Create 5 Equal Outcomes By Changing Markings On 6-sided Fair Die I found this question years ago in a Times Of India Newspaper Puzzle Column called Mindsport curated by Mukul Sharma but haven't found the answer yet:
A 6-sided fair die of course has 6 equally possible outcomes because there are 6 different markings on each side. Now I could easily change this to 3 equally possible outcomes simply by  making the same 3 markings twice (1,1,2,2,3,3) - I have changed the probability of a 1,2 or 3 occurring to 1/3. I can also change this to 2 equally possible outcomes by making the same 2 markings thrice (1,1,1,2,2,2) - I have changed the probability of a 1 or 2 occurring to 1/2. 
The question is: Can I change the probability of a 6-sided fair die to 1/5. That is, the outcome must be that the die gives out a 1,2,3,4 or 5 with equal probability of 1/5.
You can use any markings, formula, calculation etc. on each of the 6 sides. For example one side could say "Add the other 5 numbers" or "Take the square root of the number on the opposite side".
tl;dr: I need 5 equal outcomes from a fair 6-sided die using any markings you wish to make on the 6 sides.
Thanks!
EDIT:
Just to clarify, there should not be any "Throw Again" cases. I should be able to throw a 6-sided die and every single time it should give me one of 5 equally probable outcomes. Of course, your outcomes don't even have to be the numbers 1-5. As long as they are unique one can easily map your 5 unique outcomes to 1,2,3,4 and 5.

Paint faces 1 through 5 with the numbers 1 through 5.
Paint the 6th face with the math expression: ceil(rand() * 5),
where rand() produces rational numbers in the range (0, 1]

