
If you draw a card with a value of four or less from a standard deck of cards, I will pay you $18
.  If not, you pay me $9
.  (Aces are considered the highest card in the deck.)

Step 2 of 2 :  If you played this game 968
 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.

We can solve this problem by calculating the expected value of a single game and then multiplying that value by the number of games played (968).

### Step 1: Determine the probability of drawing a card with a value of 4 or less.

In a standard deck of 52 cards:

- There are 4 suits, and each suit contains the numbers 2, 3, and 4 (total of 3 cards per suit with values 4 or less).
- Therefore, there are $ 3 \times 4 = 12 $ cards with a value of 4 or less.

The probability of drawing one of these cards is:
\[
P(\text{winning}) = \frac{12}{52} = \frac{3}{13}
\]

The probability of drawing any other card (and losing) is:
\[
P(\text{losing}) = 1 - P(\text{winning}) = 1 - \frac{3}{13} = \frac{10}{13}
\]

### Step 2: Calculate the expected value of a single game.

- If you win (draw a card with a value of 4 or less), you win $18.
- If you lose (draw any other card), you lose $9.

The expected value $ E(X) $ for a single game is given by:
\[
E(X) = (\text{Probability of winning} \times \text{Amount won}) + (\text{Probability of losing} \times \text{Amount lost})
\]
\[
E(X) = \left(\frac{3}{13} \times 18\right) + \left(\frac{10}{13} \times (-9)\right)
\]
\[
E(X) = \left(\frac{54}{13}\right) + \left(\frac{-90}{13}\right)
\]
\[
E(X) = \frac{-36}{13} \approx -2.77
\]

This means that for each game, you can expect to lose approximately $2.77 on average.

### Step 3: Multiply by the number of games.

To find the total expected value over 968 games:
\[
\text{Total expected value} = 968 \times (-2.77) \approx -2679.36
\]

So, after playing 968 times, you would expect to lose approximately **$2,679.36**.

### Final Answer:
You would expect to lose **$2,679.36**.

---

Would you like to see more details or have any questions? Here are five related questions to explore further:

1. How do you calculate expected value in different types of games?
2. What is the concept of risk and reward in probability games?
3. How do changes in the payout structure affect expected value?
4. What role does variance play in games like this?
5. Can expected value guarantee outcomes over a large number of games?

**Tip:** Always think of expected value as the average outcome over a long period, not a guaranteed result in a single trial.

If you draw a card with a value of four or less from a standard deck of cards, I will pay you $18. If not, you pay me $9. Aces are considered the highest card in the deck. If you played this game 968 times, how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.

Calculate the expected value of a card game where drawing a card with a value of four or less wins you $18 and drawing any other card loses you $9. Learn how probability and expected value determine the average outcome of this game played 968 times, and why you can expect to lose $2,679.36 over the long run.

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