Yesterday we had a bit of a squiz at using monte carlo simulations to analyse the probability that a given swing would produce some number of seats for the ALP. Today we can go one further and use this same simulation framework to look at the probability of the ALP gaining the 45 seats it need to form a government from any TPP preferred result at the coming election.
We are again using a standard deviation of 4% for these swings – the results are a chart where you can simply pick an ALP two party preferred result from the bottom, trace it up until it intersects the red line, then trace it across to the left to find the probability of the ALP gaining at least 45 seats from that two party preferred vote.
(just click to expand the chart)
The ALP still has a 50% probability of gaining 45 seats from a two party preferred vote of just over 48.6%, which again highlights just what an enourmous task it is for the LNP to win.
11 Comments
If you are right about 48.6%. Then the LNP need less than the necessary uniform swing to be a better than a 50% chance.
That’s right, because the current distribution of votes across electorates means that a uniform swing is unfavourable to the LNP. Possum is assuming historical levels of non-uniformity.
(I still think that swings should be normalised based on the TPP vote in each seat!)
Courier Mail website front page promising shock, Horror, You’ll be aghast, poll out at 5.00pm Brisbane Time.
steve,
I didn’t see horror or you’ll be aghust. I did see shock however. I think to live up to its name it would have to be at least 5% either side of the previous poll. If it is less the MoE will come into play. So 58/42 to Labor or 52/48 to LNP at a minimum.
It wouldn’t surprise me if it was some small poll of 4 or 5 marginal seats that had the LNP equal or ahead 51/49. Probably a Galaxy – we’ll see in an hour I suppose.
The more I think about this analysis the more I like it. I think the “50% probability of win” swing is a much better number to quote from the pendulum than the “uniform swing required for victory”. Perhaps even quoting the 25% and 75% probability swings too.
(I also reckon you could do this particular analysis without resorting to Monte Carlo simulations, just using the standard normal distribution function, but that’s a quibble).
I’m glad you like it Caf. I have thought about using just a standard probability function analysis but what it doesnt do as well as monte carlo simulations is account for the likely electoral effect of clusters of seats on various margins in terms of a swing.
If you look at a basic histogram of ALP margins in Qld:
http://blogs.crikey.com.au/electioncentral/files/2009/02/alpmarghist.png
It really does influence the way a swing would likely flow through in terms of seat changes. Monte Carlo seems to work better at dealing with that complication via simple brute force.
An interesting piece of trivia that you might enjoy is that the chart at the top of the page took 8.6 billion election simulations to make.
I just luuuurve modern computing power!
The method I was thinking of would have taken that into account too – but it turns out to require far more computing power than the Monte Carlo simulation – in fact an impossibly large amount. In case you’re interested, it was:
Pick a statewide swing value. Go through each seat. For each seat, calculate the probability of winning it under that statewide swing as the probability of the swing in that seat being greater than or equal to the required swing, using standard normal distribution. Calculate the overall probability of victory under that statewide swing by using the individual seat probabilities.
That looks fairly innocuous at first glance (at least to me!), but the devil is in the “calculate the overall probability of victory” step. I think it needs on the order of 2^89 calculations, which is out of the realms of anything not owned by the NSA.
Caf
There is no problem at all with modelling the overall probability of victory in a two party system.
(1) Assume swing is normally distributed. In a given seat, the vote for the party in which you are interested will then be normally distributed, with a mean equal to the starting vote plus the mean of the swing which you have chosen; the variance of the vote will be the same as the variance (which you have chosen) of the swing.
(2) The probability that the party in which you are interested wins a seat is then obtained by integrating the normal distribution to determine the probability that the vote exceeds 0.5
(3) Now define an indicator variable for a seat, which takes a value of 1 if the party in which you are interested wins, and zero otherwise. The expected value of the indicator variable is equal to the probability (call it p(i)) that the party wins the seat, as calculated at (2); and the variance of the indicator variable is equal to p(i)(1-p(i)).
(4) Assuming (as one can) that the indicator variables are independent, sum them. This produces a random variable of the total number of seats won by the party in which you are interested. It has a distribution known as the generalised binomial or sometimes the Poisson binomial distribution. (See P A P Moran, An Introduction to Probability Theory, Part 1.10). This distribution has a mean obtained by summing the expected values of the indicator variables across all seats; the variance is similarly obtained. This set up satisfies the Lindeberg condition, the central limit applies, and the sum is asymptotically normally distributed (though you can obtain exact probabilities with a bit of work, using generating functions).
I first worked on this in 1980-81, when writing a thesis at ANU on Probability and Electoral Bias, now long superseded by Simon Jackman’s modelling. In those days, running programs in GLIM, it was a bit of a pain; now it’s dead easy to set the whole thing up in an Excel spreadsheet.
In the light of all this, I can’t quite see the point of running Monte Carlo simulations; but it would be quite interesting to compare this modelling with the Monte Carlo approach. Seems to me that the Monte Carlo results should converge to the outcome of the model I’ve described.
Comments?
That all makes sense to me, and I agree that the Monte Carlo results should converge to the exact solution.
It sounds like it would be worth running the model through and comparing the results, even if just to verify the Monte Carlo model.
One point of pedantry: in my post above, at (2), I should have said that you need to integrate the normal density, not the normal distribution.