Negative Marking and the Decision to Guess
A candidate is taking a computer-based admission test with four options per question. The marking scheme is +3 for a correct answer, −1 for a wrong answer, and 0 for an unattempted question. With 8 minutes left, six questions remain unattempted: Q1 and Q2, where the candidate can confidently eliminate two options; Q3 and Q4, where the candidate can eliminate one option; and Q5 and Q6, where the candidate has no useful idea. The candidate can attempt all six in the remaining time. A friend advises, “Never guess under negative marking.” The candidate wants to maximize expected marks but is also uneasy about a large negative swing, since a scholarship cutoff is close. The candidate’s current score is 2 marks above the estimated cutoff.
1. Compute the expected gain from guessing in each of the three categories. Decide which questions the candidate should attempt and evaluate the friend’s advice.
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Let p be the probability of a correct guess. For one question, expected score = 3p + (−1)(1 − p) = 4p − 1. The break-even value is p = 0.25.
Q1–Q2: two options eliminated, so p = 1/2. Expected score per question = 4(1/2) − 1 = +1. For two questions, expected gain = +2.
Q3–Q4: one option eliminated, so p = 1/3. Expected score per question = 4(1/3) − 1 = +1/3. For two questions, expected gain = +2/3.
Q5–Q6: no elimination, so p = 1/4. Expected score per question = 4(1/4) − 1 = 0. Guessing is neutral in expectation.
The friend’s rule is too absolute. Any guess with p > 0.25 has positive expected value. The candidate should definitely attempt Q1–Q4 and is indifferent on Q5–Q6. If the candidate’s utility is risk-neutral, leaving Q5–Q6 blank or guessing them gives the same expected total; attempting them adds variance without adding expected marks.
2. The candidate is risk-averse because the cutoff is close. Compare the risk of attempting only Q1–Q4 with attempting all six. Use the variance of the score change for one guessed question.
Read the worked answer
For one question with correct probability p, the score X is +3 with probability p and −1 with probability 1 − p. E[X] = 4p − 1. Also E[X2] = 9p + 1(1 − p) = 1 + 8p. Hence
Var(X) = (1 + 8p) − (4p − 1)2 = 16p(1 − p).
For p = 1/2, Var = 4. For p = 1/3, Var = 32/9 ≈ 3.56. For p = 1/4, Var = 3.
Attempting Q1–Q4: expected gain = 2(1) + 2(1/3) = 8/3 ≈ 2.67; variance = 2(4) + 2(32/9) ≈ 15.11, standard deviation ≈ 3.89.
Attempting all six adds two questions with p = 1/4: expected gain unchanged at 2.67, but variance increases by 2(3) = 6 to about 21.11, standard deviation ≈ 4.59.
The extra two guesses do not improve the expected score, but they widen the distribution and increase the chance of dropping below the cutoff. A risk-averse candidate should stop after Q1–Q4 unless there is a tie-break or other reason to prefer a chance of a higher score.
3. Suppose the marking scheme changes to +4 for correct and −1 for wrong. Recalculate the break-even probability and state how the decision changes.
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With +4 and −1, expected score = 4p − 1(1 − p) = 5p − 1. Break-even occurs at 5p − 1 = 0, so p = 0.20.
Now Q1–Q2 (p = 1/2) give 5(1/2) − 1 = +1.5 each. Q3–Q4 (p = 1/3) give 5(1/3) − 1 = +2/3 each. Q5–Q6 (p = 1/4) give 5(1/4) − 1 = +0.25 each.
All three categories now have positive expected value, including a blind guess. If the candidate is still risk-neutral, attempting all six is optimal. The break-even threshold has fallen because the reward for a correct answer has increased relative to the penalty. This shows that the “never guess” rule is scheme-dependent, not a universal principle.