A few weeks ago I ran into the amazing Linda Hut, the teacher and coordinator over at City Hall School, and she mentioned that her class was asking questions about the election. One student asked what would happen if there was a tie. The answer, which can be found in the Local Authorities Election Act (Section 99), might surprise you:

If it appears on the calculation of the votes that 2 or more candidates for any office have received the same number of votes, and if it is necessary for determining which candidate is elected, the returning officer shall write the names of those candidates separately on blank sheets of paper of equal size and of the same colour and texture, and after folding the sheets of paper in a uniform manner and so that the names are concealed, shall deposit them in a receptacle and direct some person to withdraw one of the sheets, and the returning officer shall declare the candidate whose name appears on the withdrawn sheet to have one more vote than the other candidate.

That’s right, months of campaigning and the decision could all come down to one piece of paper drawn from a hat. Fortunately, the chance of a tie happening is pretty tiny!

To start, what’s the likelihood that all six mayoral candidates could receive the same number of votes? Well, technically it’s impossible as there are 619,138 eligible voters, but let’s round that up to 619,140 for the sake of argument. That would mean each candidate would need to get 103,190 votes exactly for all six to be tied. According to Wolfram Alpha, there’s a 1 in 735 chance (about 0.14%) of a candidate getting that number. If we reduce that to just the 34% of voters that actually turned up last election, the chances improve slightly to 1 in 429 (about 0.23%). But actually, that has to happen six times in order for all of them to tie, which is just incredibly unlikely.

Now that’s a fairly unrealistic scenario as we know that there are a lot of factors that make it unlikely our full slate of mayoral candidates will all receive the same support. In the 2010 election, the top three candidates (out of seven) received about 95% of the vote. So let’s assume this year’s election will be similar. Continuing on with the 34% turnout, there would be 199,983 votes up for grabs by the top three. The chance that a candidate will get exactly one third of those votes is about 1 in 528 (0.19%). And the odds that all three will tie? Again, it’s not going to happen.

What if one of the three receives a quarter of the vote and the other two split the remaining votes? The likelihood of that is much better than a three-way tie at about 1 in 485 (0.21%), but still, it’s just very unlikely!

With the number of votes we’re talking about, ties are just not very realistic. Even assuming 34% turnout in Ward 9 where there are just two candidates, there’s still a less than 1% chance of a tie (1 in 178, 0.56%).

This is all just simple math for the purposes of illustration, and it ignores the many, many factors that go into why someone will vote for candidate A over candidate B. Still, now you know how unlikely a tie is!

Amazing that it could all come down to the drawing of a name from a hat. I hope that conclusion was drawn once they looked at the probabilities.

Come to think of it…maybe we should try that, just for one election…

Think of the money we could save just drawing names from a hat

Thanks for a breakdown on the math Mack. Can I call dibs on being the guy to draw the name from the hat?

And there was a name drawn out of a bucket in Edson!

http://www.edmontonjournal.com/news/Draw+from+plastic+determined+last+councillor+Edson/9073503/story.html