Assume a bell-shaped, normal distribution. Imagine you could insert a candidate into the race with completely random traits, name recognition, likability, etc. (I know that makes no sense, but just bear with me here and suspend disbelief). Let's call him John Q. Random.
The column on the right shows the percent likelihood that the random candidate could have surpassed that candidate's vote total. In other words, there's only a 2.3 percent chance that the random candidate surpasses Mercier's total, but it's better than 97 percent that he beats Fred Doyle. The people closest to the mean are at the top of the curve, which is why Mr. Random would be about 50/50 to be on either side of their vote total.
** EDIT: I've already caught some flak for this, which is good (hey, someone was reading it!)...please let me clarify what I mean by 'random.' I don't mean 'random' in the sense of a name plucked from thin air and placed on the ballot. I don't mean 'random' in the colloquial usage of something or someone out of place, unexplained, etc. What I mean is that it would be a candidate whose strengths and weaknesses -- the things that would make a person vote or not vote for him or her -- were randomized.
For you Gaussian fans, I'm saying that with the area under the entire curve being 1.00, Rita's vote total falls on the right-side tail, with only .023 of the area not covered. All the candidates except Mercier, Elliott, Pech, and the Doyles fall inside the one-standard deviation portion of the bell. That's a strong statement about the competitiveness of city elections!
As an active voter, I only have nine votes. I tried to keep it diverse, but that meant not voting for all the candidates I liked in such a strong candidate pool. As someone who normally votes blank on state/congressional elections, I need to be grateful for the wide variety of choices.
ReplyDeleteFrustrating to have three of your choices come in 10th, 11th, & 12th.