The question is asking to find the expected value of a game where the odds of tossing three heads in a row are 3 to 2 against you. The outcomes are as follows:

• If you succeed, you win $3.
• If you fail, you lose $2.

To calculate the expected value of this game, we'll use the formula:

$E(X) = (P(\text{win}) \times \text{win amount}) + (P(\text{lose}) \times \text{lose amount})$

Step 1: Determine the probabilities

The odds of winning are 3 to 2 against, meaning the probability of success (getting three heads) is:

$P(\text{win}) = \frac{2}{5}$

and the probability of failure (not getting three heads) is:

$P(\text{lose}) = \frac{3}{5}$

Step 2: Apply to the expected value formula

Now, substitute the values into the expected value formula:

$E(X) = \left(\frac{2}{5} \times 3\right) + \left(\frac{3}{5} \times (-2)\right)$

Step 3: Calculate

$E(X) = \left(\frac{6}{5}\right) + \left(\frac{-6}{5}\right)$

$E(X) = 0$

Conclusion:

The expected value of the game is 0. This means, on average, you neither win nor lose money in the long run for one game. Over 100 games, the expected outcome is still 0, so you would expect to break even.

Would you like more details or have any questions about the process? Here's a list of related questions to further explore:

1. What happens if the odds were in your favor (e.g., 2 to 3)?
2. How do probabilities affect the expected value calculation?
3. Can expected value change if you alter the win or lose amounts?
4. What is the role of variance in determining how much you win or lose in repeated games?
5. How would you calculate the expected value if the probabilities were given as percentages instead of odds?

Tip: In gambling or game theory, understanding expected value is crucial for determining whether a game is favorable or not in the long run.