To calculate the expected value of this game, we analyze each possible outcome and its probability, accounting for the rules about receiving the highest prize. Then, we compute the weighted average of the prizes minus the cost to play.

Step 1: Total Possible Outcomes

Two dice are rolled, each with 6 sides, so there are $6 \times 6 = 36$ possible outcomes.

Step 2: Prizes and Rules

1. Prize for sum > 10 ($40)

A sum greater than 10 can occur for the following rolls:

• $(5,6), (6,5), (6,6)$.
• Total: $3$ outcomes.

2. Prize for sum < 6 ($35)

A sum less than 6 can occur for the following rolls:

• $(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)$.
• Total: $6$ outcomes.

3. Prize for two odd numbers ($8)

Two odd numbers occur for the following rolls:

• $(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)$.
• Total: $9$ outcomes.

Overlap Between Prizes

A roll can only receive the highest-value prize, so we account for overlaps:

• Sum > 10 overlaps with no other prize. All $3$ outcomes receive $40.
• Sum < 6 overlaps with two odd numbers. Specifically:
  • Odd rolls with sum < 6 are $(1,1), (1,3), (3,1)$ (3 outcomes).
  • These $3$ outcomes would normally receive $8 but are upgraded to $35.
  • Remaining $6 - 3 = 3$ outcomes with sum < 6 receive $35.
• Two odd numbers:
  • $3$ odd outcomes already counted in the sum < 6 category.
  • Remaining odd rolls: $9 - 3 = 6$ outcomes receive $8.

Outcomes Summary

• $3$ rolls receive $40.
• $6$ rolls receive $35.
• $6$ rolls receive $8.
• $36 - 3 - 6 - 6 = 21$ rolls receive $0.

Step 3: Expected Value Calculation

The expected value is calculated as: $E(X) = \sum (\text{Prize} \times \text{Probability}) - \text{Cost to Play}$ $E(X) = (40 \times \frac{3}{36}) + (35 \times \frac{6}{36}) + (8 \times \frac{6}{36}) + (0 \times \frac{21}{36}) - 5$ $E(X) = \left(3.33 + 5.83 + 1.33 + 0\right) - 5 = 10.49 - 5 = 5.49$

Final Answer:

The expected value of the game is $5.49.

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Would you like a detailed breakdown of any part? Here are some related questions:

1. How are the probabilities for each prize calculated in detail?
2. Why do we count overlaps separately in expected value problems?
3. How does the expected value change if the cost to play increases to $10?
4. What is the probability of winning any prize in this game?
5. Can you simulate the results of this game over multiple plays?

Tip: In expected value problems, always carefully separate outcomes, calculate probabilities, and consider overlaps to avoid double-counting.