To calculate the player's expectation for this bet, we need to analyze the possible outcomes, the probabilities of each outcome, and the amounts won or lost.

Problem Analysis

1. Total numbers on the roulette wheel: 18 black + 18 red + 2 green = 38.
2. Probability of winning (landing on a black number): $P(\text{Black}) = \frac{18}{38} = \frac{9}{19}$
3. Probability of losing (landing on red or green): $P(\text{Not Black}) = 1 - P(\text{Black}) = 1 - \frac{9}{19} = \frac{10}{19}$

Winnings and Losses

• If the player wins: They win the amount they bet, which is $3.
• If the player loses: They lose the amount they bet, which is $3.

Expected Value Calculation

The expected value (E) can be found by summing the products of each outcome's probability and the respective win/loss amount.

$E = (P(\text{Black}) \times \text{Winnings}) + (P(\text{Not Black}) \times \text{Loss})$ Plugging in the values:

$E = \left(\frac{9}{19} \times 3\right) + \left(\frac{10}{19} \times (-3)\right)$

Calculating Each Term

1. Winning term: $\frac{9}{19} \times 3 = \frac{27}{19} \approx 1.42$
2. Losing term: $\frac{10}{19} \times (-3) = \frac{-30}{19} \approx -1.58$

Summing These Values

$E = 1.42 - 1.58 = -0.16$

Conclusion

The player's expectation is approximately -0.16. This means that, on average, the player loses 16 cents for every $3 bet on black.

Final Answer

$\boxed{-0.16}$

Would you like a deeper explanation on any part of this solution?

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1. What is the expected value if the player bets a different amount, say $10?
2. How would the expectation change if the game were fair, with an equal number of black, red, and no green slots?
3. How does expectation relate to probability and outcomes in games of chance?
4. Why do casinos set up games with negative expected values for players?
5. How would the expectation change if there were no green slots (only red and black)?

Tip: Understanding expected value helps you assess the long-term outcome of repetitive bets or games of chance.