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I have six six-sided dice. Five of their faces are blank and identical. One face on each die contains the number 1,2,3,4,5, or 6. Suppose I roll them together, the output from this random event is an unordered set for example: {B,B,3,1,B,6} (blank side results, for our purposes, are indistinguishable.)

What is a simple way to use this result like an ordinary, fair six-sided die?

My first thought was to keep rolling until the only non-blank results are permutations of

{A,B,B,B,B,B}

{A,A,B,B,B,B}

{A,A,A,B,B,B}

{A,A,A,A,B,B}

{A,A,A,A,A,B}

{A,A,A,A,A,A}

where A is some repeated number 1-6 and B is blank.

This would work but it is not efficient. Obviously the many {B,B,B,B,B,B} results can’t be used (probably?) but, what about mixed results? For any result where one number is repeated more than the rest I could take the most repeated number as the result... but how can I interpret {1,2,1,2,4,B,6}?

Basically, I’m looking for a simple function that a human* could remember easily that maps the results of this random event to the numbers 1-6 with equal frequency.

*A table is another solution, but I would like fewer than 5 or 6 dead simple rules that use all of the results that are not {B,B,B,B,B,B}. If there is a way to use {B,B,B,B,B,B} that is even better.

a photo of the dice

futurebird
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    Why not place the six dice in a bag, pick one at random and read the number? Then there will be a $\frac{1}{6}$ probability for each outcome just like a fair die. – John Douma Mar 17 '18 at 02:39
  • That is a good out of the box solution. – futurebird Mar 17 '18 at 03:45
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    It’s only tangential to the question, but I’m curious about how you acquired these dice and why someone would ever manufacture them. Is the coloration significant? – gen-ℤ ready to perish Mar 17 '18 at 17:41
  • I bought them from aliexpress for about $1 shipping included. I bought them because I wanted to know what they are for. I still have no idea. – futurebird Mar 18 '18 at 02:25

2 Answers2

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Roll the dice and eliminate any that show up blank. Continue rolling until one die remains. If all of your remaining dice get eliminated in one throw, then reroll.

Edit (thanks Rahul): Eliminate whichever group (i.e., blank or numbered) is less numerous, provided the groups aren’t the same size.

It’s fair and it shouldn’t take long, because each die has a $5/6$ chance of being eliminated each throw.

gen-ℤ ready to perish
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    Small efficiency improvement inspired by Ross's answer: If fewer than half the dice show up blank, eliminate the others. –  Mar 17 '18 at 03:19
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You need to think about whether there is a way to pick one number out of the ones you see on top. Clearly if you get six blanks you need to roll again. If you get five blanks you can use the one number showing. If you get four blanks you can use the higher number showing counting the short way around, so $1$ is higher than $5$ and $6$ and $2$ is higher than $6$. If you get four blanks and two numbers that differ by $3$, reroll because it isn't easy to pick one of the two. If you get three blanks, take the highest of the ones that are neighboring, including having $1$ higher than $6$. That works for any three blanks except $135$ and $246$, which you reroll. If you get less than three blanks, do the preceding with the missing numbers.

This has you rerolling six blanks, $14,25,34, 135, 246$ and their complements, which is not many. I think you could come up with a rule to cut the rerolls a small amount, but at the cost of complexity.

The dominating contribution to the rerolls is when you get six blanks, which is $\left(\frac 56\right)^6$ slightly over $\frac 13$ of the time. You average about $1.5$ rolls to get each number.

Ross Millikan
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