To calculate the expected value of the bet, we need to take into account the possible outcomes, their probabilities, and the corresponding payouts. The expected value (EV) formula is:

$EV = (P(\text{both diamonds}) \times \text{payout for both diamonds}) + (P(\text{not both diamonds}) \times \text{payout for not both diamonds})$

Step 1: Calculate the probability of drawing two diamonds

A standard deck has 52 cards, and there are 13 diamonds in the deck. The probability of drawing two diamonds without replacement can be calculated as follows:

• The probability of the first card being a diamond is $\frac{13}{52}$.
• After the first card is drawn, there are 12 diamonds left, and 51 cards remaining. So, the probability of the second card being a diamond is $\frac{12}{51}$.

The probability of both cards being diamonds is the product of these probabilities:

$P(\text{both diamonds}) = \frac{13}{52} \times \frac{12}{51}$

Now, let's calculate this value:

$P(\text{both diamonds}) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} \approx 0.0588$

Step 2: Calculate the probability of not drawing two diamonds

The probability of not drawing two diamonds is simply the complement of the probability of drawing two diamonds:

$P(\text{not both diamonds}) = 1 - P(\text{both diamonds}) = 1 - 0.0588 = 0.9412$

Step 3: Calculate the expected value

The payouts are as follows:

• If both cards are diamonds, you win $810.
• If not both cards are diamonds, you lose $49 (you pay your friend $49).

So, the expected value is:

$EV = (0.0588 \times 810) + (0.9412 \times -49)$

Now, let's calculate this:

$EV = (0.0588 \times 810) + (0.9412 \times -49)$

$EV = 47.628 + (-46.125)$

$EV = 47.628 - 46.125 = 1.503$

Final Answer:

The expected value of your bet is $1.50.

Would you like further clarification on any part of the solution?

Additional Questions:

1. How would the expected value change if the bet amount for a win was increased to $1,000?
2. What would happen if you were to draw three cards instead of two?
3. How does the probability of getting two diamonds change if the deck contained only 40 cards, with 10 diamonds?
4. What are the concepts of "risk" and "variance" in gambling, and how do they relate to expected value?
5. How would this situation change if your friend was the one making the bet instead?

Tip: The expected value is a useful tool to assess the fairness of a wager, but it doesn’t guarantee specific outcomes in the short term. It's more of a long-term average.