One of our students, Saul Cunow, currently in Brazil sends along this interesting tidbit:

In every municipal election there are 1-2 municipalities [among Brazil’s 5000+] where the leading candidates for mayor tie. Per the constitution, the tie-breaker is age — the older candidate wins. Apparently, age is also used as a tie-breaker in civil service exams and elsewhere as well.

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Probably as good a criterion as any when the votes are tied. Unlike a coin-toss, it is stable if repeatedly verified (contrast random transfers and Gregory fractional transfers under STV).

Of course, if numbers are very small then ties become more likely! 5,000 municipalities is a large divisor even when the numerator is as big as the population of Brazil. (c/f how the same rule – “referendum must pass a majority of States” – translates into 12/23, or 52.17%, in Switzerland but 4/6,or 66.667%, in Australia…).

I note that in several legislatures (Bundestag and Knesset, IIRC), the oldest member presides pending the election of a Speaker/ President. I imagine this could be easily ascertained without a pre-ballot ballot (and the risk of infinite regress…) -a “duel of the driving licences”, perhaps?

There is some logic behind the preference for the elder: the younger has slightly more chance to participate the next time, for the elder it may be his last opportunity.

But I stil prefer a lottery in the event of a tie: both candidates are treated equal and when a tie can be expected (when a small council votes) no-one can manipulate the outcome (by proposing an elder candidate than the opponent)

I also suggest a new category ‘tie-braking rules’ with plants numbers 391, 1467, 1723, 4225, 4356,… (in some however, ties are only mentioned in some branches of the discussion)

Bancki, I respectfully dissent. If a coin toss, the question is then

whosetoss, and why Senator Reid’s calling “heads” should trump Senator McConnell’s calling “tails”. Age, by contrast, is transitive.Your point that an older candidate has fewer opportunities is a good one, though I would prefer to see it from the voter’s point of view – an older candidate has more experience to work for their interests. Of course, s/he could be more corrupt and/or hidebound.

Randoms election can be done in more elegant ways than tossing a coin. The Commonwealth Electoral Act gives the divisional returning officer a casting vote but contains very detailed provisions for random selection of candidate positions on the ballot paper.

The part of my brain that likes cryptography suggests:

Before each election, randomly select a number S for each district where counting occurs. The number should be large, and chosen from a large range, say 2^64 < S = 2 candidates/parties, the parties are ordered alphabetically, and then the ((S mod P) + 1)’th party is chosen. Then we remove the used portion of S, setting S’ := S / P .

Now the counting process is repeatable and stable, but still random. Some might not like that the winner of a tie-break can be predicted in advance, however. This process is also limited to a certain number of ties (64 in my example), but that can be easily addressed by choosing a larger number.

And then there is Robson Rotation…

I think it would be a serious error to have the tie-winner known in advance. In some elections and post-election challenges (think states starting with F) the strategy would be to get to a tie by excluding ballots.