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 $$\Delta = p_1 - q_1 = p_1 - (1- p_1) = 2p_1 - 1 = 2 \frac{21}{32} - 1 = \frac{5}{16} = 32.125\%.$$