A person gets dollars equal to the square of the number which comes up when a balanced die, with faces marked 1, 2, 3, 4, 5, 6 is rolled once. If the game is repeated an indefinitely large number of times, how much money can he expect in the long run per game?

Answer

To solve the problem, we need to calculate the expected value of the amount of money a person earns when rolling a balanced die. The amount earned is equal to the square of the number that comes up on the die.

### Step-by-Step Solution:

1. **Define the Random Variable**:
Let \( X \) be the random variable representing the amount of money earned per game. The possible outcomes when rolling a die are 1, 2, 3, 4, 5, and 6.

2. **Calculate the Values of \( X \)**:
The values of \( X \) based on the die roll are:
- If the die shows 1, \( X = 1^2 = 1 \)
- If the die shows 2, \( X = 2^2 = 4 \)
- If the die shows 3, \( X = 3^2 = 9 \)
- If the die shows 4, \( X = 4^2 = 16 \)
- If the die shows 5, \( X = 5^2 = 25 \)
- If the die shows 6, \( X = 6^2 = 36 \)

Thus, the possible values of \( X \) are \( 1, 4, 9, 16, 25, 36 \).

3. **Determine the Probability of Each Outcome**:
Since the die is balanced, the probability of each outcome is the same:
\[ P(X = 1) = P(X = 4) = P(X = 9) = P(X = 16) = P(X = 25) = P(X = 36) = \frac{1}{6} \]

4. **Calculate the Expected Value \( E(X) \)**:
The expected value \( E(X) \) is calculated using the formula:
\[ E(X) = \sum (X \cdot P(X)) \]
Substituting the values we have:
\[ E(X) = 1 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 9 \cdot \frac{1}{6} + 16 \cdot \frac{1}{6} + 25 \cdot \frac{1}{6} + 36 \cdot \frac{1}{6} \]
\[ E(X) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} \]

5. **Calculate the Sum**:
Now, calculate the sum of the values:
\[ 1 + 4 = 5 \]
\[ 5 + 9 = 14 \]
\[ 14 + 16 = 30 \]
\[ 30 + 25 = 55 \]
\[ 55 + 36 = 91 \]

6. **Final Calculation of Expected Value**:
Substitute back into the expected value formula:
\[ E(X) = \frac{91}{6} \approx 15.1667 \]

### Conclusion:
The expected amount of money the person can expect to earn per game in the long run is approximately \( 15.17 \) dollars.