What is the ideal number of guesses to maximize your marks in MCQs?

[kshitij]

Homework Statement

A test has 60 questions. Each question has 4 options with only one correct answer and their marking scheme is, +4 for correct response, 0 for no response and -1 for incorrect response. A student knows the correct answer to only 45 questions. What is the ideal number of guesses the student should make, such that the probability for the student to benefit/profit (i.e, score 180 or more marks) from these guesses is maximum?

Relevant Equations

Basic Probability Formulas

To approach this, I first assumed the case when the students attempts all the remaining questions.

Probability that they gain 4 marks for a guess = $\frac 1 4$
Probability that they lose 1 for a guess = $\frac 3 4$

Now let us say the number of correct guesses = $r$

Now we should have at least 3 correct guesses,
because in this case they will score
45⋅4+3⋅4-12⋅1=180+0 marks

So our required probability = $\sum_{r=3}^{15}\binom {15} r \frac 1 4^r \frac {3} {4}^{15-r}$

this gives us

So, there is approx 76% chance that they can benefit from attempting all the remaining questions.

Now, just to get a better idea of the situation, let's consider another scenario where the student knew the correct answers to 50 questions.
Here it might seem that if the student knew more,
then the risk of attempting rest of the questions should be lower.

For this is the case, our limiting value is 2 correct guesses as 2⋅4-8⋅1=0,

Surprisingly, here we have only 75% chance to benefit!
So, its actually less than the previous case. So, this definitely doesn't work as I thought it should.

For now, let's stick to our original question,

In the above cases I assumed that the student attempted all the questions which they couldn't solve,
but they don't necessarily have to, so my answer is incomplete.

Now, this question is not out of any textbook so I don't even know that whether this is a valid question,

Of course, even though I don't have an idea that which cases will have a higher probability, I can simply just do all the cases one by one, i.e., we can assume that they attempts all the questions and then find its probability of profit and then similarly we can do this by assuming that they attempted only 14 and then 13... and so on. And Finally we can compare out of all these cases which one has the highest probability of profit.

But I wanted a more generalized approach,

First, because this way we can extend this question to lots of other cases like, what happens when the student knew only 40 out of 60 questions?
We can still do the above process here, but there must be some way to form a function which could altogether give us the probability for all such possible cases.

Second reason is because what I actually want to know is how to maximize your marks in MCQs?
And to answer this question we will have to consider the marks scored by the student as well.
As in the original (made up) question, I only talked about "benefit/profit". Which means that even if the student scored 180 marks or 181 marks, this is considered as profit.
But that does not "maximize" the student's marks.

But again to answer such questions, one doesn't have a definite answer. All we can have is probabilities. This becomes a much harder problem.

So, in the end, I believe that if we could answer the simpler problem (the one in the homework statement), and we could generalize this situation, then maybe we could have a better idea on how to maximize our marks by comparing the situations, this time by taking both the marks scored in a case and the probability to score that marks in that case, into account simultaneously.

This way we could finally say that, yes, even though the probability to "X" marks is low, but the profit here is so high that it should be the best situation!

P.S., I know that the case when the student scores 180 marks is not actually a profit, but I considered this as benefit/profit because if the student doesn't lose marks, they can take this risk and attempt even the questions they couldn't solve. Its a win-win situation for them.

Edit: The question should also include that at least one guess is made by the student.

 

[PeroK]

Homework Statement:: A test has 60 questions. Each question has 4 options with only one correct answer and their marking scheme is, +4 for correct response, 0 for no response and -1 for incorrect response. A student knows the correct answer to only 45 questions. What is the ideal number of guesses the student should make, such that the probability for the student to benefit/profit (i.e, score 180 or more marks) from these guesses is maximum?
Relevant Equations:: Basic Probability Formulas

To approach this, I first assumed the case when the students attempts all the remaining questions.

Probability that they gain 4 marks for a guess = $\frac 1 4$
Probability that they lose 1 for a guess = $\frac 3 4$

Now let us say the number of correct guesses = $r$

Now we should have at least 3 correct guesses,
because in this case they will score
45⋅4+3⋅4-12⋅1=180+0 marks

So our required probability = $\sum_{r=3}^{15}\binom {15} r \frac 1 4^r \frac {3} {4}^{15-r}$

this gives us
View attachment 278087
So, there is approx 76% chance that they can benefit from attempting all the remaining questions.

I don't understand the question.

First, if the student has 45 questions correct, then he/she already has 180 marks, so they are certain to score the required 180 marks with no guesses. The optimum number of guesses, therefore, should be zero. But, I suspect this is not what the question setter intended.

What is the question asking? To optimise the probability of getting strictly more than 180 marks?

What you have calculated is the probability that they answer 3 or more questions correctly if they guess the remaining 15 questions, hence either benefit or break even by guessing all 15. How do you know you can't do better than this by guessing fewer questions? And, you can do better by not guessing at all.

 

[mitochan]

Taking a look at anyone from 15 questions, expectation value of score by guessing is
$$+4*\frac{1}{4}-1*\frac{3}{4}=\frac{1}{4}>0$$
So guess is better than no response. we should guess for all the 15 questions to get expected score
$$\frac{1}{4}*15=3.75$$

 

[PeroK]

Taking a look at anyone from 15 questions, expectation value of score by guessing is
$$+4*\frac{1}{4}-1*\frac{3}{4}=\frac{1}{4}>0$$
So guess is better than no response. we should guess for all the 15 questions to get expected score
$$\frac{1}{4}*15=3.75$$

The question does not ask to optimise the expected score, but to optimise the probability of getting 180+.

I think the question should be to optimise the probability of getting 181+.

 

[kshitij]

I don't understand the question.

First, if the student has 45 questions correct, then he/she already has 180 marks, so they are certain to score the required 180 marks with no guesses. The optimum number of guesses, therefore, should be zero. But, I suspect this is not what the question setter intended.

What is the question asking? To optimise the probability of getting strictly more than 180 marks?

What you have calculated is the probability that they answer 3 or more questions correctly if they guess the remaining 15 questions, hence either benefit or break even by guessing all 15. How do you know you can't do better than this by guessing fewer questions? And, you can do better by not guessing at all.

Originally the intent of this question was to find whether its better to guess or to leave no response for the remaining questions. Thus, while framing the question I always had in my mind that the student will guess even though the net profit after that is zero.

So, yeah the question should also include that number of guesses should be greater than or equal to zero.

 

[kshitij]

How do you know you can't do better than this by guessing fewer questions?

I did say that my solution is incomplete because I didn't account for the cases where they guessed fewer than 15 questions.

 

[PeroK]

So, yeah the question should also include that number of guesses should be greater than or equal to zero.

It can't be otherwise.

Originally the intent of this question was to find whether its better to guess or to leave no response for the remaining questions. Thus, while framing the question I always had in my mind that the student will guess even though the net profit after that is zero.

First, the net profit is not zero, but $+1/4$ for every guess. You would need only +3 for a correct answer and -1 for a wrong answer to have zero net profit.

To optimise your score, therefore, you guess all remaining questions.

If we imagine, however, that to get an A you need 181+ (and you need/want to get an A), and you have only 45 x 4 = 180 points, then you are forced to guess. Then it's an interesting and unsual question. Do you guess all 15 or is there a better strategy?

 

[PeroK]

I did say that my solution is incomplete because I didn't account for the cases where they guessed fewer than 15 questions.

It's not the solution in that case!

 

[kshitij]

Do you guess all 15 or is there a better number to guess?

You don't necessarily need to guess all 15. I mentioned that.
I assumed that that student guessed all 15 because then I knew what to do for this case.

It's not the solution in that case!

It is not the desired solution but one can reach an answer if they want with this method.
As I said,

Homework Statement:: A test has 60 questions. Each question has 4 options with only one correct answer and their marking scheme is, +4 for correct response, 0 for no response and -1 for incorrect response. A student knows the correct answer to only 45 questions. What is the ideal number of guesses the student should make, such that the probability for the student to benefit/profit (i.e, score 180 or more marks) from these guesses is maximum?
Relevant Equations:: Basic Probability Formulas

Of course, even though I don't have an idea that which cases will have a higher probability, I can simply just do all the cases one by one, i.e., we can assume that they attempts all the questions and then find its probability of profit and then similarly we can do this by assuming that they attempted only 14 and then 13... and so on. And Finally we can compare out of all these cases which one has the highest probability of profit.

 

[PeroK]

It can't possibly be 14 guesses, because you still need 3 correct answers. And that's less likely than it is with 15 guesses.

The optimum strategy is stil zero guesses, because you are certain to get 180. As soon as you start guessing, the probability cannot be 100%. You need to change the question to 181+. What you have is a non-question.

 

[kshitij]

It can't possibly be 14 guesses, because you still need 3 correct answers. And that's less likely than it is with 15 guesses.

The optimum strategy is stil zero guesses, because you are certain to get 180. As soon as you start guessing, the probability cannot be 100%. You need to change the question to 181+. What you have is a non-question.

Think of this as the student in the middle of this test.

You did all you could and you're sure that you got 45 of these questions definitely correct.
But now your thinking, what should I do with the remaining question,
Should I attempt some and leave some?
Should I attempt all and leave none? or
Should I leave all and attempt none?

Of course statistically when you'll leave all you'll get exactly 180 marks (with a probability of 100%).

But if I told you that if you attempted 5 of the remaining question and left 10, you have a probability of 80% to score 220 marks (I don't know the exact numbers, this is just an example), then what will you do?

Well I don't know about you but if those are the exact numbers, I would definitely take that risk!

That is what I'm asking, what are those exact numbers?

That is why I want you to assume that you aren't willing to leave all of those remaining questions, you want to take risk. But how should you attempt your remaining responses such that you risk is minimum and reward is maximum?

 

[PeroK]

Think of this as the student in the middle of this test.

You did all you could and you're sure that you got 45 of these questions definitely correct.
But now your thinking, what should I do with the remaining question,
Should I attempt some and leave some?
Should I attempt all and leave none? or
Should I leave all and attempt none?

Of course statistically when you'll leave all you'll get exactly 180 marks (with a probability of 100%).

But if I told you that if you attempted 5 of the remaining question and left 10, you have a probability of 80% to score 220 marks (I don't know the exact numbers, this is just an example), then what will you do?

Well I don't know about you but if those are the exact numbers, I would definitely take that risk!

That is what I'm asking, what are those exact numbers?

That is why I want you to assume that you aren't willing to leave all of those remaining questions, you want to take risk. But how should you attempt your remaining responses such that you risk is minimum and reward is maximum?

That's a different question. I answered that in post #7 above. You get a $+1/4$ net profit for every question you guess. The more you guess the better, on average.

The way to make the question interesting is to have targets - like 181 for an A; or, perhaps a penalty if you get 175 or less. Then it's about the probability of hitting these targets. This is the basis of "derivatives" in the trading markets.

In your question: if all you need is 180 for an A, then why take any chances by guessing? If you need 181, then you must take a chance by guessing.

 

[kshitij]

Think of this as the student in the middle of this test.

You did all you could and you're sure that you got 45 of these questions definitely correct.
But now your thinking, what should I do with the remaining question,
Should I attempt some and leave some?
Should I attempt all and leave none? or
Should I leave all and attempt none?

Of course statistically when you'll leave all you'll get exactly 180 marks (with a probability of 100%).

But if I told you that if you attempted 5 of the remaining question and left 10, you have a probability of 80% to score 220 marks (I don't know the exact numbers, this is just an example), then what will you do?

Well I don't know about you but if those are the exact numbers, I would definitely take that risk!

That is what I'm asking, what are those exact numbers?

That is why I want you to assume that you aren't willing to leave all of those remaining questions, you want to take risk. But how should you attempt your remaining responses such that you risk is minimum and reward is maximum?

If you followed my solution,
you'll see that if I attempted all those remaining questions, then there is a 76% that I will score 180 or more. Of course I can also score less than 180 with a chance of 24%, But, if you look closely, you'll see that, 76% is the chance for you to score anywhere between 180 and 240, and 24% is the chance of scoring anywhere between 165 and 180.

Now again, I don't know about you, but I like those numbers and I will definitely attempt all the remaining questions

 

[kshitij]

That's a different question. I answered that in post #7 above. You get a $+1/4$ net profit for every question you guess. The more you guess the better, on average.

The way to make the question interesting is to have targets - like 181 for an A; or, perhaps a penalty if you get 175 or less. Then it's about the probability of hitting these targets. This is the basis of "derivatives" in the trading markets.

In your question: if all you need is 180 for an A, then why take any chances by guessing? If you need 181, then you must take a chance by guessing.

What if I told you that the test is not for grades but for ranks.

That means you don't even know whether 180 is a good score or bad. That depends on how others perform as well.

So, in this case the more you score, the better.

 

[PeroK]

PS did you know that you can do cumulative binomial distributions in MS Excel? For example:

=BINOMDIST(2, 15, 0.25, TRUE)

Will give you the probability of getting 0, 1 or 2 successful outcomes in 15 trials, each with a probability of 0.25 of success. This is the complement of your answer in your original post.

(Putting FALSE in the last parameter gives you the probability of exactly 2 successful outcomes.)

 

[PeroK]

What if I told you that the test is not for grades but for ranks.

That means you don't even know whether 180 is a good score or bad. That depends on how others perform as well.

So, in this case the more you score, the better.

Then you guess all 15. You'll average nearly 184 marks. That's a no-brainer, as they say!

 

[kshitij]

PS did you know that you can do cumulative binomial distributions in MS Excel? For example:

=BINOMDIST(2, 15, 0.25, TRUE)

Will give you the probability of getting 0, 1 or 2 successful outcomes in 15 trials, each with a probability of 0.25 of success. This is the complement of your answer in your original post.

(Putting FALSE in the last parameter gives you the probability of exactly 2 successful outcomes.)

That is very cool. I was thinking that there must be some site like that as well like the integration calculator.

 

[PeroK]

That is very cool. I was thinking that there must be some site like that as well like the integration calculator.

To go back to your question. The other possibilities are 5 guesses, where you need only 1 correct, and 10 guesses, where you need only 2 correct. Interestingly, 5 is better than 10, but then 15 is very slightly better than 5 - so you were right after all!

 

[kshitij]

Then you guess all 15. You'll average nearly 184 marks. That's a no-brainer, as they say!

But how are you sure that attempting 5 and leaving 10 will not give a better average?
Or maybe some other cases where you didn't attempt all of them.

 

[kshitij]

To go back to your question. The other possibilities are 5 guesses, where you need only 1 correct, and 10 guesses, where you need only 2 correct. Interestingly, 5 is better than 10, but then 15 is very slightly better than 5 - so you were right after all!

But there are still other possibilities

 

[kshitij]

Interestingly, 5 is better than 10

Also how did you conclude that so quickly?

I think I am missing something.

 

[PeroK]

But there are still other possibilities

Those are the only three candidates for the best probability. 5 is "obviously" better than 1-4 and 10 is obviously better than 6-9 etc.

 

[PeroK]

Also how did you conclude that so quickly?

I think I am missing something.

I calculated the probabilities using Excel.

 

[kshitij]

Those are the only three candidates for the best probability. 5 is "obviously" better than 1-4 and 10 is obviously better than 6-9.

Can you please elaborate? I can't understand how did you reach those results?

 

[PeroK]

Can you please elaborate? I can't understand how did you reach those results?

The aim is not to lose. If you guess 1-5 questions, the you need only 1 correct answer. If you guess 6-10 questions, then you need 2 correct answers. And, if you guess 11-15 questions you need 3 correct answers.

Clearly, the best chance is the highest number of guesses in each case: 5, 10, 15.

If we go for the variation where the aim is to gain (i.e. get 181+), then the candidates are 4, 9 and 14 guesses. With 15 guesses, you need 4 correct.

 

[kshitij]

You'll average nearly 184 marks

One final question, How did you calculate that?

 

[kshitij]

One final question, How did you calculate that?

Let's say if the situation was, that I knew 40 out 60 question, then what is the average marks now if I attempt all?

Also this way I'm getting a feeling that even if I knew 0 out of 60 questions, I would still score some positive marks?

 

[PeroK]

One final question, How did you calculate that?

The basic point is that with +4 for a correct answer (probability $1/4$) and -1 for a wrong answer (probability $3/4$), the expected gain from a guess is $+1 - 3/4 = +1/4$.

The expected gain from 15 guesses is $15/4$.

 

[PeroK]

Also this way I'm getting a feeling that even if I knew 0 out of 60 questions, I would still score some positive marks?

With that scoring system, yes. If you guessed all 60 questions you would expect to score 15 marks on average.

It would be a bit sad, therefore, if a genuine student scored less than 15 points. I'm sure it's happened!

 

[kshitij]

The basic point is that with +4 for a correct answer (probability $1/4$) and -1 for a wrong answer (probability $3/4$), the expected gain from a guess is $+1 - 3/4 = +1/4$.

The expected gain from 15 guesses is $15/4$.

Okay, yes know I get it!

Also this way I'm getting a feeling that even if I knew 0 out of 60 questions, I would still score some positive marks?

yes I would end up with some positive average marks!

But why this isn't as simple as it looks is because it assumes that there is an equal distribution in the answer key, i.e., these are 15 questions whose answer is A, 15 whose answer is B and so on.

But in real life that isn't the case.

Me and my friends have once discussed this point before when we were making fun of someone who score 3 in that test

We said to him, "bruh, why didn't you just attempted all the responses as B "

But, then we realized, its more complicated than that.