To find the expected value of this proposition, we need to compute the weighted average of all possible outcomes based on their probabilities and payoffs.

Step 1: List the possible sums when throwing two dice

The sums of two dice can range from 2 to 12. We need to find the probability of each sum occurring.

Here are the possible sums of two dice and the number of ways each can occur:

• Sum of 2: (1,1) → 1 outcome
• Sum of 3: (1,2), (2,1) → 2 outcomes
• Sum of 4: (1,3), (2,2), (3,1) → 3 outcomes
• Sum of 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
• Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
• Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
• Sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
• Sum of 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
• Sum of 10: (4,6), (5,5), (6,4) → 3 outcomes
• Sum of 11: (5,6), (6,5) → 2 outcomes
• Sum of 12: (6,6) → 1 outcome

There are 36 possible outcomes in total when rolling two dice (since each die has 6 faces, so $6 \times 6 = 36$).

Step 2: Identify the win and loss conditions

• Win: You win $19 if the sum is 8 or more. This includes the sums of 8, 9, 10, 11, and 12.
• Loss: You lose $17 if the sum is less than 8 (i.e., sums of 2, 3, 4, 5, 6, and 7).

Step 3: Calculate the probabilities of winning and losing

The number of ways to win is the total number of outcomes for sums 8, 9, 10, 11, and 12:

• 8 → 5 outcomes
• 9 → 4 outcomes
• 10 → 3 outcomes
• 11 → 2 outcomes
• 12 → 1 outcome

So, the total number of winning outcomes = $5 + 4 + 3 + 2 + 1 = 15$.

The number of ways to lose is the total number of outcomes for sums 2, 3, 4, 5, 6, and 7:

• 2 → 1 outcome
• 3 → 2 outcomes
• 4 → 3 outcomes
• 5 → 4 outcomes
• 6 → 5 outcomes
• 7 → 6 outcomes

So, the total number of losing outcomes = $1 + 2 + 3 + 4 + 5 + 6 = 21$.

Thus, the probabilities of winning and losing are:

• Probability of winning: $\frac{15}{36} = 0.4167$
• Probability of losing: $\frac{21}{36} = 0.5833$

Step 4: Calculate the expected value (EV)

The expected value is given by the formula:

$\text{EV} = (P(\text{win}) \times \text{win payoff}) + (P(\text{lose}) \times \text{lose payoff})$

Substituting the values: $\text{EV} = (0.4167 \times 19) + (0.5833 \times (-17))$

$\text{EV} = (7.9163) + (-9.9161) = -2.00$

Conclusion:

The expected value of this proposition is $-2.00.

Would you like to dive deeper into how this works, or do you have any further questions? Here are some related questions to consider:

1. How do probabilities and payoffs affect the expected value in gambling scenarios?
2. How can expected value help in decision-making under uncertainty?
3. What happens to the expected value if the winning or losing amounts change?
4. How can you use the expected value to compare different betting scenarios?
5. How would the expected value change if the dice had a different number of sides?

Tip: Always remember that the expected value represents the long-term average outcome, which may differ from the result of a single trial.