The expected value (EV) of the game is calculated by taking the weighted average of all possible outcomes, where the weights are the probabilities of each outcome.

Problem Breakdown:

1. Possible outcomes:

   • Win: The player wins $400 if their chosen number is selected.
   • Loss: The player wins $0 if any other number is selected.

2. Probabilities:

   • Probability of winning: $P(\text{Win}) = \frac{1}{100}$.
   • Probability of losing: $P(\text{Loss}) = 1 - P(\text{Win}) = \frac{99}{100}$.

3. Net gains/losses:

   • If the player wins, they receive $400, but since they paid $17 to play, their net gain is $400 - 17 = 383$.
   • If the player loses, they gain nothing and lose their $17, so the net loss is $-17$.

4. Expected value formula: $EV = P(\text{Win}) \times \text{Net Gain (Win)} + P(\text{Loss}) \times \text{Net Loss (Loss)}$

Calculation:

$EV = \left(\frac{1}{100}\right) \times 383 + \left(\frac{99}{100}\right) \times (-17)$

Step-by-step computation:

1. First term: $\frac{1}{100} \times 383 = 3.83$
2. Second term: $\frac{99}{100} \times (-17) = -16.83$
3. Combine: $EV = 3.83 - 16.83 = -13$

Conclusion:

The expected value of the game is $-13$ dollars. This means, on average, the player loses $13 per game.

---

Would you like further details or clarification? Here are five related questions to consider:

1. How would the expected value change if the prize amount were increased?
2. What is the probability of winning if the number pool changes (e.g., to 1,000 numbers)?
3. How does the expected value inform whether the game is fair for players?
4. Could changing the bet amount impact the expected value? Why or why not?
5. What is the house edge in this game?

Tip: When analyzing games of chance, consider the expected value to determine whether the game favors the player or the house.