21
votes
Accepted
21
votes
Accepted
9
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7
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Making two fair dice
Surprisingly, this works if and only if you divide the pips evenly between the dices!
Let us take one of the dices and denote its numbers by $\sigma_1,\ldots,\sigma_6$ (in increasing order). Then, the ...
6
votes
Accepted
Making fair dice
You can't do it for
Or for
It is possible for
Finally, it's not possible for
So the answer is
4
votes
Making two fair dice
Sneftel's sufficient condition is in fact not necessary. A different construction to achieve 50% win rate is the following.
This construction does include some cases that Sneftel's doesn't (and vice ...
3
votes
Accepted
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