How do you generate a random number in real life from $1$ to $k$. $k\leq 4$ 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. 
 A: 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. 
A: 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.
A: 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!
A: Count the number of letters in a "randomly" selected portion of the question, and take $(count\mod k) + 1\ $ as your "random" number.
