Final answer:
Without the exact probabilities of winning, we cannot determine the precise expected returns for game 111. Game 222's expected return is likely to be negative due to the very low probability of winning, suggesting game 111 might be a better choice. To maximize returns, one should focus on the game with the higher expected return.
Step-by-step explanation:
To calculate the expected return from game 111, where you flip 444 coins and win $5 if 111 or fewer are heads (P(win)) and lose $2 if more than 111 are heads (P(lose)), we use the formula for expected value (EV): EV = (P(win) × amount won) + (P(lose) × amount lost). However, without the probabilities given, we cannot compute the exact expected return. Nevertheless, generally, the calculation would involve using the binomial probability distribution to find P(win) and P(lose).
As for the expected return from game 222 which requires flipping 666 coins and winning $150 only if all 666 are heads, the chance of that occurring is incredibly low (0.5^666), making this game highly unfavorable. Since you only win if all coins come up heads, the EV is approximately: EV = (1 × $150) × (0.5^666) + (-$2) × (1 - (0.5^666)), which is almost guaranteed to be negative.
The overall strategy to maximize expected return when playing both games a combined total of 30 times depends on the specific probabilities and payouts of each game. Ideally, we would compare the expected returns of both games and play the one with the higher expected return more frequently. However, based on the information provided, it seems game 111 may have a better chance at offering a positive expected return than the extremely unlikely game 222. Good analysis must include consideration of probabilities and outcomes.