# Expected Value in Guessing: Elimination Problem Let's break down the problem step by step: - **Correct answer:** +7 points - **Incorrect answer:** points - **Blank:** +2 points - **Total choices per question:** 5 You want to know: **How many choices must be eliminated for guessing to be better than leaving the question blank?** --- ## 1. Setting Up the Expected Value Suppose you can eliminate \( n \) incorrect choices. - Choices remaining to guess from: \( 5 - n \) - Probability of guessing correctly: \( \frac{1}{5-n} \) - Probability of guessing incorrectly: \( \frac{5-n-1}{5-n} = \frac{4-n}{5-n} \) ### Expected Value if You Guess \[ \text{EV}_{\text{guess}} = (\text{prob of correct}) \times 7 + (\text{prob of incorrect}) \times \] \[ = \frac{1}{5-n} \times 7 + \frac{4-n}{5-n} \times \] \[ = \frac{7}{5-n} \] ### Expected Value if You Leave Blank \[ \text{EV}_{\text{blank}} = 2 \] --- ## 2. When is Guessing Advantageous? Set the expected value of guessing to be greater than blank: \[ \frac{7}{5-n} > 2 \] Solve for \( n \): \[ \frac{7}{5-n} > 2 \] \[ 7 > 2(5-n) \] \[ 7 > 10 - 2n \] \[ 2n > 3 \] \[ n > 1.5 \] --- ## 3. Interpreting the Result \( n \) must be an integer (you can't eliminate half a choice). So, **you must eliminate at least 2 choices**. | Choices eliminated (\( n \)) | EV when guessing | Is it > 2? | |------------------------------|------------------|------------| | | \( \frac{7}{5} = 1.4 \) | No | | 1 | \( \frac{7}{4} = 1.75 \) | No | | 2 | \( \frac{7}{3} \approx 2.33 \) | Yes | | 3 | \( \frac{7}{2} = 3.5 \) | Yes | | 4 | \( 7 \) | Yes | --- ## 4. **Final Answer** > **You must eliminate at least 2 out of 5 choices for guessing to be advantageous (expected value greater than 2).** --- **Summary Table:** | Number of choices eliminated | Remaining choices | Guessing is advantageous? | |-----------------------------|-------------------|--------------------------| | | 5 | No | | 1 | 4 | No | | 2 | 3 | Yes | | 3 | 2 | Yes | | 4 | 1 | Yes |

# Expected Value in Guessing: Elimination Problem Let's analyze how to determine the number of choices that must be eliminated for guessing to be advantageous. ## 1. Problem Setup - **Correct answer:** +7 points - **Incorrect answer:** 0 points - **Blank:** +2 points - **Total choices per question:** 5 The goal is to find out how many incorrect choices must be eliminated so that guessing yields an expected value (EV) greater than leaving the question blank. ## 2. Expected Value Calculation Assume you eliminate \( n \) incorrect choices. This leaves you with \( 5 - n \) choices to guess from. - **Probability of guessing correctly:** \[ P(\text{correct}) = \frac{1}{5-n} \] - **Probability of guessing incorrectly:** \[ P(\text{incorrect}) = \frac{4-n}{5-n} \] ### Expected Value if You Guess The expected value (EV) when guessing can be calculated as follows: \[ \text{EV}_{\text{guess}} = \left(\frac{1}{5-n} \times 7\right) + \left(\frac{4-n}{5-n} \times 0\right) = \frac{7}{5-n} \] ### Expected Value if You Leave Blank The expected value for leaving the question blank is simply: \[ \text{EV}_{\text{blank}} = 2 \] ## 3. Setting Up the Inequality To determine when guessing is advantageous, we set up the inequality: \[ \frac{7}{5-n} > 2 \] ### Solving the Inequality 1. Multiply both sides by \( 5 - n \) (assuming \( 5-n > 0 \)): \[ 7 > 2(5-n) \] 2. Expand and rearrange: \[ 7 > 10 - 2n \] \[ 2n > 3 \] \[ n > 1.5 \] Since \( n \) must be an integer, the smallest possible value for \( n \) is 2. ## 4. Conclusion To ensure that guessing is advantageous, you must eliminate at least **2 choices** out of the original 5. This means, if you can eliminate 2 or more choices, the expected value of guessing will exceed the expected value of leaving the question blank.

# Expected Value in Guessing: Detailed Step-by-Step Explanation In this problem, we need to determine how many choices must be eliminated for guessing to become advantageous compared to leaving a question blank, given the scoring rules of a test. ## 1. Problem Setup ### Scoring Rules - **Correct answer:** +7 points - **Incorrect answer:** 0 points - **Blank answer:** +2 points - **Total choices per question:** 5 ### Objective We want to find the minimum number of incorrect choices, denoted \( n \), that must be eliminated so that the expected value (EV) of guessing exceeds 2 points (the score for leaving the answer blank). ## 2. Expected Value Calculation ### Choices After Eliminating Incorrect Answers If \( n \) incorrect choices are eliminated, the number of choices left for guessing becomes: \[ \text{Choices remaining} = 5 - n \] ### Probabilities for Remaining Choices - **Probability of guessing correctly:** \[ P(\text{correct}) = \frac{1}{5-n} \] - **Probability of guessing incorrectly:** \[ P(\text{incorrect}) = \frac{(5 - n) - 1}{5 - n} = \frac{4 - n}{5 - n} \] ### Expected Value When Guessing To find the expected value when guessing, we combine the probabilities and their corresponding scores: \[ \text{EV}_{\text{guess}} = \left(\text{Probability of correct} \times \text{Score for correct}\right) + \left(\text{Probability of incorrect} \times \text{Score for incorrect}\right) \] Substituting the values we have: \[ \text{EV}_{\text{guess}} = \left(\frac{1}{5-n} \times 7\right) + \left(\frac{4-n}{5-n} \times 0\right) = \frac{7}{5-n} \] ### Expected Value When Leaving Blank The expected value for leaving a question blank is straightforward: \[ \text{EV}_{\text{blank}} = 2 \] ## 3. Setting Up the Inequality To determine when guessing is advantageous, we set up the inequality: \[ \text{EV}_{\text{guess}} > \text{EV}_{\text{blank}} \] Substituting the expected values: \[ \frac{7}{5-n} > 2 \] ## 4. Solving the Inequality ### Step 1: Clear the Fraction Multiply both sides by \( 5-n \) (assuming \( 5-n > 0 \) to avoid division by zero): \[ 7 > 2(5-n) \] ### Step 2: Expand the Right Side Distributing the 2 gives: \[ 7 > 10 - 2n \] ### Step 3: Rearranging the Inequality Rearranging the terms leads to: \[ 7 - 10 > -2n \] \[ -3 > -2n \] ### Step 4: Dividing by -2 When dividing by a negative number, we reverse the inequality: \[ \frac{3}{2} < n \] This simplifies to: \[ n > 1.5 \] ### Step 5: Finding Integer Solutions Since \( n \) must be an integer (you cannot eliminate half a choice), the smallest integer satisfying this condition is: \[ n \geq 2 \] ## 5. Conclusion To conclude, you must eliminate at least **2 choices** out of the original 5 for guessing to be advantageous, meaning that the expected value of guessing will be greater than leaving the question blank. In summary, the exact steps and calculations confirm that eliminating at least 2 choices increases the likelihood of a better score through guessing compared to opting for a blank answer.

# Expected Value in Guessing: Detailed Explanation In this guide, we will explore how to determine the number of answer choices that must be eliminated in a test scenario where guessing becomes advantageous compared to leaving a question blank. ## 1. Problem Setup ### Scoring System - **Correct answer:** +7 points - **Incorrect answer:** 0 points - **Blank answer:** +2 points - **Total choices per question:** 5 ### Objective Our goal is to identify the minimum number of incorrect choices, denoted as \( n \), that must be eliminated so that the expected value (EV) of guessing exceeds 2 points (the score for leaving the question blank). ## 2. Expected Value Calculation ### Choices After Eliminating Incorrect Answers If \( n \) incorrect choices are eliminated, the number of choices remaining for guessing becomes: \[ \text{Choices remaining} = 5 - n \] ### Probabilities for Remaining Choices 1. **Probability of guessing correctly:** The probability of selecting the correct answer from the remaining choices is given by: \[ P(\text{correct}) = \frac{1}{5-n} \] 2. **Probability of guessing incorrectly:** The probability of guessing incorrectly is the number of incorrect answers left divided by the total remaining choices: \[ P(\text{incorrect}) = \frac{(5 - n) - 1}{5 - n} = \frac{4 - n}{5 - n} \] ### Expected Value When Guessing The expected value (EV) when guessing can be calculated using the formula: \[ \text{EV}_{\text{guess}} = \left(\text{Probability of correct} \times \text{Score for correct}\right) + \left(\text{Probability of incorrect} \times \text{Score for incorrect}\right) \] Substituting the probabilities and scores gives: \[ \text{EV}_{\text{guess}} = \left(\frac{1}{5-n} \times 7\right) + \left(\frac{4-n}{5-n} \times 0\right) = \frac{7}{5-n} \] ### Expected Value When Leaving Blank The expected value for leaving the question blank is simply: \[ \text{EV}_{\text{blank}} = 2 \] ## 3. Setting Up the Inequality To determine when guessing is advantageous, we need to set up the following inequality: \[ \text{EV}_{\text{guess}} > \text{EV}_{\text{blank}} \] Substituting the expected values gives: \[ \frac{7}{5-n} > 2 \] ## 4. Solving the Inequality ### Step 1: Clear the Fraction Multiply both sides by \( 5 - n \) (assuming \( 5-n > 0 \)): \[ 7 > 2(5-n) \] ### Step 2: Expand the Right Side Distributing the 2 results in: \[ 7 > 10 - 2n \] ### Step 3: Rearranging the Inequality Rearranging the terms leads to: \[ 7 - 10 > -2n \] \[ -3 > -2n \] ### Step 4: Dividing by -2 When dividing by a negative number, we must reverse the inequality: \[ \frac{3}{2} < n \] This simplifies to: \[ n > 1.5 \] ### Step 5: Finding Integer Solutions Since \( n \) must be an integer (you cannot eliminate half a choice), the smallest integer satisfying this condition is: \[ n \geq 2 \] ## 5. Conclusion To summarize, you must eliminate at least **2 choices** out of the original 5 for guessing to be advantageous. This means that the expected value of guessing will be greater than leaving the question blank. ### Summary of Findings - **Minimum choices to eliminate:** 2 - **When guessing is advantageous:** Expected value of guessing exceeds 2 points. This conclusion ensures that by removing 2 incorrect choices, the probability of guessing correctly improves significantly, justifying the decision to guess rather than leave the answer blank.