In this game, you roll a die, and the outcome $X$ determines whether you pay or receive money. Let’s break it down by analyzing the possible outcomes and their probabilities.

Step 1: The possible outcomes of a die roll

A fair six-sided die has the following outcomes: $1, 2, 3, 4, 5, 6$. Half of these outcomes are odd, and half are even:

• Odd outcomes: $1, 3, 5$
• Even outcomes: $2, 4, 6$

Step 2: Defining the payoffs

• If $X$ is odd, you pay him $X$.
• If $X$ is even, he pays you $X$.

Thus, for each roll:

• If $X = 1$, you lose $1$
• If $X = 2$, you gain $2$
• If $X = 3$, you lose $3$
• If $X = 4$, you gain $4$
• If $X = 5$, you lose $5$
• If $X = 6$, you gain $6$

Step 3: Calculating the expected profit

We now calculate the expected value (long-run average) of the game. The expected value $E[X]$ is the sum of the payoffs, each multiplied by its probability.

Since the die is fair, the probability of any outcome is $\frac{1}{6}$.

The expected profit is:

$E[\text{Profit}] = \left( \frac{1}{6} \times (-1) \right) + \left( \frac{1}{6} \times 2 \right) + \left( \frac{1}{6} \times (-3) \right) + \left( \frac{1}{6} \times 4 \right) + \left( \frac{1}{6} \times (-5) \right) + \left( \frac{1}{6} \times 6 \right)$

Simplifying this:

$E[\text{Profit}] = \frac{-1 + 2 - 3 + 4 - 5 + 6}{6} = \frac{3}{6} = 0.5$

Conclusion:

The long-run average profit for you in this game is $0.5$ dollars per roll. That means, over many rolls, you can expect to win 50 cents on average per roll.

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Do you want more details or have any questions?

Here are 5 related questions to consider:

1. What would the long-run profit be if the die were biased?
2. How does the expected profit change if different payouts are used for odd and even numbers?
3. What would the expected profit be if we used a 12-sided die instead of a 6-sided one?
4. What is the variance of the profit in this game?
5. How many rolls would you need to be reasonably confident that you’re making a profit?

Tip: When calculating expected value, always multiply each outcome by its probability and sum them up!