Below is a detailed description of the game:
The player starts with an initial balance of zero. On each turn, the player fips a fair coin - if it landson heads (H). the pot balance increases by 1 dollar, if it lands on tails (T), the potbalance decreases by 2 dollars. The game ends when the player flips three heads in arow (HHH), at which point an ertra 8 dollars is added to the balance. At the end of the game, the player gets to take their pot home (if the balance is positive) or is obligated to make up for the shortfall (if the balance is negative). The game cannot be terminated early by the player. On purely mathematical grounds, what is the maximum amount you would pay to enter this game?
What do you think is the solution? In theory we need $2^{3+1}-2=14$ flips to get HHH, which terminates the game. With an expected flip of 14, 7 of them is expected to be H, the rest 7 expected to be T. Is the payoff equal to $7*1+7*(-2)+8=1$?
Let $X_1,\ldots$ denote iid random variables modeling the outcomes of the tosses. Let $S$ model the total payoff.
Conditioning on the first toss, $E[S] = \frac 12 E[S|X_1=T] + \frac 12E[S|X_1=H]$. Given that $X_1$ is Tails, the same game starts over, but with initial balance $-2$ thus $E[S|X_1=T] = E[S-2] = E[S]-2$. Consequently, $$E[S] = \frac 12 (E[S]-2) + \frac 12E[S|X_1=H].$$
Further conditioning on the second toss, $$E[S|X_1=H] = \frac 12 E[S|X_1=H, X_2=T] + \frac 12E[S|X_1=H, X_2=H],$$ and for similar reasons we obtain $$E[S|X_1=H] = \frac 12 E[S+1-2] + \frac 12E[S|X_1=H, X_2=H].$$
Conditioning on the third toss, $$\begin{aligned} E[S|X_1=H, X_2=H] &= \frac 12 E[S|X_1=H, X_2=H, X_3=T] + \frac 12E[S|X_1=H, X_2=H, X_3=H] \\&= \frac 12 E[S+1+1-2] + \frac 12 (1+1+1+8). \end{aligned} $$
Plugging everything back, we obtain $E[S] = \frac{7E[S]+1}8$, thus $E[S]=1$.