Suppose you have to take a test with ten questions, each with four different options (no multiple answers), and a wrong-answer penalty of half a correct answer. Blank questions do not score neither positively nor negatively.

Supposing you have not studied specially hard this time, what's the optimal number of questions to try to answer so the probabilities to pass the exam (having at least five points).


Let's work this all the way through. Suppose you answer $n$ questions. Let $X$ be the number you get correct. Assuming $\frac{1}{4}$ chance of getting an answer correct, $X$ is binomial$(n,1/4)$. Let $Y$ be the actual score on the exam, including the penalty. Then $Y = X - \frac{1}{2}(n-X) = \frac{3}{2}X-\frac{1}{2}n$. To maximize the probability of passing the exam, you want to choose the value of $n$ that maximizes $P(Y \geq 5)$. This probability is $$P\left(\frac{3}{2}X - \frac{n}{2} \geq 5\right) \Longleftrightarrow P\left(X \geq \frac{10+n}{3}\right) = \sum_{k = \lceil(10+n)/3\rceil}^n \binom{n}{k} \left(\frac{1}{4}\right)^k \left(\frac{3}{4}\right)^{n-k}$$ $$ = \left(\frac{3}{4}\right)^n \sum_{k =\lceil(10+n)/3\rceil}^n \binom{n}{k} \frac{1}{3^k}.$$ This can be calculated quickly for the possible values of $n$. I get, via Mathematica,

5   0.000976563
6   0.000244141
7   0.00134277
8   0.00422668
9   0.00134277
10  0.00350571

Thus you maximize your probability of passing by answering eight questions.

  • $\begingroup$ Of course, it's extremely unlikely that you're going to pass no matter what your strategy is. Which, if you haven't studied, is as it should be. :) $\endgroup$ – Mike Spivey Nov 2 '11 at 16:43
  • $\begingroup$ If $8$ questions are answered at random, the expected number of correct answers is $2$ and so the expected exam score is $2 - 6\times \frac{1}{2} = -1$. Some days it doesn't pay to get out of bed (and go to the exam).... $\endgroup$ – Dilip Sarwate Nov 2 '11 at 18:04
  • $\begingroup$ Piggybacking off of @Dilip's comment, the expected score for any $n > 0$ is negative. So, if you haven't studied, you maximize your expected score by skipping the exam. (I believe some of my students have tried this strategy, now that I think of it...) $\endgroup$ – Mike Spivey Nov 2 '11 at 18:48

If you answer 5, you need them all right, so your chance (assuming $\frac{1}{4}$ chance of a correct answer) is $(\frac{1}{4})^5=\frac{1}{1024}$. If you answer 7, you need at least 6 right, with chance $\binom {7}{6}(\frac{1}{4})^6\frac{3}{4}+(\frac{1}{4})^7$, which I make to be about $\frac{1}{745}$. You can do trying 9, but I think you are in a world of hurt.


Let's say that you have a probability $p$ of getting the correct answer to each question. Then if you answer $n$ of the questions, the number that you answer correctly is distributed as a binomial random variable $X\sim B(n,p)$.

Your score on the exam, $Y$, is related to this. You score 1 for each correct answer and lose 1/2 for each incorrect answer, which means that

$$Y = X - \tfrac{1}{2}(n-X) = \tfrac{1}{2}(3X-n)$$

You are asking for the probability that $Y\geq 5$, which reduces to

$$\begin{align} P(Y\geq 5) & = P\left(\frac{3X-n}{2} \geq 5\right) \\ & = P\left(X \geq \frac{10+n}{3}\right) \end{align}$$

This probability can be easily calculated from the binomial density, or (cheating) using the following R code:

> f <- function(n,p) {pbinom((10+n)/3 - 0.001, n, p, lower.tail=FALSE)}
> plot(0:10, f(0:10, 0.25), type='l')

which generates this chart, from which you can see that you should answer eight questions:

[image upload not working at the moment, I'll come back and edit the image in later]


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