To solve this problem, we first need to calculate the expected value of a single bet. The expected value helps determine how much you would expect to win or lose over a large number of trials (in this case, 914).

Step 1: Calculate the probability of drawing two diamonds

A standard deck of cards has 52 cards, and 13 of them are diamonds.

• The probability of drawing a diamond on the first card is: $P(\text{1st card is a diamond}) = \frac{13}{52} = \frac{1}{4}$

• After drawing one diamond, there are 12 diamonds left out of the remaining 51 cards. So, the probability of drawing a diamond on the second card (given the first card was a diamond) is: $P(\text{2nd card is a diamond | 1st card is a diamond}) = \frac{12}{51}$

Thus, the probability of drawing two diamonds in a row is: $P(\text{both cards are diamonds}) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$

Step 2: Calculate the expected value of one bet

The payout structure is as follows:

• If both cards are diamonds, you win $810.
• If either card is not a diamond, you lose $49.

The expected value (EV) of a single bet is the sum of the possible outcomes weighted by their probabilities. It can be calculated as: $\text{EV} = (P(\text{both diamonds}) \times \text{win amount}) + (P(\text{not both diamonds}) \times \text{lose amount})$

We already know that the probability of both cards being diamonds is $\frac{1}{17}$, so the probability of not both cards being diamonds is $1 - \frac{1}{17} = \frac{16}{17}$.

Now, we calculate the expected value: $\text{EV} = \left(\frac{1}{17} \times 810\right) + \left(\frac{16}{17} \times (-49)\right)$ $\text{EV} = \left( 47.65 \right) + \left( -46.59 \right)$ $\text{EV} = 47.65 - 46.59 = 1.06$

So, the expected value of a single bet is $1.06.

Step 3: Calculate the expected total for 914 bets

If the same bet is made 914 times, the expected total win or loss is: $\text{Total expected value} = 914 \times 1.06 = 969.84$

Final Answer:

You would expect to win $969.84 if the bet is made 914 times.

Let me know if you need further clarification or have any additional questions!

Follow-up questions:

1. What is the probability of drawing two cards of the same suit?
2. How does the expected value change if the payout amounts are adjusted?
3. What would happen to the expected value if the number of diamonds in the deck increased?
4. How would the expected value differ if you replaced the deck with a non-standard deck (e.g., a deck with 60 cards)?
5. What does the expected value represent in the context of repeated trials?

Tip:

In probability problems, always ensure you're considering both the favorable and unfavorable outcomes and their associated probabilities to calculate the expected value correctly.