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?
Best Answer
Using the Law of Total Probability: $$\mathsf P(\mathrm{Win}) = \mathsf P(\mathrm{case}_1)\,\mathsf P(\mathrm{Win}\mid \mathrm{case}_1)+\mathsf P(\mathrm{case}_2)\,\mathsf P(\mathrm{Win}\mid \mathrm{case}_2)$$
You have evaluated $\mathsf P(\textsf{case}_1)$ and the two conditional probabilities, of a win given the case played. Put it together and evaluate the expected value of return.