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Juan is playing a game in which he spins a spinner with 6 equal-sized slices numbered 1 through 6 . The spinner stops on a numbered slice at random. This game is this: Juan spins the spinner once. He wins \( \$ 1 \) if the spinner stops on the number \( 1, \$ 4 \) if the spinner stops on the number 2 , \( \$ 7 \) if the spinne stops on the number 3 , and \( \$ 10 \) if the spinner stops on the number 4 . He loses \( \$ 11 \) if the spinner stops on 5 or 6 . (a) Find the expected value of playing the game. (b) What can Juan expect in the long run, after playing the game many times? Juan can expect to gain money. He can expect to win \( \square \) dollars per spin. He can expect to lose \( \square \) dollars per spin. O Juan can expect to break even (neither gain nor lose money).

Ask by Sherman Jimenez. in the United States
Jan 06,2025

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Tutor-Verified Answer

Answer

Juan can expect to break even (neither gain nor lose money) per spin.

Solution

Let's analyze Juan's game step by step. ### Part (a): Finding the Expected Value The **expected value (E)** of a game is the average amount Juan can expect to win or lose per spin if he plays the game many times. It is calculated by multiplying each possible outcome by its probability and then summing all these values. Here are the possible outcomes and their corresponding payouts: - **Spin stops on 1:** Win \$1 - **Spin stops on 2:** Win \$4 - **Spin stops on 3:** Win \$7 - **Spin stops on 4:** Win \$10 - **Spin stops on 5 or 6:** Lose \$11 each Since the spinner has 6 equal-sized slices, the probability of landing on any particular number is \( \frac{1}{6} \). Now, calculate the expected value: \[ E = \left( \frac{1}{6} \times 1 \right) + \left( \frac{1}{6} \times 4 \right) + \left( \frac{1}{6} \times 7 \right) + \left( \frac{1}{6} \times 10 \right) + \left( \frac{1}{6} \times (-11) \right) + \left( \frac{1}{6} \times (-11) \right) \] \[ E = \frac{1 + 4 + 7 + 10 - 11 - 11}{6} = \frac{0}{6} = 0 \] So, the **expected value is \$0**. ### Part (b): Long-Run Expectation Since the expected value is \$0, this means that **on average, Juan neither gains nor loses money** per spin in the long run. Over many spins, the gains and losses balance each other out. **Summary:** - **Expected Value:** \$0 - **Long-Run Expectation:** Juan can expect to break even (neither gain nor lose money) per spin. **Answer:** Problem (b) Answer: Juan can expect to break even (neither gain nor lose money).

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To find the expected value of playing the game, we can calculate it using the outcomes and their respective probabilities. The outcomes and their associated wins/losses are as follows: - Winning \$1 with probability \( \frac{1}{6} \) (stopping on 1) - Winning \$4 with probability \( \frac{1}{6} \) (stopping on 2) - Winning \$7 with probability \( \frac{1}{6} \) (stopping on 3) - Winning \$10 with probability \( \frac{1}{6} \) (stopping on 4) - Losing \$11 with probability \( \frac{1}{6} \) (stopping on 5) - Losing \$11 with probability \( \frac{1}{6} \) (stopping on 6) Now, let's calculate the expected value \( E \): \[ E = \left(\frac{1}{6} \cdot 1\right) + \left(\frac{1}{6} \cdot 4\right) + \left(\frac{1}{6} \cdot 7\right) + \left(\frac{1}{6} \cdot 10\right) + \left(\frac{1}{6} \cdot (-11)\right) + \left(\frac{1}{6} \cdot (-11)\right) \] \[ E = \frac{1 + 4 + 7 + 10 - 11 - 11}{6} \] \[ E = \frac{0}{6} = 0 \] (a) The expected value of playing the game is \( \$0 \). (b) In the long run, after playing the game many times, Juan can expect to break even since the expected value is \( \$0 \) per spin. Thus, he will neither gain nor lose money overall.

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