A Happy Probability Question

Today in a literature class, we were given a 28 question, machine-graded, multiple choice, single correct answer test.
All questions had 4 possible choices.
We were instructed to answer only 25 out of the 28, as the teacher would only be grading out of 25, but wanted to give us a break.
Bull-hockey, said I. I’ll answer all 28, most likely get at least 3 wrong, and let the machine decide which three I skipped.
No no, said the prof. You must answer only 25. Otherwise you increase your chances of getting a good grade, and that wouldn’t be fair to the other students.
So, the queston: What is the difference in probabilities (of a perfect score, say) between these two systems? Also, in the average case, what is the difference in score? Is this a solvable problem?

The answer depends on a number of things. For example, it depends on how much you know about literature (i.e., how likely you are to answer a given question correctly). It also depends on how your quiz is graded if you do decide to answer all 28 questions.

Let’s assume for the moment that your odds of answering each of the 28 questions is equal (it’s an unrealistic assumption, but bear with me). Then if the grader just adds up the number of correct answers, and lowers the score to 25 if it was greater than 25, then yes you’d do better to answer all 28.

On the other hand, the grader may decide that if you answer more than 25 questions then only your first 25 answers count. In this case, answering all 28 questions doesn’t improve your situation. Mathematically speaking it doesn’t harm your situation either, but that’s only because of that assumption we made earlier that you have an equal chance of getting the right answer on each of the 28 questions. If in fact you know the answers to some questions better than others, then you’re better off only answering the 25 questions that you’re the most confident about. In other words, if you really know the answer to question 28 but are only guessing on question 3, then you don’t want the grader to count your answer for question 3 instead of your answer to question 28.

You can’t work out an exact numerical probability of a perfect score without more data or more assumptions, I’m afraid. In particular you need to know your odds of answering each particular question correctly (25%? 90%?).

If you choose the 25 questions to do “randomly” and
then select one of the 4 answers for each one “randomly”,
the probability of a perfect score is
If you select one of the 4 answers for each of 28 questions
“randomly”, then the probability that your score
is greater than or equal to 25 is
(14926)(1/4)^25(3/4)^3 + (1427)(1/4)^26*(3/4)^2

  • 28*(1/4)^27*(3/4) + (1/4)^28
    Since (14926)*(3/4)^3>1, the second probability is
    clearly larger.

Pretty tough to answer and not enough data.

If one assumes that there are at least 3 questions that you have no idea about with 3 choices each then chances are pretty good (80%) that you would get at least one extra mark just by guessing at the answers rather than not answering the toughest 3. Usually with multiple guess you can eliminate at least one wrong answer so that would increase the odds accordingly.

It depends very heavily on how the machine chooses the ones you skipped, as well as what assumptions we make about your probabilities of answering them correctly. If it will be kind enough to decide that you skipped 3 wrong answers, you are gaining an advantage over playing fair, given that you actually guessed on more than three questions, as suggested by your prof. If it will simply randomly pick 3 to throw out, it could be to your disadvantage, as it could throw out some of your known correct answers.
If we assume that you purely guess on every question, and it throws out three wrong:

you play fair - you answer 25, expect to get 6.25 correct.

you answer 28 - you expect to get 7 correct, it will throw out 3 wrong ones, still leaving you with 7 correct. Not fair.

Hmm…I should have known that would be important.
Using my previous test scores, I find I have a 76% chance of getting any single question correct.
Also, based on previous experience, the grader will count the test out of 25, no matter how many questions were answered.
From the random case calculations curiousgeorgeordeadcat gave, I can see I’m in some trouble :wink:
Now I’m just wondering how much…

Oh, and if you aren’t guessing one of 4 answers
“completely randomly”, but it is reasonable to assume
that your probability of getting any given question
correct is the same probability p, then you can
replace 1/4 by p above and 3/4 by 1-p. Then it
isn’t so obvious the second probability is larger.
But it is, since for the second method,
P(>=25 correct)>P(first 25 correct, any performance
on last 3)=p^25.

I just recalled you asked for the average scores.
With probability p of getting each question correct,
assuming your performance on one question is independent
of your performance on another, and assuming you are
equally likely to pick any subset of 25 questions from
the 28 for method one, your average scores are:
For method 1 (answering only 25):
For method 2 (answering all 28, getting the smaller
of your score or 25 as your total score):

Unless the other students all answer 28, and only you and a couple others answer only 25. The you’re the one who loses. Better answer all 28.

As for the probability, that’ll depend on how many questions you know the answer to, and how many you’re guessing on. To simplify, assume you have no clue on N of the questions, but are sure about the other 28-N questions.

If N <= 3, it makes no difference, since you know at least 25 answers.
If N = 4, then if you answer only 25, you have 1 chance in 4 of getting 25 correct, but if you answer 28, you have a chance of (3/4)^4 = 0.3164 of missing all four, so your chance of getting at least 1 right is 0.6836. Your expected number of correct answers (out of 25) has increased by 0.4336.

As N gets bigger, your expected increase in the number of correct answers should approach 0.75.