How do you generate a random number in real life from $1$ to $k$. $k\leq 4$

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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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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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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.