In American​ roulette, the wheel has the 38​ numbers, 00,​ 0, 1,​ 2, ...,​ 34, 35, and​ 36, marked on equally spaced slots. If a player bets ​$77 on a number and​ wins, then the player keeps ​$77 and receives an additional ​$245245. ​Otherwise, the player is awarded​ nothing, and the casino takes the​ player's ​$77. Find the expected value​ E(x) to the player for one play of the game. If x is the gain to a player in a game of​ chance, then​ E(x) is usually negative. This value gives the average amount per game the player can expect to lose.
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Part 1
The expected value is ​$enter your response here

To solve this, we will calculate the expected value (E(x)) for the player in the game of American roulette.

The **expected value** is a weighted average of all possible outcomes, where each outcome is weighted by its probability.

There are two possible outcomes for the player in this scenario:
1. The player **wins**: They keep the $77 bet and receive an additional $245,245.
2. The player **loses**: They lose the $77 bet.

### Probabilities:
- The player wins if the ball lands on the number they bet on, which is just 1 out of the 38 possible outcomes (since there are 38 numbers on the wheel). So, the probability of winning is:

\[
P(\text{win}) = \frac{1}{38}
\]

- The player loses if the ball lands on any other number (37 out of 38 numbers), so the probability of losing is:

\[
P(\text{lose}) = \frac{37}{38}
\]

### Outcomes:
- If the player wins, their net gain is **$245,245**.
- If the player loses, their net gain (or rather, loss) is **-$77** (they lose the amount they bet).

### Expected Value Formula:

\[
E(x) = (\text{gain when winning} \times P(\text{win})) + (\text{loss when losing} \times P(\text{lose}))
\]

Substituting the values:

\[
E(x) = (245245 \times \frac{1}{38}) + (-77 \times \frac{37}{38})
\]

Now, let's calculate each term:
- **First term**: $245245 \times \frac{1}{38} = 6453.82$
- **Second term**: $-77 \times \frac{37}{38} = -74.97$

### Final Expected Value:
\[
E(x) = 6453.82 - 74.97 = 6378.85
\]

Therefore, the expected value for the player in one play of this game is **$6378.85**. This means, on average, the player can expect to gain $6378.85 per game.

Would you like a breakdown of this process, or have any other questions? 

Here are 5 related questions:
1. What is the probability of winning multiple rounds in a row in roulette?
2. How do casinos balance their expected revenue against player winnings?
3. What are other common bets in American roulette, and how do their expected values compare?
4. How does the expected value differ in European roulette, which has only 37 numbers?
5. What strategies do players use to maximize their chances of winning in roulette?

**Tip:** In games of chance like roulette, understanding expected value helps you see why the odds are often in favor of the casino in the long run.

In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, ..., 34, 35, and 36, marked on equally spaced slots. If a player bets $77 on a number and wins, then the player keeps $77 and receives an additional $245,245. Otherwise, the player is awarded nothing, and the casino takes the player's $77. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

Discover how to calculate the expected value (E(x)) in American roulette, where a $77 bet can lead to a $245,245 payout. This detailed solution walks through probability concepts, the expected value formula, and the game's mechanics. Perfect for advanced high school or college-level students exploring probability and game theory.

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