How to calculate the expected return in this coin toss game? In a coin tossing game that is made of two rounds, there are two cases.
In first round, if heads comes up, case 1 will be played in second round, if tails comes up, case 2 will be played in second round.
In case 1, two coins are tossed. If at least one of them is heads, player wins.
In case 2, two coins are tossed:
HH: Player wins,
HT and TH: Player loses
TT: Coin gets tossed again with the rules of case 2.
Now the probabilities of the first round are easy to calculate:
$$P(case_1) = P(case_1) = \frac{1}{2}$$
Probabilities of each individual cases are also easy:
For case 1:
$$P(win) = \frac{3}{4}$$
For case 2:
$$P(win) = \frac{1}{3} $$
Winning makes a profit of 1 dollar, and losing makes a loss of 1 dollar. What is the expected return of the game?

If we land on case 1 in round two, our game is guaranteed to end with either a win or a loss. The probability of a win is simply $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}.$
However, if we land on case 2 in round two, our game could last for an infinite amount of time. To solve for the probability of a win, we sum the infinite geometric series $\frac{1}{2} \cdot \left((\frac{1}{4})^{1} + (\frac{1}{4})^{2} + (\frac{1}{4})^{3} + ... \right) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.$
Our total win probability is $\frac{3}{8} + \frac{1}{6} = \frac{13}{24}.$
The probability of loss is just one minus the probability of a win. This is just $1 - \frac{13}{24} = \frac{11}{24}.$
Our expected return is $\frac{13}{24} \cdot 1 - \frac{11}{24} \cdot 1 = \boxed{+\frac{1}{12}}.$
