## What probability your vote decides?

Recently Andrew Leigh posted some comments about the number of elections in Australia where a single vote has decided the result and commented that this implied there was a probability that there was a 1 in 4500 chance that your vote would be the decisive one.

I have a few objections to this number, the first being that in a preferential system there are a number of points where a single vote can have an impact on the outcome as the preferences are redistributed up to two parties, so even though the final two party outcome is decisive you may have been decisive in deciding which party won. Whether this has ever occurred, I don’t know but it would seem possible.

A second objection is that a victory by two votes is also decisive, as otherwise you could have driven the result to a tie. I don’t know whether this has ever occurred.

Main objection though is with the calculating probabilities based on the historical outcome rather than the expected outcome for an electorate. Voters know when they go to the booth whether they are in a safe seat or not and whether their vote is likely to count and I would presume that this effects the way they view their vote.

I’m sure that Andrew is using 1 in 4500 as just a rough estimate of the sort of likelihood we are facing on average, and it is perhaps in that sense not unrealistic and acceptable for the sort of economic modelling of voting behaviour he mentions. However it got me thinking that you can perhaps get a better estimate for how likely it is for your vote to count if you have some idea of how safe your electorate is before you vote.

So the question is how can this be quantified? If we know the polls say that the two candidates each have a 50% two-party preferred. This means that each voter has an expected probability of 50% of voting for a particular party and we can calculate rough odds for the outcome being potentially decisive. The results fall according to the binomial distribution, which is effectively equivalent to normal distribution for such large numbers. For a given expected result, and number of voters we can calculate the chance of a tie and hence the approximate chance of your vote counting.

For an electorate with 85000 formal voters, under this model, you get the probability of the result being tied as 0.27% or a 1 in 365 chance of your vote deciding the two party result. For an electorate with 45000 like the federal NT seats, you get a 0.376% chance or 1 in 266, for state electorates like in the SA or 20000, you get 0.564% chance or 1 in 177.

It should be noted that because of the high numbers of voters (trials) the standard deviation of the distribution of voting outcomes is small and if the prior probabilities are 50.5% of voting one way then, for the 85000 seat electorate the probability of a tie drops to 1 in 1057. However, any real poll has a sampling error much larger error than this, typically a few percent, and so when we consider 50% may also mean 51% or 49% we get a considerable flattening of the real outcome. This reduces the probability of a tie for a poll outcomes near 50% but increases it for poll outcomes further away from 50%. I can’t be bothered calculating the exact effect of this (integrating two normal distributions) but the sort of reduction of the outcome for a 50% split is likely to be of the order of one half, while the wider results will increase slightly.

Still it tells us that for marginal seats voters should be expecting something more like a 1 in 500, to 1 in 1000 chance of their vote being decisive, or even higher odds for the smaller seats. Of course for non-marginal seats the probability implied for a tie is much lower than 1 in 4500 indicating that its probably not a bad overall estimate across all electorates and times. Its just a question of whether the this average is meaningful for any particular voter who realistically knows more about the individual electorate than the average for the entire country over time tells them.

1. ozrisk says:
2. Steve says: