To find the expected value $E(x)$ for a player in American roulette, we use the formula for expected value:

$E(x) = \sum [P(x) \cdot x]$

Step 1: Define the Outcomes

• The player bets $4 on a single number.
• There are 38 numbers on the roulette wheel.
• If the player wins, they receive $140 plus the original $4, for a total gain of $144.
• If the player loses, they get nothing and lose their $4.

Step 2: Compute Probabilities

• Probability of winning (betting on 1 specific number out of 38):

  $P(\text{win}) = \frac{1}{38}$

• Probability of losing:

  $P(\text{lose}) = \frac{37}{38}$

Step 3: Compute the Expected Value

$E(x) = P(\text{win}) \cdot \text{winnings} + P(\text{lose}) \cdot \text{loss}$

$E(x) = \left(\frac{1}{38} \times 144\right) + \left(\frac{37}{38} \times (-4)\right)$

$E(x) = \frac{144}{38} + \frac{-148}{38}$

$E(x) = 3.79 - 3.89 = -0.10$

Final Answer:

$E(x) = -0.10$

So, the expected value is $-\$0.10$ per play. This means, on average, a player loses 10 cents per bet.

Would you like a more detailed explanation or have any questions?

Related Questions:

1. What is the expected value if the player bets on multiple numbers at once?
2. How does the expected value change in European roulette (which has 37 numbers instead of 38)?
3. What is the house edge in American roulette, and how is it calculated?
4. How does the payout structure impact the expected value in different roulette variations?
5. What are the odds of winning in other common roulette bets, such as red/black or even/odd?

Tip:

Expected value is useful in determining long-term profitability. In games of chance, a negative expected value means the casino has an edge!