To determine the expected value of your bet, we will compute the expected payout based on the possible outcomes.

Step 1: Define the probabilities

• A standard deck contains 52 cards.
• There are 13 diamonds in the deck.

When drawing two cards without replacement, the probability that both cards are diamonds can be calculated as follows:

$P(\text{both diamonds}) = \frac{13}{52} \times \frac{12}{51}$ $P(\text{both diamonds}) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$

Thus, the probability that both cards are diamonds is $\frac{1}{17}$.

The probability that not both cards are diamonds is the complement of this, which is: $P(\text{not both diamonds}) = 1 - \frac{1}{17} = \frac{16}{17}$

Step 2: Define the payouts

• If both cards are diamonds, your friend pays you $801.
• If not, you have to pay your friend $48.

Step 3: Compute the expected value

The expected value $E$ is calculated as:

$E = \left(\text{payout if both diamonds} \times \text{probability of both diamonds}\right) + \left(\text{payout if not both diamonds} \times \text{probability of not both diamonds}\right)$

Substituting the values:

$E = (801 \times \frac{1}{17}) + (-48 \times \frac{16}{17})$

Now compute each part:

$801 \times \frac{1}{17} = 47.12$ $-48 \times \frac{16}{17} = -45.18$

Now, summing these:

$E = 47.12 - 45.18 = 1.94$

Final Answer:

The expected value of your bet is $1.94.

Would you like any further details or clarifications?

Here are 5 related questions:

1. How would the expected value change if the payout for both diamonds were higher?
2. What is the expected value if you draw three cards instead of two?
3. What if the wager was with a different suit, say hearts—how would that change the expected value?
4. How does removing one or more diamonds from the deck affect the outcome?
5. Can we calculate the variance or standard deviation of the outcomes?

Tip: The expected value tells you the average outcome if the bet were repeated many times, not the outcome of a single trial.