Optimal number of answers for a test with wrong-answer penalty 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.
