To calculate the expected value $E(x)$ for the player in one game of American roulette, we’ll use the following steps:

1. Identify the possible outcomes and probabilities:

   • The player wins if the ball lands on the chosen number (1 out of 38 possible slots).
   • The player loses if the ball lands on any of the other 37 slots.

2. Calculate the probabilities of each outcome:

   • Probability of winning: $P(\text{win}) = \frac{1}{38}$
   • Probability of losing: $P(\text{lose}) = \frac{37}{38}$

3. Define the outcomes:

   • Winning outcome: The player gains $140140 in winnings and keeps the $44 bet. This makes the total gain $140140$.
   • Losing outcome: The player loses their $44 bet, so the total gain is $-44$.

4. Calculate the expected value: The expected value $E(x)$ is calculated as: $E(x) = \left(\frac{1}{38} \times 140140\right) + \left(\frac{37}{38} \times -44\right)$

   Let’s plug in the values:

   • Winning part: $\frac{1}{38} \times 140140 = 3688.95$
   • Losing part: $\frac{37}{38} \times -44 = -42.79$

   Combining these results: $E(x) = 3688.95 - 42.79 = 3646.16$

Answer:

The expected value $E(x)$ to the player for one play of the game is approximately $3646.16.

Would you like a detailed breakdown of each calculation step, or do you have specific questions?

Relative Questions:

1. How would the expected value change if the payout were different?
2. What happens to $E(x)$ if there are fewer or more numbers on the wheel?
3. Why is the expected value generally negative in most casino games?
4. How does the house edge affect the probability of winning in roulette?
5. What other betting strategies could impact the expected value in roulette?

Tip: Expected values give a long-term average but don’t predict short-term results.