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Continuing from this question, except I've changed the statement so that no two questions have the same answer.

You are given a multiple choice exam. It has twelve questions, and each question has four possible answers labelled (a)-(d).You didn't study for the exam at all, so you might as well just guess the answer to each question. But you do have one important piece of information: the exam was designed so that no two consecutive correct answers have the same label. So if one answer is (c), the next one cannot be (c).

What strategy should you adopt to maximize the probability that you get

  1. at least half the questions correct,
  2. at least one question correct.

I'm asking this question because I really have no intuition on how these two questions are that different from the previous (linked) question.

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    Remark: take a simpler problem, say there are just two questions and the constraint is that they must have different answers. Then you should guess the same answer twice. That has a $\frac 12$ chance of being right once. Absent the constraint, you have only a $\frac 7{16}$ chance of being right at least once (though of course you might be right both times). The constraint tends to mix the answers up...so guessing a single letter has a higher chance of hitting at least once. – 2017-01-23
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    Can we make the assumption that it's extremely unlikely that you've forgotten *everything* on the exam? – 2017-01-23
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    @Travis If you did not study at all, and you did not learn anything at all, it's very reasonable to assume that you did not forget anything at all ... you can't lose what you don't have, after all :) – 2017-01-23
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    I have to agree with lulu: I'd rephrase the question. It's easier to develop an intuition if you look at the case when no two consecutive correct answers are the same. – 2017-01-23
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    "So if two consecutive correct answers are (c)...": this is invalid in the new question, and you should change it. I would have changed it myself, but the obvious change is just to remove the whole sentence, which I didn't want to do unilaterally. – 2017-01-23

1 Answers 1

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To maximize the chance of getting at least half the answers right, answer a,b,c,d,a,b,c,d,a,b,c,d. To maximize the chance of getting at least one answer right, answer a,a,a,a,a,a,a,a,a,a,a,a.

As has been noted elsewhere, every strategy will yield the same expected number of correct answers (one quarter of 12, i.e. 3). If you ensure that every answer of yours equals the preceding $3$ as often as possible, you decrease the probability of it being right if they are right - and by the same token then also the probability of it being wrong if they are wrong. Thus, you increase the probability your score is closer to the average, and in particular that it's at least $1$.

Conversely, if you ensure that every answer differs from the preceding $3$ as often as possible, you decrease the probability of it being wrong if they are right, and decrease the probability of getting just an "average" result - you are more likely to get an improbably bad result, but also an improbably good one such as getting at least $6$ answer right!

It's easier to reason about this type of problem looking at the case where the answer to the first and second questions are guaranteed to be different, and so are those to the third and fourth, to the fifth and sixth etc. and to ask what's the best strategy to avoid getting them all wrong, or to succeed in getting them all right.

If you answer all questions randomly, the probability of getting any given pair of answers both wrong is $(1-1/4)(1-1/4)=9/16$, and the probability of getting all answers wrong is then $(9/16)^6=0.03167...$. If you give all identical answers, the probability of getting both answers of a pair wrong is $(1-1/4)(1-1/3)=1/2$, which is less than $9/16$ obtained above! The probability of not getting a single answer right is $(1/2)^6=0.01562...$, or about half that with random answers. But of course, getting them all right is impossible, since if one is right, the other of the pair is certainly wrong.

If you answer all questions randomly, the probability of getting any given pair both right is $(1/4)(1/4)=1/16$. If you give different answers, however, the probability becomes a slightly larger $(1/4)(1/3)=1/12$. Note that the probability of getting the second right is $1/3$ if the first is right and the two are different! It turns out that the probability of getting all answers right is tiny in both cases, but in the second case it's $(4/3)^6=5.61...$ times larger. But you could also show that with all answers different, the probability of getting all wrong increases compared to the random baseline.