To find the expected payoff, we need to consider the probabilities and corresponding payoffs for each possible outcome.

Step 1: Determine the probabilities

In the game, the outcome is a random integer between 1 and 50, which means all outcomes are equally likely.

• Total possible outcomes: 50 integers (from 1 to 50).

• Odd numbers between 1 and 50: The odd numbers are 1, 3, 5, ..., 49. These are all numbers of the form $2n - 1$, for integers $n$. The odd numbers form an arithmetic sequence: 1, 3, 5, ..., 49. The number of odd numbers can be calculated as: $\text{Count of odd numbers} = \frac{50 + 1}{2} = 25$ So, there are 25 odd numbers.

• Even numbers between 1 and 50: Since there are 25 odd numbers, there must also be 25 even numbers.

Step 2: Payoffs

• If the outcome is odd: You win $18.
• If the outcome is even: You win $0.

Step 3: Expected Payoff Formula

The expected value $E(X)$ is the sum of all possible outcomes, weighted by their respective probabilities: $E(X) = P(\text{odd}) \times 18 + P(\text{even}) \times 0$ Where:

• $P(\text{odd}) = \frac{25}{50} = 0.5$,
• $P(\text{even}) = \frac{25}{50} = 0.5$.

Step 4: Calculate the expected payoff

Now, we can calculate the expected payoff: $E(X) = 0.5 \times 18 + 0.5 \times 0 = 9 + 0 = 9$

Thus, the expected payoff for playing the game is $9.

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Do you have any questions or want further details?
Here are 5 related questions to think about:

1. How would the expected value change if the payoff for odd numbers was $20 instead of $18?
2. What would the expected value be if there were more even numbers than odd numbers in the game?
3. How does the law of large numbers apply if you play this game many times?
4. What is the variance of the payoff for this game?
5. What is the probability of winning more than $50 if you play this game 10 times?

Tip: When calculating expected values, always ensure you account for all possible outcomes and their associated probabilities.