Part 1 of 2 Two coins are tossed. You win \( \$ 5.00 \) if both coins are heads and \( \$ 6.00 \) if both coins are tails. You lose \( \$ 5.00 \) if the coins do not match. What is the 0 of 1 expected value of this game? Is this a fair game?

To calculate the expected value of this game, we first determine the possible outcomes when two coins are tossed. There are four equally likely outcomes: HH (both heads), HT (one head, one tail), TH (one tail, one head), and TT (both tails). - Winning \( \$ 5.00 \) for HH: Probability = \( \frac{1}{4} \) - Losing \( \$ 5.00 \) for HT and TH: Combined Probability = \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \) - Winning \( \$ 6.00 \) for TT: Probability = \( \frac{1}{4} \) Now, we calculate the expected value (EV) using the formula: \[ EV = (5 \times \frac{1}{4}) + (-5 \times \frac{1}{2}) + (6 \times \frac{1}{4}) \] Calculating each term gives: \[ EV = \frac{5}{4} + \left(-\frac{5}{2}\right) + \frac{6}{4} \] \[ EV = \frac{5}{4} - \frac{10}{4} + \frac{6}{4} \] \[ EV = \frac{1}{4} \] The expected value is \( \$ 0.25 \). Since the expected value is positive, this indicates that it is not a fair game, as on average, you would expect to make a profit over time. --- The game is a simple example of probability in action! If you analyze the probabilities closely, you’ll find that the rewards for HH and TT contribute favorably to the overall value, making this game a bit more enticing. Just keep in mind that luck plays its role—there's no guarantee you'll always win even with a positive expected value! And if you ever want to delve deeper into the fascinating realm of probability and expected values, consider exploring concepts like variance and standard deviation! These help you understand not just what to expect on average, but also how much those expectations can fluctuate.