{align*}

Answer

$$\boxed{P_s(4) = \left(\frac{1}{8} + \frac{2}{16} + \frac{2}{32}\right) = \frac{5}{16}}$$

Extra Credit

Instead of a best-of-seven series, now suppose the series is much, much longer. In particular, the first team to win N games wins the series, so technically this is a best-of-(2N−1) series, where N is some very, very large number.

In the limit of large N, what is the probability swing for Game 1 in terms of N? (For full credit, I’m expecting an answer that is rather concise!)

Solution

I will use induction to show

$$\sum\limits_{L=0}^{M} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{M-1} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}} = \boxed{\frac{N-1+M \choose M}{2^{N-1+M}}}$$

M = 1

\begin{align*} &\sum\limits_{L=0}^{1} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{0} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}} \\ =& \left[\frac{(N-2+0)!}{(N-2)!0!} \cdot \frac{1}{2^{N+0-1}}\right] + \left[\frac{(N-2+1)!}{(N-2)!1!} \cdot \frac{1}{2^{N+1-1}}\right] - \left[\frac{(N-1+0)!}{(N-1)!0!} \cdot \frac{1}{2^{N+0}}\right] \\ =& \left[\frac{(N-2)!}{(N-2)!} \cdot \frac{1}{2^{N-1}}\right] + \left[\frac{(N-1)!}{(N-2)!} \cdot \frac{1}{2^{N}}\right] - \left[\frac{(N-1)!}{(N-1)!} \cdot \frac{1}{2^{N}}\right] \\ =& \left[\frac{2}{2^{N}}\right] + \left[\frac{N-1}{2^{N}}\right] - \left[\frac{1}{2^{N}}\right] \\ =& \frac{2+N-1-1}{2^{N}} = \boxed{\frac{N-1+1 \choose 1}{2^{N-1+1}}} \\ \end{align*}

M = K-1

Assume

$$\sum\limits_{L=0}^{K-1} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{K-2} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}} = \boxed{\frac{N-1+K-1 \choose K-1}{2^{N-1+K-1}}}$$

M = K

\begin{align*} & \sum\limits_{L=0}^{K} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{K-1} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}} \\ \end{align*}

Separate the $M$ and $M-1$ terms from the respective sums,

\begin{align*} =& \left[\frac{(N-2+K)!}{(N-2)!K!} \cdot \frac{1}{2^{N+K-1}} + \sum\limits_{L=0}^{K-1} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}}\right] - \left[\frac{(N-1+K-1)!}{(N-1)!(K-1)!} \cdot \frac{1}{2^{N+K-1}} + \sum\limits_{L=0}^{K-2} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}}\right] \\ \end{align*}

Combine the remaining sums identical to expression from $M=K-1$, combine the separated $M$ and $M-1$ terms,

\begin{align*} =& \left[\sum\limits_{L=0}^{K-1} \frac{(N-2+L)!}{(N-2)!L!} \cdot \frac{1}{2^{N+L-1}} - \sum\limits_{L=0}^{K-2} \frac{(N-1+L)!}{(N-1)!L!} \cdot \frac{1}{2^{N+L}}\right] + \left[\frac{(N-2+K)!}{(N-2)!K!} \cdot \frac{1}{2^{N+K-1}} -\frac{(N-1+K-1)!}{(N-1)!(K-1)!} \cdot \frac{1}{2^{N+K-1}}\right] \\ =& \left[\frac{N-1+K-1 \choose K-1}{2^{N-1+K-1}}\right] + \frac{1}{2^{N+K-1}} \cdot \left[\frac{(N+K-2)!}{(N-2)!K!} - \frac{(N+K-2)!}{(N-1)!(K-1)!}\right] \\ =& \frac{1}{2^{N+K-1}} \cdot \left[2\cdot\frac{(N+K-2)!}{(N-1)!(K-1)!} + \frac{(N+K-2)!}{(N-2)!K!} - \frac{(N+K-2)!}{(N-1)!(K-1)!}\right] \\ =& \frac{1}{2^{N+K-1}} \cdot \left[\frac{(N+K-2)!}{(N-1)!(K-1)!} + \frac{(N+K-2)!}{(N-2)!K!}\right] \\ \end{align*}

Combine using Pascal's Formula,

\begin{align*} =& \frac{1}{2^{N+K-1}} \cdot \left[\frac{(N+K-1)!}{(N-1)!(K)!}\right] \\ =& \frac{N+K-1 \choose K}{2^{N+K-1}} = \boxed{\frac{N-1+K \choose K}{2^{N-1+K}}} \\ \end{align*}

Answer

Thus,

$$\boxed{P_s(N) = \frac{2N-2 \choose N-1}{2^{2N-2}}}$$

The swing probability is not less than 1% until a best-of-6,367 series.

Rohan Lewis

2025.11.03

Code can be found here.