Expected value for multiple-choice tests picking the best result out of multiple tests.

Consider a multiple-choice (a, b, c, d) test with 10 questions

• If you choose answers randomly, the expected grade is $$25\%$$
• If you fill out two tests randomly and pick the best grade, the expected grade is $$33\%$$.
• If you take the best among 10 random tests, the expected grade is $$\approx47\%$$.
• If you take the best among 100 random tests, the expected grade is $$\approx62\%$$.
• If you take the best among 1000 random tests, the expected grade is $$\approx72\%$$.
• If you take the best among 10000 random tests, the expected grade is $$\approx82\%$$.
• But on new questions, the random choice accuracy is still $$25\%$$.

I am confused about how $$33\%$$ and the rest of the expected grades are calculated. It is obvious the problem assumes questions are iid. For a single test, it is a Bernoulli trial. the number of questions isn't a factor here, so $$E=\frac{1}{4}$$.

It is unclear the setup of the experiment, I assume we are in a $$\binom{N}{2}$$ scenario, where $$N$$ is the number of all combinations, in this case, $$4^{10}=1048576$$, which mean the random answers cannot be the same. However, this does not explain how $$33\%$$ is computed.

This problem came from the context of optimization bias, but no formula was given.

• Is the problem missing some conditions? If so, what are the missing conditions, and how to make the problem more concrete?
• Which distribution does this question fit into?
• How these expected values are computed?
• The calculation of the expectation can be found with powers of the cumulative distribution function. For example in R, you can use sum((1-pbinom(0:10,10,1/4)^2))/10 to get 0.3256158 rounding to $33\%$ and changing the 2 to however many tests there are will give you the other results Mar 11, 2022 at 14:07
• @Henry Could you explain a little bit of where this equation comes from? And why is it /10 instead of /11, since there are 11 possible outcomes 0:10? Mar 12, 2022 at 10:21
• If you do not divide by $10$ then you get the expected number of correct answers in the best attempt, $3.256158$ in the two test case with $10$ questions, so dividing by the number of questions turns this into a proportion. Mar 12, 2022 at 10:33
• The rest of the code uses a complementary CDF-related method of calculating the mean of a non-negative integer random variable $\mathbb E[X] = \sum_n \mathbb P(X>n)$ and combining this with the fact that the probability none of the tests exceed a given mark is the the power of the probability a given test does not exceed it Mar 12, 2022 at 11:02

Each question is a Bernoulli random variable. Each test is a Binomial random variable. We are looking at the maximum of a number of tests. That is, if $$B_i\sim\text{Binomial}(10, 0.25)$$ are i.i.d. then the values given are $$\mathbb{E}(\max\{B_1,B_2,\dots,B_m\})$$ for different values of $$m$$ ($$1,2,10,100,1000,10000$$).
The only assumptions made are that we are filling out every question on every test independently and uniformly at random (each option has a $$25$$% chance of being picked).
The random variable $$\max\{B_1,B_2,\dots,B_n\}$$ is very messy: see here for the distribution function (in the slightly more general case that the success probability $$p$$ can vary).
As we have been given $$n=10,p=0.25$$ and values of $$m$$ explicitly, you can do explicit calculation to find exact values of the expectation. Or, you might just run each experiment some large number of times (on a computer) and take the mean. (For instance, for $$m=10$$, you could run $$10^6$$ trials where you take $$10$$ i.i.d. $$\text{Binomial}(10,0.25)$$ r.v.s and find the maximum of those $$10$$, and compute the mean across all trials.)
To give an example of a calculation: in the case $$m=10$$, we see that the probability of scoring $$\le 3$$ is the probability that each of the ten tests have a score $$\le 3$$. One test has a score of $$\le 3$$ w.p. $$0.25^{10}+10\times 0.75\times 0.25^9+{10\choose 2}\times 0.75^2\times 0.25^8+{10\choose 3}\times 0.75^3\times 0.25^7$$: call this number $$p$$. Then the probability that this occurs $$10$$ times without fail is $$p^{10}$$. We could do similar calculations to get the probability of scoring $$\le 0, \le 1, \dots, \le 9$$ and then $$\mathbb{P}(\text{we score}\ i\ \text{exactly})=\mathbb{P}(\text{we score}\le i)-\mathbb{P}(\text{we score}\le i-1)$$.