1. **Problem statement:** We analyze a game where a coin is tossed repeatedly up to 5 times. Each time "Pile" (heads) appears, the payout doubles starting from 2. The game ends either when "Face" (tails) appears or after the fifth toss. 2. **Formula and rules:** The payout after $n$ consecutive "Pile" is given by $$2^n$$ where $n$ is the number of consecutive heads. 3. **Expected value calculation:** Let $X$ be the payout random variable. The game stops at the first "Face" or after 5 tosses. The probability of heads or tails is $\frac{1}{2}$ each. 4. **Calculate probabilities and payouts:** - $P(X=2) = P(\text{first toss is Face}) = \frac{1}{2}$ - $P(X=2^2) = P(\text{first toss Pile, second toss Face}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ - $P(X=2^3) = P(\text{first two tosses Pile, third toss Face}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ - $P(X=2^4) = P(\text{first three tosses Pile, fourth toss Face}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$ - $P(X=2^5) = P(\text{first four tosses Pile, fifth toss Face or end}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}$ 5. **Expected value formula:** $$E(X) = \sum_{n=1}^5 2^n \times P(X=2^n) = \sum_{n=1}^5 2^n \times \left(\frac{1}{2}\right)^n = \sum_{n=1}^5 1 = 5$$ 6. **Interpretation:** The expected payout of this game is 5 units. This means on average, you expect to win 5 units per game played under these rules.