The chances of Julia Gillard winning in September are probably about 15% – better than many of use would think. That is if results in Australian elections are in any way similar to US elections.
I’ve taken my odds from a table in The Signal and the Noise by Nate Silver where he gives the probabilities of a senate candidate winning based on size of lead in polling average. While I don’t think the odds would be exactly the same, I don’t think they’d be all that dissimilar either.
So here is the table.
|Size of lead|
|Time until election||1 Point||5 Points||10 Points||20 Points|
There are reasons to think that there could be a difference, for example that voting is voluntary in the US, but hard to hypothesise why it would be that significant in terms of the result.
It’s true that US polling can be more volatile than our own, but that would tend to suggest that the 20 point or so two-party preferred lead that the federal Liberals enjoyed in the last Newspoll is even more significant.
When I think of Australian elections I can’t think of a government that has lost, or an opposition that hasn’t won, when it was ahead by 20 points at this stage, nor a state government or state opposition. But we don’t have a lot of elections here, so US statistics will probably capture the low probabilities inherent in very extreme situations better than trying to do the exercise on our figures.
Can anyone else think of a situation where an Australian state or federal government has been ahead by 12 to 20 points six months out in the polls and failed to win the ensuing election?