Question

your friend offers you a choice of 111 of 222 coin games to play with him. game 111: he proposes that you will flip 444 coins. if you flip 111 or fewer heads he will pay you \$5$5dollar sign, 5; otherwise you will pay him \$2$2dollar sign, 2. game 222: he proposes that you will flip 666 coins. if you flip all 666 heads he will pay you \$150$150dollar sign, 150; otherwise you will pay him \$2$2dollar sign, 2. what is your expected return from game 111? round your answer to the nearest cent. \$$dollar sign what is your expected return from game 222? round your answer to the nearest cent. \$$dollar sign your friend says he is willing to play both games a combined total of 303030 times. if you want to maximize your expected return, what should you do?

Answer 1

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.

Answer 2

To calculate the expected return from game 111, use the binomial probability formula for 111 coin flips. The same method can be used to calculate the expected return from game 222. Compare the expected returns from both games to maximize your expected return.

Step-by-step explanation:

The expected return from game 111 can be calculated by multiplying the probability of winning by the amount gained and subtracting the probability of losing multiplied by the amount lost. In this game, there are 111 coin flips and you win if you get 111 or fewer heads. The probability of getting exactly 111 heads can be calculated using the binomial probability formula.

The expected return from game 222 can be calculated in the same way. In this game, there are 666 coin flips and you win if you get all 666 heads. Again, the probability of getting exactly 666 heads can be calculated using the binomial probability formula.

To maximize your expected return, you should compare the expected returns from both games and choose the one with the higher expected return.