"Facts are meaningless. You could use facts to prove anything that's even remotely true!"
-Homer J. Simpson

Wednesday, October 04, 2006

A Mathematical Model for Predicting the Liberal Leadership

Problem: Can we create a mathematical model for predicting the outcome of the Liberal Leadership race in December?

Answer: Yes (with some important caveats)

Since the Liberal Leadership numbers have been preying on my mind lately I decided to go back and re-read some sections of Do Conventions Matter : Choosing National Party Leaders in Canada by a professor of Political Studies at the University of Saskatchewan named John Courtney. (Full disclosure: I took a class from Mr. Courtney and was not overly impressed with him, but I think his work in this field is stellar) He looks at the 19 leadership races (of all Federal parties) between 1919 and 1993 and draws some statistical interferences.

There are a couple of statistical trends that he examines in his book that will help us shed some light on the Liberal’s December convention.

Note: I am NOT a liberal and am not part of one leadership camp or another. I am doing this as an exercise in political prognostication and for fun, not to promote one candidate ahead of the others. Also note, we are dealing with STATISTICS and PROBABILITY here people, not crystal balls and tarot decks. If you want certainty, go read a blogging tory blog, I deal with reality :-)

Point 1: Number of Ballots

Assuming that there are 8 people on the 1st ballot then according to table 10-2 in the book (page 353) there is a 19% chance of there only being 2-3 ballots and an 81% chance of their being more than 3 ballots. Assuming that Volpe drops out because he can’t pay his 20,000 fine and/or some of the other “can’t wins” decide they are going to throw their support behind somebody while it still makes a difference (i.e. before convention starts) then we could have a few as 6 people on the ballot at convention. If that is the case there is 51% chance of a 2-3 ballot affair and a 49% chance of a 4-5 ballot affair.

Conclusion: It is safe to say that there will be between 3 to 5 ballots

Point 2: Initial Support of Eventual Winner

Since we can conclude above that there will be between 3 to 5 ballots, let’s look at what support is need on the first ballot to win. Courtney breaks conventions up into categories based on the number of ballots. We will look at his analysis from table 10-1 (page 352) for the cases of 3 ballots, 4 ballots and 5 ballots.

In the case of a 3 ballot convention, the eventual winners had had an average of 43.7% support on the first ballot and ended up with 54% of the final support.

In the case of a 4 or 5 ballot convention, the eventual winners had an average of 27.9% on the first ballot and ended up with 55.3% of the final support

Conclusion 1: The winner will only need approximately 55% of the 1st, 2nd and 3rd choice support to win.

Conclusion 2: Only candidates who have close to 27.9% on the first ballot can win.

Point 3: Initial Ranking and Final Ranking

If you look at table 10-6 on page 356 you will find a very interesting statistic. Of all the leadership conventions that were discussed in the book; 85% of eventual winners were in first place on the first ballot and only 15% of winners were not first on the first ballot. However, the one factor that remains constant is that “the candidate who gains the largest share of votes from the first ballot to the second ballot is the eventual winner” (page 231)

Conclusion: There is an 85% change the person who is first place going in will win and if he does not then it will be due to another candidate getting a huge bump in-between the 1st and 2nd ballots.

Final Conclusion

The race will take between 3-5 ballots and thus the eventual winner needs approximately 27% of the vote on the first ballot to win. There is an 85% chance that the person in first place will win and the only way that that will not happen is if one of the candidates with a significant amount of support drops out and throws his support behind another candidate to create the largest increase in support between ballots.

So, after all that, we come across Paul Wells who just throws it all down without doing any sort of statistical analysis:

So Ignatieff doesn't need anyone's endorsement; he just needs support to bleed to him at the rate of one delegate in four. And he's been getting that all through this piece: when Hedy Fry and Carolyn Bennett went to Bob Rae, they failed to bring all their support with them. In the normal course of events, Ignatieff can expect to lure one previously unsympathetic delegate in four. Which means he can expect to win this.

The only way to stop him is to interrupt the normal course of events.

One of the second-tier candidates (Rae Kennedy Dion) would have to turn this race into a referendum on whether it is acceptable for Michael Ignatieff to become the Liberals' next leader. And the only way to demonstrate that the whole campaign should turn on that single question would be to pull out of the race immediately and throw to another second-tier candidate.


Kennedy's going to throw to him? Rae's going to throw to him? Nope. The near-perfect three-way Mexican standoff among Rae, Kennedy and Dion gives each man reason to hope, and therefore to stay in. And therefore to ensure Ignatieff's momentum isn't braked.

For each of the Second Three, and I am sad to say for Dion especially, the only question now is: Is it acceptable for Michael Ignatieff to become the next Liberal leader? If it is acceptable -- not ideal, not one's fondest wish, but simply a result that falls short of catastrophe -- Dion, Rae and Kennedy should stay in the race and try their chances. By staying in, they will probably ensure his eventual victory.

But if any of the Second Three believes Ignatieff must be stopped, they need to get out to make it happen.

I guess that’s why he writes for Macleans and I’m just a blogger , eh?

Vote for this post on progressive bloggers