If a gambling game is played with expected value \( \$ 0.40 \), then there is a \( 40 \% \) chance of winning. true false

The statement is false. The expected value of a game is calculated by multiplying each possible outcome by its probability and then summing those products. For example, even if a game has an expected value of $0.40 per play, that does not mean there is necessarily a 40% chance of “winning” in the sense of receiving a positive payout. As an illustration, consider a game with only two outcomes: • Outcome 1: Win $1.00 • Outcome 2: Win $0.00 If the probability of winning $1.00 is p, the expected value EV is calculated as:  EV = p × $1.00 + (1 − p) × $0.00 = p To have an expected value of $0.40, p would indeed be 40%. However, many gambling games have multiple outcomes or payoffs that differ from this simple binary example. For instance, a game might give:  • A win of $4.00 with a 10% chance, and  • A win of $0.00 with a 90% chance, which also results in an expected value of $0.40 (0.10 × $4.00 = $0.40). Thus, the expected value of $0.40 can be achieved with a win probability of 10% (if the win is $4), or any number of other combinations. Therefore, you cannot automatically conclude that an expected value of $0.40 implies a 40% chance of winning.