So, as Your Humble Blogger mentioned, this Town elects a council of nine members, at large, every two years. A Party can nominate up to six candidates, and voters can vote for up to six candidates. The top nine vote-getters are in. The bottom three are out.
Does this seem totally bizarre to you? I mean, aside from the always perplexing business of a constituency voting to restrict its ability to vote for whoever the hell it wants.
Let’s take a town of a hundred, of whom sixty are in the Sock Party, twenty are in the Rock Party, and twenty are unaffiliated. I don’t know the split in my home town, but it would have to be something like that to make the limitation make sense in the first place, wouldn’t it? OK, the Socks candidates are Steve, Schmuel, Sarah, Sally, Soren, and Sully; the Rocks candidates are Raul, Randy, Richard, Renee, Rowland, and Rakhi. If all the Sock Party members vote the straight Sock ticket, and all the Rock Party members vote the straight Rock Party ticket, then it’s up to the twenty unaffiliated voters to decide which three of the six Rocks get to be on the council. These free voters know, presumably, that they can’t turf any of the Socks, but they can, by choosing which Rocks to support, swing those seats.
Except, of course, that the Sock voters know what’s going on. And we (for I am, in this scenario, one of the sixty) may well have preferences for which Rocks we have to deal with. True, none of their candidates are sufficiently hosacious, and they are all far too concerned with silicate matters, but in truth only one of them is just batshit insane. Remember, by the way, that since there are only twenty Rocks in the whole town, it’s far easier for the crazy one to get onto the ballot. I could just vote my own ticket and trust that the unaffiliated voters will do the right thing, but if they were really trustworthy, wouldn’t they be in the Sock Party? Maybe the thing to do is to vote for the three most Sock-like Rocks, and the three best Socks. Only, I don’t want to run the risk of one of the fringe Socks not getting enough votes. Maybe I should vote for the three weakest Socks, and the three best Rocks, and not vote for my favorite three candidates at all.
See, it’s perfectly plausible to me the final tally will be something like
| Steve | Schmuel | Sarah | Sally | Soren | Sully | Raul | Randy | Richard | Renee | Reese | Rakhi |
| 70 | 68 | 65 | 58 | 48 | 38 | 30 | 26 | 26 | 25 | 24 | 24 |
That’s figuring that there were four unaffiliated voters who were just determined to throw the bums out, some mild picking and choosing, and a lot of staying home. And there are only two points difference between Rakhi and Randy, but Rakhi goes home to paint the garage, and Randy sits on the town council. And I can’t remember which one was the crazy one, but if it was Randy, then I’m screwed, and really should have voted for Rakhi. Who seemed nice, you know, for a Rock.
Tolerabimus quod tolerare debemus,
-Vardibidian.
Odd, indeed.
FWIW, I’m totally with you on Socks. Socks are great. Are you of the Argyle persuasion, or the Plain? Tube, or Fitted?
peace
Matt
You should vote for only the candidates you like, because there could always be a third party someday. That third party is more likely to step up if there isn’t artificial support for the second party. Also, no mandate.
It’s not much easier here voting for a simple 7 out of 10, with one psycho, two great, and seven dull. If there is no psycho, vote just for the two great. If there are no great, vote for as many non-psychos as possible. But with the current pattern, we end up with the greats losing to the dull because we had to vote for the dull in addition to the great to keep the psycho off.
The longer-term view makes the decision easier — by only voting positive, I both feel better and have more hope of improving the candidate pool over the long term.