Suppose a student can take a test at most thrice. Once a test concludes, the student has to either accept the score or reject it and take the next test (until he takes the second re-test or the third & final test). The university has a rule that he can take the third test only if he gets less than $40$ in the second one. Taking the second test has no such restriction. At what score in the first test should a student consider retaking to maximize the expected final score? Assume that the score always comes from $\text{Uniform}[0,100]$ each time.
The expected score is $50$ each time. The student will consider retaking if he gets $< 50$ in the second test, so if he gets $< 40$ in the second test, he will definitely re-take the test. There's a $60\%$ chance that he'll keep the (expected) score of $50$ and $40\%$ chance that he will re-take after the second test.
I don't know how to proceed from here. I am looking for a more verbal and less mathematical explanation.
Like you said, we know that $m\ge 50$, and the student should retake if their score is below 50 and they are allowed to. We are trying to find the probability density function $p(y)$. Let $y\in[0,100]$ be a score.
If $y\in [0, 40)$, then the only possible way to obtain a score of $y$ is first getting a score below $m$, and then getting a score below 40, and finally getting a score of $y$. Then $p(y)=\frac{m}{100} \frac{40}{100} \frac{1}{100}$ in this case.
If $y\in [40, m)$, then there are two ways to obtain a score of $y$:
Then $p(y)=\frac{m}{100}\frac{1}{100}+\frac{m}{100}\frac{40}{100}\frac{1}{100}$ in this case.
If $y\in [m, 100]$, then there are 3 ways to obtain a score of $y$:
Then $p(y)=\frac{1}{100}+\frac{m}{100}\frac{1}{100}+\frac{m}{100}\frac{40}{100} \frac{1}{100}$ in this case.
The expected score is then: $$ \begin{align*} \int_0^{100} y p(y)\, dy = -\frac{1}{200}m^2 + \frac{31}{50}m + 50. \end{align*} $$ It is easy to compute that $m=62$ maximizes the expected score.