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I've been in some multiple choice exams (4 choices, no penalty for incorrect answers) where I have $2$ minutes on the clock, and $10$ questions to go. According to probability, if I randomly chose one of the $4$ answers in each question, on expectation, I should get somewhere in the $2-3$ extra marks with a fairly good probability (Unless Karma is against me).

Now my question is, during an exam, with no access to a computer or a programming library, how can I efficiently and quickly generate a random number from $1$ to $4$?

I guess the question can also generalize for generating a number from $1$ to $k\leq 4$ in the condition that I can get rid of some of the choices in a question.

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    $\begingroup$ I recommend that you just write down whatever answer feels good to you and not worry about the quality of its randomness. If the answers themselves were ordered randomly, your selection method doesn't matter. If they weren't, it only matters if your lack-of-randomness interacts badly with the test-writer's lack-of-randomness (and it's just as likely to interact well). And you might subconsciously know a little more than you think you do, which could raise your score a bit. $\endgroup$ – Micah Mar 5 '18 at 23:21
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    $\begingroup$ You could toss a coin twice. TT = 1, TH = 2, Ht = 3, HH = 4. On the other hand, most teachers I've known try to randomize the answers, at least informally, so that a student who just picks 1 on all the questions won't get an undeserved great. In that case, just guess 1 gives you about 25%. $\endgroup$ – saulspatz Mar 5 '18 at 23:23
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    $\begingroup$ Assuming the teacher tries to vary the correct answer choices so that they are dispersed well: One thing to keep in mind is that guessing $A$ for every question is probably a safe choice to ensure you'll get at least a couple right. However, picking randomly would give you the potential for a higher score, but also the potential for a very low score. $\endgroup$ – Remy Mar 5 '18 at 23:24
  • $\begingroup$ If you truly don't have a clue (and if the correct answers are randomly distributed), then just take the first digit between $1$ and $4$ that appears on the exam sheet and cycle from there e.g. $2$-$3$-$4$-$1$-$2$-$\ldots$ $\endgroup$ – dxiv Mar 5 '18 at 23:31
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Given that the desired numbers are integers. Then you can define an operator say XOR and make a table to solve variations of $k_1$ XOR $k_2$ where $k_1$, $k_2$ belongs to $[1,4]$. They you can choose randomly afterwards.

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  • $\begingroup$ Of course, composing the table will take some time! Reminds me of Rimmer from Red Dwarf who spent so much time organising a time table for his revision, he never left enough time to revise. $\endgroup$ – jim Mar 5 '18 at 23:33
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To actually generate the number find some event that occurs without your interaction, example person in a certain row moves their pencil or second hand on the clock is odd/even. If you did something active like flipping a coin it would be counted as distracting and might bring a reprimand.

There are tests that allow calculators, if your question doesn't object to the use of one, you can randomly hit a key; many calculators have a mod() key and a random key also.

As for actually answering the question it's best to exclude answers you know are incorrect, increasing your chances beyond 25%.

When pressed for the last few seconds just tick away, it's sufficiently random since if it were not you could draw a straight line and get everything right or wrong.

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I would agree with @Remy. I believe that choosing the 3rd option as the correct one for all remaining questions is a better strategy. The reason I feel is that the answers seldom have a "random" pattern!

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Count the number of letters in a "randomly" selected portion of the question, and take $(count\mod k) + 1\ $ as your "random" number.

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