kitchen table math, the sequel: help desk

## Saturday, May 5, 2012

### help desk

2 seats open; 5 candidates running; maybe 1600 people will vote. Two candidates are running as a slate, with posters around town saying "Two seats, two votes."

There will be some bullet-voting. No idea how much.

What's the fewest votes a candidate would need to win if the race is close?

ChemProf said...

How does the election work? A plurality, I'd assume, so that the two candidates with the most votes wins? And again, how many can you vote for? Pick two of the five?

In that case, if the votes were distributed randomly, each candidate would get 40% of the vote or 640 votes. If it were really close, so close to random, you'd need at least 641 votes to win.

(By the way, I wasn't sure about the 40% number, which is just 1/5 of the vote times two, but when I worked through the permutations of five pick two, there are sure enough ten possibilities and each candidate appears in four of the ten).

Glen said...

The problem states that "maybe 1600 people will vote." So, maybe 1600 and maybe not. So what are the constraints on the total number of votes? None stated, but it's reasonable to assume that the word "vote" itself implies non-negative integers, as it usually does. So the fewest allowable number of votes that would qualify as a "close race" with "maybe 1600 votes" would be a 5-way tie of zero votes each. That wouldn't produce a winner in a 2-seat contest, though, so give one candidate a single vote, and that one vote becomes the fewest votes a candidate would need to win an election in the specified set of elections: "2-seat, 5-candidate close race with a total number of votes of maybe 1600."

Catherine Johnson said...

Everyone can vote for two candidates; some will bullet vote, but I have no idea how many.

Ed gave me the 1600 figure - I'll ask what he meant. I THINK he was talking about last year's election, where one seat was open & there were 1600 votes altogether.

Catherine Johnson said...

The other question we've been puzzling over concerns the two candidates running as a slate ("Two seats, two votes").

Several years ago, two PTSA women ran as a slate in order to defeat a reform-and-accountability candidate, which they did. I have now forgotten why running as a slate was more effective than running singly, but that was the first time it had been done, and the women who ran came up with it explicitly as a means to keep an academic accountability person out.

Anyway, that's the situation again this year ... and I absolutely can't figure out whether that was a good idea in this case (a good idea for them).

ChemProf said...

I think the thing about running as a slate is that it makes the voting a lot less random -- with this kind of election, name recognition is key, and if you can link two of the names, you can drive a lot of votes.

Glen said...

Ah, I thought it was a tricky math problem you found in a puzzle book somewhere, not a real-life situation you were trying to deal with. I was doing what you do with tricky math problems (spot the angle), not being a smart aleck.

The question has too many degrees of freedom--too much that is unspecified--for an answer to be both simple and useful. Here are a couple of thoughts, for what they're worth:

I don't think the two running as one makes much difference. If they're tied together, each will get roughly the same number of votes as his partner, but if "close election" means the other three candidates do as well as the partners, then the other three will get about the same number as each member of the joined pair. In other words, the votes will be about the same as if the pair weren't partnering.

So, by "close," do you mean that all five get roughly the same number of votes? If so, then the minimum needed to win is just a little over N/5, where N is the total number of votes cast which, by your description, sounds as though it would be about N = (maybe 1600) x 2 - (some bullet votes) = maybe 3000 or so? If that's about right, then you could potentially win with as few as about 500 votes if it's just a bit more than the third-place winner.

But if, by "close," you mean that the two winners aren't obvious, which could mean a 2-, 3-, 4-, or 5-person tie for second place, for example, then the minimum number needed to win wouldn't be a simple formula but would depend on the details.

I hope that's a more useful answer.

Catherine Johnson said...

Ah, I thought it was a tricky math problem you found in a puzzle book somewhere, not a real-life situation you were trying to deal with.

I know!

I was laughing when I read --- would that it WERE a tricky math problem.

Nope, I'm trying to get my candidate re-elected, and she's opposed by a former PTSA president and a former BOE president....

You really see the GOTO issue. My candidate, I think, has the support of the broader community, but the broader community doesn't vote.

Parents vote, and parents want other parents on the board (it seems).

Catherine Johnson said...

Thanks Glen & chemprof - both of these are terrifically useful (and I do realize that this is a problem with too many degrees of freedom, assuming I know what that means --)

Glen said...

Catherine, just another note in case it helps. If your candidate can get one vote more than one-third of the total votes, he/she/they can't lose, regardless of how many candidates there are, how many total votes there are, who partners with whom, etc. (If you have >1/3 of the votes, there can be one person who gets as many or more votes than you, but there can't be two who do.)

It's possible to win with only a single vote in the (extremely unlikely) event that #1 gets all the votes but one (so all but one are bullet votes), you get the one remaining vote, and the other three get none. The more votes taken by #1, the fewer votes needed to win #2. So, even though winning with one vote is an extreme example, I'm mentioning it here, because it is POSSIBLE for you to win with any number of votes >0 if the front runner is popular enough (so, unlike one-winner elections, don't give up if you're behind), and you are GUARANTEED to win with anything more than one-third.