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There is the well-known method of unbiasing of bit sequences due to von Neumann. Are there similar schemes applicable to other sequences, e.g. the result of throwing a normal die?

Raphael
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Mok-Kong Shen
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3 Answers3

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The following simpler adaptation of von Neumann's trick is more efficient than the one described in HdM's answer. Throw your die twice. If the two answers $x,y$ are different, write $0$ if $x<y$ and $1$ if $x>y$. This way, you get an unbiased bit source. If you are so inclined, you can use this bit source to simulate fair die throws.

Yuval Filmus
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You can generalize the von-Neumann procedure canonically. Assume you have a discrete RV $X$ which you want to make fair, i.e., each event in the event space $\Omega$ of $X$ shall occur with the same probability. Then you can redraw $X$ exactly $n:=| \Omega|$ times, yielding $x_1, \ldots, x_n$. In case $x_1 \not = \ldots \not = x_n$, you return $x_1$. Otherwise, you repeat the above procedure.

Since the probability of drawing $x_1, \ldots, x_n$ differently is just a product of probabilities, the order of the elements does not matter. Hence, the probability of drawing an element $x$ at position $1$ is just $1/n$.

HdM
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It seems to me that presumably the only viable satisfactory solution would be: Obtain a bit sequence from the sequence from the die, apply the von Neumann unbiasing, then generate from that a sequence for the die.

Mok-Kong Shen
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