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I will like to play a dice game but my dice is lost. Fortunately, I have a coin. Is it possible to replace a dice by the coin such that I will get equal probabilities for $1, 2,\ldots,6$ if I can't decline throws?

Without declining it would be easy:

0,0,0 -> 1
0,0,1 -> 2
0,1,0 -> 3
0,1,1 -> 4
1,0,0 -> 5
1,0,1 -> 6
1,1,0 -> new three throws
1,1,1 -> new three throws

But it is possible that this will lead to infinite loop.

So, is it possible that for some given $n$ that if I throw coin $n$ times, I will get equal changes to get every number from the set $\{1,2,\ldots 6\}$ without declining throws? First I thought it is not possible because $6\nmid 2^n$ but I'm not sure in general for example if I for example distribute the sample of $n$ throws somehow to make equal probabilities.

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  • $\begingroup$ The fact that $6$ does not divide $2^n$ proves that it is impossible to "distribute the sample of $n$ throws somehow to make equal probabilities." $\endgroup$ – Did Mar 26 '14 at 11:37
  • $\begingroup$ Yep I think this is correct ; you can reduce the probability to make a new series of throws but you wont be able to get away with a complete series as 3 will always be a problem $\endgroup$ – T_O Mar 26 '14 at 11:39
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Your first thought is right. You throw $n$ times, so the are $2^n$ possibilities. Let $A$ the set of all of them. You must define a partition of six subsets $B_1, B_2,\ldots B_6$, in $A$ with the same probability, that is, $1/6$. Since $1/6=p(B_j)=\#B_j/2^n$, we have $\#B_j=2^n/6$, which is impossible since a finite cardinal must be a natural number.

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Pragmatic (but silly) answer:

Let $f(\theta) = \left\lfloor\dfrac{3\theta}{\pi}\right\rfloor + 1$, where $\theta \in [0,2\pi)$ is the orientation of the coin as it lands on the hand/surface.

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