Extra Credit swinging the probabilities, or ... Hey, how'd the Catalan numbers show up here???

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 $N$?

Applying the same logic used in the Classic Fiddler problem, we want to first find $p_{1,N} = \mathbb{P} \{ \text{win best of (2N-1) series} \mid \text{win game 1} \},$ from which we get the probability swing of game 1 in a best of $(2N-1)$ series as $\Delta_N = 2p_{1,N} - 1.$ Again following in the Classic Fiddler's solution's footsteps, if you win games $1$ and $k$, then there are $\binom{k-2}{N-2}$ ways of arranging another $N-2$ wins in the other $k-2$ games, so $$p_{1,k,N} = \mathbb{P} \{ \text{winning a best of $(2N-1)$ series in $k$ games} \mid \text{win game 1} \} = \binom{k-2}{N-2} \frac{1}{2^{k-1}}.$$ Summing over all possible values of $k = N, N+1, \dots, 2N-1,$ we get $$p_{1,N} = \sum_{k=N}^{2N-1} \binom{k-2}{N-2} \frac{1}{2^{k-1}}.$$ We could try to go further and define some generating function f_N, but this would lead to some escalating number of derivatives, that gets messy fast.

Instead let's set up a recursive formula. We note that $p_{1,1} = 1,$ which makes sense since it is a winner-takes-all one game playoff. For some $N \geq 1,$ let's take a look at $$p_{N+1} = \sum_{k=N+1}^{2N+1} \binom{k-2}{N-1} \frac{1}{2^{k-1}}.$$ The standard binomial coefficient recursion formula (which comes from the Pascal triangle) gives $$\binom{k-2}{N-1} = \binom{k-3}{N-1} + \binom{k-3}{N-2},$$ so we have \begin{align*} p_{N+1} & = \sum_{k=N+1}^{2N+1} \left(\binom{k-3}{N-1} + \binom{k-3}{N-2} \right) \frac{1}{2^{k-1}} \\ &= \left( \sum_{k=N}^{2N} \binom{k-2}{N-1} \frac{1}{2^k} \right) + \left( \sum_{k=N}^{2N} \binom{k-2}{N-2} \frac{1}{2^k} \right) \\ &= \frac{1}{2} \left( \sum_{k=N+1}^{2N} \binom{k-2}{N-1} \frac{1}{2^{k-1}} \right) + \frac{1}{2} \left( \sum_{k=N}^{2N-1} \binom{k-2}{N-2} \frac{1}{2^{k-1}} \right) + \binom{2N-2}{N-2} \frac{1}{2^{2N}} \\ &= \frac{1}{2} p_{1,N+1} - \binom{2N-1}{N-1} \frac{1}{2^{2N+1}} + \frac{1}{2} p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N}}.\end{align*} Pulling the copy of $\frac{1}{2}p_{1,N+1}$ back onto the lefthand side and then multiplying by 2, we get the recursion formula \begin{align*} p_{1,N+1} &= p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N-1}} - \binom{2N-1}{N-1} \frac{1}{2^{2N}} \\ &= p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N-1}} - \left( \binom{2N-2}{N-1} + \binom{2N-2}{N-2} \right) \frac{1}{2^{2N}} \\ &= p_{1,N} - \frac{1}{4^N} \left( \binom{2N-2}{N-1} - \binom{2N-2}{N-2} \right) \\ &= p_{1,N} - \frac{1}{4^N} C_{N-1}, \end{align*} where $C_n = \frac{1}{n+1} \binom{2n}{n} = \binom{2n}{n} - \binom{2n}{n+1},$ for $n \in \mathbb{N}$ is the standard Catalan number.

Since we start with $p_{1,1} = 1,$ we then see that $$p_{1,N} = 1 - \sum_{k=1}^{N-1} \frac{C_{k-1}}{4^k} = 1 - \frac{1}{4} \sum_{k=0}^{N-2} \frac{C_k}{4^k}, \,\, \forall N \in \mathbb{N}.$$ We can rely on the fact that the generation function of the Catalan numbers is $$c(x) = \sum_{n=0}^\infty C_n x^n = \frac{1 - \sqrt{1-4x}}{2x},$$ so that $$\frac{1}{4} \sum_{k=0}^\infty \frac{C_k}{4^k} = \frac{1}{4} c(\frac{1}{4}) = \frac{1}{4} \frac{1 - \sqrt{1 - 4 \cdot \frac{1}{4}}}{2 \cdot \frac{1}{4}} = \frac{1}{2}.$$ Therefore, we see that $$p_{1,N} = 1 - \frac{1}{4} \sum_{k=0}^{N-2} \frac{C_k}{4^k} = 1 - \frac{1}{4} \sum_{k=0}^\infty \frac{C_k}{4^k} + \frac{1}{4}\sum_{k=N-1}^\infty \frac{C_k}{4^k} = \frac{1}{2} + \frac{1}{4} \sum_{k=N-1}^\infty \frac{C_k}{4^k},$$ for all $N \in \mathbb{N}.$ Now when $k$ is large we have $$C_k \sim \frac{4^k}{k^{3/2} \sqrt{\pi}},$$ from repeated applications of Stirling's approximation, so when $N$ is sufficiently large we have $$\frac{1}{4} \sum_{k=N-1}^\infty \frac{C_k}{4^k} \approx \frac{1}{4\sqrt{\pi}} \sum_{k=N-1}^\infty k^{-3/2} \approx \frac{1}{2\sqrt{\pi (N-1)}},$$ where the last approximation is due to the fact that $\sum_{k=N}^\infty k^{-p} \sim \int_N^\infty x^{-p} \,dx.$ Therefore, in a fairly concise way, we have $$p_{1,N} \approx \frac{1}{2} + \frac{1}{2\sqrt{\pi(N-1)}},$$ when $N$ is large, so the probability swing of winning the first game is $$\Delta_N = 2p_{1,N} -1 \approx \frac{1}{\sqrt{\pi(N-1)}}$$ when $N$ is large.

Swinging the probabilities

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?

Let's break this down as follows. Let $p_1 = \mathbb{P} \{ \text{win series} \mid \text{win game 1} \}.$ In order to win the series, you must win it in $k$ games for some $k=4, 5, 6, 7,$ so let's further let $$p_{1,k} = \mathbb{P} \{ \text{win series in $k$ games} \mid \text{win game 1} \},$$ where here we see that $p_1 = \sum_{k=4}^7 p_{1,k}.$ Now, the total number of ways to win the first and $k$th games and two others somewhere in games $2$ through $k-1$ is given by $\binom{k-2}{2}$ and the overall probability of any particular combination of $k$ games is $\frac{1}{2^k},$ so $$p_{1,k} = \frac{\mathbb{P} \{ \text{win series in $k$ games and win game $1$} \}}{\mathbb{P} \{ \text{win game 1 } \}} = \binom{k-2}{2} \frac{1}{2^{k-1}}.$$ Therfore, $$p_1 = \sum_{k=4}^7 \binom{k-2}{2} \frac{1}{2^{k-1}} = \frac{1}{2} \sum_{k=4}^7 \binom{k-2}{2} \frac{1}{2^{k-2}} = \frac{1}{2} \sum_{k=2}^5 \binom{k}{2} \frac{1}{2^k}.$$

Now one way of computing $p_1$ would be using some generating function wizardry. Define the function $f(x) = \sum_{k=2}^5 \binom{k}{2} x^k,$ in which case, $p_1 = \frac{1}{2} f(\frac{1}{2}).$ Now we also see that \begin{align*} f(x) &= \frac{1}{2} x^2 \sum_{k=2}^5 k(k-1) x^{k-2} \\ &= \frac{1}{2} x^2 \frac{d^2}{dx^2} \left( \frac{1-x^6}{1-x} \right) \\ &= \frac{1}{2} x^2 \frac{d}{dx} \left( \frac{1-6x^5+5x^6}{(1-x)^2} \right) \\ &= \frac{1}{2} x^2 \frac{2(1-15x^4+24x^5-10x^6}{(1-x)^3} \\ &= \frac{ x^2 ( 1 - 15 x^4 + 24x^5 -10x^6) }{(1-x)^3}.\end{align*} So we have $$p_1 = \frac{1}{2} f(\frac{1}{2}) = \frac{1}{2} \frac{ \frac{1}{4} \left( 1 - \frac{15}{16} + \frac{24}{32} - \frac{10}{64} \right) }{ \frac{1}{8} } = \frac{ 42}{64} = \frac{21}{32}.$$

Now from symmetry, we see that the probability of you winning having lost the first game, let's say $q_1 = \mathbb{P} \{ \text{win series} \mid \text{lose game 1} \}$ is the same as the probability of you winning the series having lost the series having won the first game. That is $q_1 = 1 - p_1.$ So the proabbility swing of the first game is $$\Delta = p_1 - q_1 = p_1 - (1- p_1) = 2p_1 - 1 = 2 \frac{21}{32} - 1 = \frac{5}{16} = 32.125\%.$$