Question

1.1.5 how much can i expect to win?

take a spin
3 - 47. consider the following game: after you spin the wheel at
right, you win the amount spun.

a. if you play the game 10 times, how much money
would you expect to win? what if you played the
game 30 times? 100 times? explain your process.

b. if you were to play only once, what would you expect to earn accordi
your answers in part (a)? is it actually possible to win that amount?
explain why or why not.

Explanation:

Step1: Calculate the probability of each outcome

The wheel is divided into 2 equal parts. The probability of landing on $0$ is $P(0)=\frac{1}{2}$, and the probability of landing on $4$ is $P(4)=\frac{1}{2}$.

Step2: Calculate the expected value for one spin

The formula for the expected value $E(X)$ of a discrete - random variable is $E(X)=\sum_{i}x_{i}P(x_{i})$. Here, $x_1 = 0$ and $x_2 = 4$.

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Step3: Calculate the expected value for multiple spins

• For $n = 10$ spins: The expected value $E_{10}=n\times E(X)=10\times2 = 20$.
• For $n = 30$ spins: The expected value $E_{30}=n\times E(X)=30\times2=60$.
• For $n = 100$ spins: The expected value $E_{100}=n\times E(X)=100\times2 = 200$.

Step4: Answer part (b)

If you play only once, the expected value is $2$. But when you play once, you can only win either $0$ or $4$. Since $2$ is not one of the possible outcomes ($0$ or $4$), it is not actually possible to win $2$ in a single play. The expected value is a long - term average.

Answer:

a. If you play 10 times, you expect to win $20$; if you play 30 times, you expect to win $60$; if you play 100 times, you expect to win $200$.
b. If you play only once, the expected value is $2$. But it is not actually possible to win $2$ in a single play because the possible outcomes are $0$ or $4$. The expected value is a long - term average, not an outcome of a single trial.