# Strategies to guess choices for multiple choice questions

Multiple choice questions (MCQs) are common in examinations over here in Singapore. A set of, say, $40$ questions are given to students, and each is accompanied with a list of $4$ choices of answer, $A, B, C$ or $D$.

Suppose a student did not study for a particular test. Out of desperation, he decides to "guess" a choice for each question to secure at least some points. He considers two possible strategies:

1. For each question, he picks a random choice as his answer.
2. He picks a random choice, say, $B$ and uses it for all his answer.

Assuming the correct answers are randomly distributed and his guess is completely random, which strategy would give a higher probability of securing more correct answers than the other?

Intuitively, the second strategy would give a higher probability of securing more marks. However, I am unable to come up with a mathematical proof (or disproof).

For the second strategy, we only need consider the probability that the correct answer is the choice that was picked. Under the assumption that the correct answers are randomly distributed, the probability of a particular choice being the correct answer is $\frac{1}{4}$. Hence, the student would get approximately $25\%$ of his guesses correct.

However, we can also use the same argument for the first strategy to say that the student would also get approximately $25\%$ of his guesses correct. This would imply that both strategies are equally effective, but I am pretty sure the second strategy is more effective.

EDIT: In order to prevent psychological factors from distorting reality, I decided to write a program (in C#) that simulates the aforementioned MCQ tests. I configured the program to simulate the taking of $1000$ randomly generated MCQ tests with $40$ questions each using both strategies.

It turns out that the percentage scores for both strategies have the same average ($\approx25\%$) and the same standard deviation ($\approx 6.84$)!

Code:

static Random rng = new Random((int)DateTime.Now.Ticks);

{
for (int i = 0; i < answers.Length; i++)
{
}
}

{
int score = 0;
{
{
score++;
}
}
return score;
}

{
int choice = rng.Next(4);
return answers.Count(x => x == choice);
}

• Your intuition is incorrect. – vadim123 Aug 27 '14 at 14:27
• If the generation of answers is random and uniform, then all guessing strategies give the same likelihood of success. If the guesser knows that some answers are favored over others (e.g. if you knew answer C will be correct $30\%$ of the time), then there is a profitable guessing strategy--guess that every answer is the most likely answer. – paw88789 Aug 27 '14 at 14:30
• From a mathematical point of view, the answer is simple: assuming the correct answers were chosen uniformly and at random, there is no difference between the two strategies. However, people are notoriously bad at simulating randomness. There's some evidence, for example, that over a large number of exams with choices A, B, C, and D, the B and C answers slightly predominate. – Rick Decker Aug 27 '14 at 15:10
• @RickDecker they often do this since they don't want the correct answer to be one of the extremes hence giving them the feel that it would be choosen easily, so we need to realise that our world is not purely real. – RE60K Apr 7 '15 at 17:31

Multiple choice questions (MCQs) are common in examinations over here in Singapore. A set of, say, $40$ questions are given to students, and each is accompanied with a list of $4$ choices of answer, $A, B, C$ or $D$.

Suppose a student did not study for a particular test. Out of desperation, he decides to "guess" a choice for each question to secure at least some points. He considers two possible strategies:

1. For each question, he picks a random choice as his answer.
2. He picks a random choice, say, $B$ and uses it for all his answer.

Assuming the correct answers are randomly distributed and his guess is completely random, which strategy would give a higher probability of securing more correct answers than the other?

The variable here is number of correct answers, i.e. let it be $X$.

1) $$\rm E[X]=\sum_{k=0}^{40} P(X=k)k=\sum_{k=0}^{40}\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}k=10$$ This is $25\%$. Now the standard deviation is: $$\rm \sigma[X]=\sqrt{\sum_{k=0}^{40}(k-10)^2\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}}=\sqrt{\frac{15}2}\approx2.73861$$

2) $$\rm E[X]=\sum_{k=0}^{40} P(X=k)k=\sum_{k=0}^{40}\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}k=10$$ This is $25\%$. Now the standard deviation is: $$\rm \sigma[X]=\sqrt{\sum_{k=0}^{40}(k-10)^2\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}}=\sqrt{\frac{15}2}\approx2.73861$$

Sorry, but do you know why I did that? Because $\rm P(X=k)$ is constant for both because of inter-independence of questions.I would further say that any strategy you device would all get you same results. I can't prove it, but you can call it my intution. I think like publishing this theory of independent results! :D

Both will give an the same probability of securing marks. (Both will give you an expected marks of 25%.)

The key thing to note is that each question is independent of another. This means that if the answer to the first question is B, then it does not make the answer to the second question more likely to be B (or any other choice, for that matter).

Thus, you can consider each question separately: the 25% chance of getting the first question correct does not affect the 25% chance that you'll get the second question correct, so whatever you choose for the first question isn't going to make a certain answer for the next question more correct than another.

Note: However, psychology of test setters might come into play, where test setters unknowingly avoid having consecutive answers that are both C, or other biases, and this might affect the resultant probability.

The Theoretical probability for both strategies is the same. If in your experience, you find one strategy works better than the other, you are talking about Empirical probability. The dice roll is probably the classic example. The theoretical probability of a 6 showing on a roll is $\frac{1}{6}$. However, if you roll a dice 6 times and you get 2 sixes, you may conclude that the empirical probablity is $\frac{2}{6}$ which is similar to your conclusion that one strategy seems to work better based on your experience.

Suppose four unprepared students sit for the test, say $W,X,Y,Z$. They decide that $Z$ will always guess $A$, $Y$ will always guess $B$, $X$ will always guess $C$, and $W$ will always guess $D$.

On any given test, it is unlikely that all four students will get the same score. At random one of them, say $Y$, may get slightly more than 25%, while another, say $W$, will get slightly less than 25%. Afterwards $Y$ will go on math.stackexchange with intuition that always picking $B$ is the winning strategy, because it leads to good results. But also $W$ will go on math.stackexchange with intuition that picking at random is a better strategy, because always picking $D$ leads to bad results.

OP is $Y$.

• Your general intuition is incorrect, as I have proved above. If your peers and you all pick different guessing strategies, then automatically some will do worse and some better than average. – vadim123 Aug 27 '14 at 14:58