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  • $\begingroup$ I was thinking of a very similar scheme while coming up with my answer. I think this has the same problem in that the person who compares the final result to the initial number will be able to determine the exact number of votes. $\endgroup$
    – 2012rcampion
    Commented Dec 27, 2022 at 18:36
  • $\begingroup$ I was editing, not sure which version of the answer you're referring to, but the last paragraph should solve that issue. Then the first person would just get back some randomly large negative number after the second go-around, revealing no information regardless of anyone's choice $\endgroup$
    – hedgedandlevered
    Commented Dec 27, 2022 at 18:39
  • 3
    $\begingroup$ Since you start with an arbitrary number X, at the end you need to subtract off X to determine if the sum of votes is positive or negative. The person who does that step will be able to tell if they are the only 'no' vote. $\endgroup$
    – 2012rcampion
    Commented Dec 27, 2022 at 18:43
  • $\begingroup$ In any information sharing scheme you use here where all participants contribute to the state, the final person, if it is the case that 1) they can break or not break unanimity based on how they modify the final state, and if 2) unanimity can be determined based on the final state, they will be able to modify the state of the final piece of information to reveal unanimity or not. Thus, they will always know if they are the lone dissenter. If they weren't, they would not be able to (within the rules) adjust the state to show unanimity. $\endgroup$
    – hedgedandlevered
    Commented Dec 27, 2022 at 19:33
  • $\begingroup$ The above only applies when you have to look at final state. In the case where you do not, it's trivial... you simply pretend to look at the final state, and announce to the group that it is not a unanimous vote yes $\endgroup$
    – hedgedandlevered
    Commented Dec 30, 2022 at 20:02