3 ms·

So the tie-breaker is essentially to toss a coin? Hmm, there might be something in that...

• dragonwriter 8y ago
  Limited to the Smith set, but sure. Same as lots of places do with FPTP and a tie; no Condorcet winner is effectively a “Condorcet tie” between the members of the Smith set, each of whom beats every candidate outside the set, but where victory within the set is undeterminable.
  • iainmerrick 8y ago
    Yes, UK elections use coin tosses on ties. It’s pretty rare to get an exact tie when thousands of people are voting, though, and it’s also hard to argue against the coin toss when literally nothing separates the two candidates. The Smith set tie would occur more often, and in complex scenarios where it really isn’t clear that it actually is a dead heat.
    • dragonwriter 8y ago
      You can do other tiebreakers in the Smith set, like plurality first-rank winner in the Smith set (proceeding down preference ranks, with random selection only if you fall off the bottom without resolving the tie.)
• baddox 8y ago
  Indeed, if you allow randomness, you can get all the other desired attributes listed in Arrow’s impossibility theorem. Random ballot voting, for instance, has nearly every desirable property. There’s no reason to vote strategically (dishonestly). The race between candidates A and B can’t be affected by the presence or absence of any third candidate (there are no spoiler candidates). At least one voter is guaranteed to be happy with the election results (ok, that one is sort of a joke). But of course, random ballot has just one pesky undesirable property: the fact that it’s random.