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Think about the following situation:

Elections are held by dropping notes with the symbol of the political party to a ballot box. In order to save on time, a new automatic counting machine is proposed to count the votes at each ballot box. But, since we don't trust machines, a small (relatively) number of ballots ($X$) is chosen randomly to also count by hand and thus validating the results of the automatic counters.

In order to select the list of sampled ballots, each party sends a representative with a list of $X$, supposedly random, numbers. The final list is selected using the following algorithm:

1. Sum up the ith entry from all the lists
2. Modolu the sum by the number of ballots
3. If the ballot already appears, drop the entry.
4. If the entry is new, validate
5. i = i+1

I know that if at least one of the lists is truly random, then all the lists will be random, but what happens if none of the lists is random?

In the case where all the actors are cheaters, and every actor has a specific list of ballots that they either want to be checked, or wants to make sure aren't checked. Will the resulting list be random, or is there a way for actors to build their list in a way that benefits them?

All the actors (political parties) are in competition with one another, but some actors are friendly to certain actors more than other, meaning, actors will cooperate with parties from the same political philosophy if it means hurting the opposing philosophy.

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  • $\begingroup$ If the lists are not random, the result in not random either, obviously. It might "look complicated", but it's absolutely not random. If a political party know the other lists beforehand, it's easy, to build a list that will lead to a given set of checked ballots, hence to cheat. Not feasible without any assumption of these other lists though. $\endgroup$ – Jean-Claude Arbaut Apr 21 at 13:56
  • $\begingroup$ @Jean-ClaudeArbaut, I understand it, but, they are competing, the question is, can they be selected in a way that benefits one side over the other? $\endgroup$ – SIMEL Apr 21 at 13:58
  • $\begingroup$ If the other lists are known, yes it's feasible. If they are not known but some bounds are known, it might be feasible to predict ballots which won't be checked (hence to cheat). If nothing is known on the lists, then for the purpose of cheating, it's like they are random: you can't predict the numbers, hence which ballots will be picked. There may be other ways to cheat though (the "usual" ones: make the dead vote, etc.) $\endgroup$ – Jean-Claude Arbaut Apr 21 at 14:03
  • $\begingroup$ @Jean-ClaudeArbaut, at least some of the other lists are unknown, you can cooperate with some of the other parties, but not with all. $\endgroup$ – SIMEL Apr 21 at 14:10

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