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I wanted to share a cute probability exercise I came up with.

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 three consecutive correct answers have the same label. So if two consecutive correct answers are (c), the next one cannot be (c).

What strategy should adopt to maximize the expected number of questions you answer correctly on the exam, and using this strategy what is the expected number of questions you will answer correctly?

I've posted a continuation of this question here, related to trying to maximize the probability of other "optimal" outcomes.

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    This is only a guess, but I think that this important piece of information is irrelevant, and the optimal strategy is simply choosing randomly the options. Paradoxically, you may answer (a) at all questions and still pass the exam. – 2017-01-23
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    It's not clear what you mean with "optimum" strategy. For example, you can show that to maximize the probability of getting right (at least) half the answers and that of getting right (at least) one answer you need two different strategies! – 2017-01-23
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    If you have trouble wrapping your head around this, start by think about a test with n>=2 qeustions and 2 possible answers where 2 consecutive answers cannot be the same. – 2017-01-23
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    @Anonymous, yeah I was just thinking about maximizing the *score* on the test, so the expected value of the number of correct answers. I really didn't think that could be ambiguous. Also, I don't have any intuition as to why the strategy for getting at least one question right would be *that* different. So I've posted [another question here on that](http://math.stackexchange.com/q/2110742/167197). – 2017-01-23
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    "Maximizing the score" is ambiguous because it doesn't say whether a strategy with a 1% chance of a perfect score and 99% chance of a zero score is better or worse than one with a 10% chance of 90% correct and 90% chance of 10% correct. Maximizing _expected_ score has a more well-defined meaning, but it turns out _every_ strategy is precisely as good as any other according to the precise meaning of expected score. – 2017-01-23

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Assuming that "optimum" just means maximizing the expected number of correct answers, this is one of those problems that is trivialized by linearity of expectation. For each individual problem, your probability of getting that problem correct is $1/4$, whatever choice you make. So your expected score on the whole test is $25\%$, no matter how you choose your answers. The correlations between the answers between the different questions don't matter if you're just looking at the expected value.

A more interesting and probably much harder question would be something like, "how can you maximize the probability that you get at least half the questions correct?".

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    Nice, although "best" might include more than EV. – 2017-01-23
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    Ah, I interpreted the question as asking just about expected value, since it said "using this strategy what is the expected number of questions you will answer correctly?" – 2017-01-23
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    I think your answer is good. Although it's sort of funny because in this case the optimum strategy is to do whatever – 2017-01-23
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    @JorgeFernándezHidalgo, yeah I was unsure exactly how to ask the question. But yeah, I want to maximize expected score (number of questions correct) on the exam. If you know how to edit the question to make that clear, I would appreciate it. – 2017-01-23
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    I don't have any intuition as to why the other question would be harder (or why it would be *that* different), so I've posted [another question here on that](http://math.stackexchange.com/q/2110742/167197). – 2017-01-23
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    @MikePierce: Well, my intuition is just that you can't use linearity of expectation and will have to get your hands dirty with understanding how the answers to different questions are correlated, so there probably won't be any simple trick that makes it easy. – 2017-01-23