Does always choosing C improve multiple-choice guessing?

Your math is correct. The expected count under uniform distribution for each answer choice on an $n$-question test is $n/4$, with standard deviation $\sqrt{n(0.25)(0.75)} = \sqrt{3n}/4$. For $n=40$: $\mu = 10$, $\sigma \approx 2.74$. Your observed 13 is $z \approx 1.1$, $p \approx 0.27$. Not significant. **But theres a psychological bias:** When constructing tests, teachers may unconsciously put correct answers in the middle (option C) because: 1. They tend to avoid A (too obvious) and E (suspicious) 2. The middle feels "balanced" 3. They may want to avoid patterns that test-takers can exploit **The "always guess C" strategy** is actually not optimal for random guessing either. If answers are random, your expected score is $n/4$ regardless. But if test-makers DO have a bias toward C, then guessing C is optimal. **A better strategy:** Look at past exams from the same teacher. If theres a measurable bias, exploit it. If not, pick randomly. **For best results:** Dont guess randomly. Eliminate obviously wrong answers first, then guess from the remaining. With 2 options left, you have 50% chance instead of 25%.