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Puzzling

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  • $\begingroup$ I had the same idea but with two cups and a clear and opaque fluid. I think shuffling under a table would be pretty hard. $\endgroup$
    – user9194
    Commented Dec 28, 2022 at 0:47
  • $\begingroup$ There is a very small non-zero chance that a series of no votes and shuffling returns all the cards upright, which would be interpreted as a unanimous yes. $\endgroup$
    – user9194
    Commented Dec 28, 2022 at 0:53
  • 2
    $\begingroup$ Yes. OK. But there is a much larger probability that a meteorite falls on the restaurant and they don't get any dessert. Ever. Accidentally forcing some people to eat dessert isn't the worst that can happen. $\endgroup$
    – Florian F
    Commented Dec 28, 2022 at 1:11
  • $\begingroup$ This is prone to election fraud: the last voter can substitute a similar looking deck while "shuffling" under the table. $\endgroup$
    – Bass
    Commented Dec 28, 2022 at 12:19
  • $\begingroup$ I was assuming a collaborative solution was asked. But the participants could sign a card each before starting. And the process can be repeated the other way round. And you can add the second method for safety. This being said, people being polite enough to not force a dessert on someone, but then cheat to do it anyway, but in a way nobody knows it, would they go to the restaurant together? $\endgroup$
    – Florian F
    Commented Dec 28, 2022 at 17:04