Study Document

By Niraj Kumar (Founder & Educator, BR DOT HUB) • Category: EXAM NOTES • Reading Time: 8 Mins • Updated: September 2026

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Date Start: 01 Oct 2026, 08:00 AM

Published: true

Mathematical Risk vs Reward Optimization for 1/3 Negative Marking in CBT

1. Abstract & Statistical Foundation

Modern competitive CBT examinations (e.g., RRB NTPC, SSC CGL, BPSC) incorporate negative marking ratios, predominantly 1/3 (-0.33 mark penalty for an incorrect response vs +1.0 mark for a correct response). This empirical study models decision thresholds for question abandonment vs strategic elimination.

2. Expected Value Model ($E(X)$)

Let $n$ be the number of choices ($n = 4$).

Let $k$ be the number of options conclusively eliminated by the candidate ($0 \le k \le 3$).

The probability of selecting the correct choice uniformly at random among the remaining $4 - k$ options is:

$$P(\text{correct}) = \frac{1}{4 - k}$$

$$P(\text{incorrect}) = \frac{3 - k}{4 - k}$$

The Expected Value formula:

$$E(X) = 1.0 \times \frac{1}{4 - k} - 0.333 \times \frac{3 - k}{4 - k}$$

Evaluation of Expected Value Matrix:

  • $k = 0$ (Pure Random Guessing):
  • $$E(X) = 1.0(0.25) - 0.333(0.75) = 0.250 - 0.250 = 0.00\text{ marks}$$

    Decision: Zero expected advantage. Indiscriminate guessing introduces high variance without mathematical gain.

  • $k = 1$ (One Distractor Eliminated - 3 options remain):
  • $$E(X) = 1.0(0.333) - 0.333(0.667) = 0.333 - 0.222 = +0.111\text{ marks}$$

    Decision: Statistically positive expectation. Attempt recommended over large test samples.

  • $k = 2$ (50-50 Choice - Two Distractors Eliminated):
  • $$E(X) = 1.0(0.500) - 0.333(0.500) = 0.500 - 0.167 = +0.333\text{ marks}$$

    Decision: Strong statistical mandate to attempt. Across 30 such questions, expected net gain is $+10.0$ marks.

    3. Operational Rule for Candidates

  • If $k = 0$: Abandon immediately. Never guess blindly.
  • If $k = 1$: Attempt if aggregate attempts fall below estimated category cutoff.
  • If $k = 2$: Always attempt. The mathematical odds yield consistent positive returns.
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