A casino offers a simple card game. There are 52 cards in a deck with the values of 2,3,4,…J, Q, K, A. Assume 2 is the smallest value and A is the largest value. Each time the cards are thoroughly shuffled, and each card has the same chance to be selected. You pick up a card from the deck and the dealer picks another one without replacement. If you have a larger number, you win with $50, if the numbers are equal or yours is smaller, the house wins. For each game, you pay $10. Questions: compute the probability of you winning the game? What’s your expected profit from the game?

### 1. Probability of Winning

Let’s define the game:

- You pick one card from the deck, and then the dealer picks another card without replacement.
- The goal is to determine the probability that your card is greater than the dealer’s card.

There are $52$ cards in the deck, and each card is unique. Since both you and the dealer pick a card without replacement, the total number of possible outcomes for the pair of cards is:
\[
\text{Total outcomes} = \binom{52}{2} = \frac{52 \times 51}{2} = 1326
\]

The number of outcomes where you win is the number of ways you can select a card that is larger than the dealer’s card. For each card you pick, the dealer has a fixed number of cards smaller than yours.

- For example, if you pick the lowest card (2), there are 0 cards lower than 2.
- If you pick a 3, there is exactly 1 card lower than 3.
- If you pick a 4, there are exactly 2 cards lower than 4.
- And so on until you pick an Ace, where there are 51 cards lower than Ace.

Thus, for any card $k$ you select (where $k = 2, 3, \dots, A$), the number of ways the dealer can pick a card smaller than yours is $k-2$. The total number of favorable outcomes is the sum of possible wins for each card, which is given by:
\[
\text{Total wins} = 0 + 1 + 2 + \dots + 50 = \sum_{k=1}^{50} k = \frac{50 \times 51}{2} = 1275
\]

Thus, the probability of winning the game is:
\[
P(\text{win}) = \frac{\text{Total wins}}{\text{Total outcomes}} = \frac{1275}{1326} \approx 0.9615
\]

### 2. Expected Profit

For each game, you pay $10. If you win, you receive $50, and if you lose or tie, you receive nothing.

Let’s define:
- $P(\text{win}) = 0.9615$
- $P(\text{lose or tie}) = 1 - P(\text{win}) = 1 - 0.9615 = 0.0385$

The expected value (profit) for each game is calculated as:
\[
E(\text{profit}) = (\text{profit if you win}) \times P(\text{win}) + (\text{profit if you lose or tie}) \times P(\text{lose or tie})
\]
Since the profit when you win is $50 minus the $10 cost to play, your net profit if you win is $40. If you lose or tie, your profit is -$10 (because you lose the $10 you paid).

Thus, the expected profit is:
\[
E(\text{profit}) = (40) \times 0.9615 + (-10) \times 0.0385 = 38.46 - 0.385 = 38.075
\]

### Conclusion

- **Probability of winning**: $ \approx 0.9615 $ (or 96.15%).
- **Expected profit**: $ \approx 38.08 $.

Do you want more details or have any questions?  
Here are some related questions to consider:

1. What happens to the expected profit if the payout for winning is reduced to $30?
2. How would the probability change if the game allowed the dealer to replace the card after picking it?
3. What is the probability of you losing the game?
4. How many games do you need to play to expect a total profit of $1000?
5. What is the variance of your profit in one game?

**Tip:** In probability games, calculating the expected value is crucial to determining whether the game is favorable to the player in the long run.

Discover how to compute the probability of winning in a card game with a 52-card deck and calculate the expected profit. Using probability theory and combinatorics, we explore the chances of winning and the long-term outcomes for players.

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