Legendary Math Warrior

Let $P$ be the probability of winning a game.

Expected star gain per game in the first two games:

$1\times P(win)+(-1)\times P(lose)$

$=P-(1-P)$

$=2P-1$

Expected star gain per game from the third game onwards:

$1\times P(win)+(-1)\times P(lose)+1\times (P(win))^3$

$=P-(1-P)+P^3$

$=P^3+2P-1$

Expected star gain for $N$ games

$2\times(2P-1)+(N-2)(P^3+2P-1)$

Expected star gain when $N=9722$ and $P=\frac{1}{2}$ :

$2\times(2\times\frac{1}{2}-1)+(9722-2)\times((\frac{1}{2})^3+2\times\frac{1}{2}-1)$

$=\textbf{1215}$

For the first two cases, you have (1/2)(1) + (1/2)(-1) for each game, bevause you have a 50% chance of losing a point and a 50% chance of gaining a point. This turns out to be 0.

For the cases from Game 3 - Game 9722, they have a chance of being a 3rd or more consecutive win.

The probability that the last two games were wins is (1/2)(1/2) = 1/4. Since you get 2 points from a 3rd or more consecutive win, you get (1/4)(1/2)(2) + (1/4)(1/2)(-1) + (3/4)(1/2)(1) + (3/4)(1/2)(-1), which becomes 1/8.

Multiple (9722-3+1) by 1/8, and get 1215.

This is called Linearity of Expectation.