A statistical prediction of the 2015 general election

Features

  • Author: John Fry
  • Date: 19 Jan 2015
  • Copyright: Image appears courtesy of iStock Photo

At the start of the New Year the campaign for the next general election has begun in earnest. As a small part of a wider picture the combined efforts of a group of academics to produce a statistical prediction of the general election was recently reported by BBC Newsnight [1]. The model in [1] incorporates past election results, current aggregate polling, historical aggregate polling, current raw polling data and individual constituency characteristics to produce a whole host of very detailed statistical predictions, and associated confidence intervals, down to even the level of individual constituencies.

thumbnail image: A statistical prediction of the 2015 general election

It is of interest to see how a much simpler purely statistical model can compare against this high-profile prediction from three specialists in quantitative political science. Historical results for the 14 post-war general elections are readily available online. One way of predicting the result of the 2015 general election is to predict the number of parliamentary seats, on the basis of an admittedly sparse historical record, converting the popular share of the vote into parliamentary seats. This can be done using either opinion polls or by using a projected share of the vote. Data of both kinds is available in [1]. However, whilst we tried to use a regression approach along these lines the results obtained do not appear to be sufficiently trustworthy. In fairness, it is not surprising that such an approach does not seem to work well. The above focus on individual constituencies reflects the fact that the 2015 general election is liable to have a pronounced regional flavour to it – one such feature being the likely strong performance of the Scottish National Party (SNP) in Scotland. One further complication may be the issue of tactical voting from an increasingly aware electorate. In view of the above using data from bookmakers’ odds seems to offer a more robust alternative [2].

The comparison website Oddschecker [3] offers a wide-ranging comparison across several major bookmakers on various aspects of the next general election. One obvious feature of interest is the number of seats that each of the three major parties will win though this market appears thinner, with fewer quoted prices, than other related questions such as what is the probability of a hung parliament.

How can we translate bookmakers’ odds relating to the number of parliamentary seats won in the 2015 general election into precise probability statements? As an illustrative example let X be the number of parliamentary seats won by Labour at the 2015 general election. We assume that X has a normal distribution with mean m and variance s2. One bookmaker gave the odds that X>287.5 as 8/11. This suggests that p=Prob(X>287.5) satisfies (1-p)/p=8/11; p=11/19. The same bookmaker also gave the odds that X<287.5 as 1/1. This suggests that p/(1-p)=1; p=1/2. The conventional approach is to estimate the probability by taking the average of the two prices. This gives p=1/2(11/19+1/2)=41/76. This suggests that Prob(X<287.5)=35/76. Continuing, standard A-level probability calculations involving the normal distribution [4] give F((287.5-m)/s)=35/76 and (287.5-m)/s=-0.099, where F(.) denotes the Cumulative Distribution Function (CDF) of the standard normal distribution. This approach leads to several parameter constraints that cannot all be satisfied simultaneously for each party. Here, as a pragmatic solution to this problem, we estimated the parameters m and s by minimising the least-squares distance.

The projected results for the 2015 general election obtained using this approach are as follows:
Labour Party: 287 seats
Conservative Party: 283 seats (95% prediction interval 269-296 seats)
Liberal Democrat Party: 28 seats
Other Parties: 52 seats (95% prediction interval 39-66 seats)

A plot of these results compared to the last election is shown below in Figure 1. The figures show Conservative and Labour support seemingly converging with minor parties seemingly squeezing support for both the Conservatives and the Liberal Democrats.

Figure 1: Left hand side shows the breakdown of parliamentary seats from the 2010 general election. Right hand side shows the projected number of parliamentary seats for the 2015 general election.

These results appear roughly in line with the predictions in [1] though our model does suggest slightly more parliamentary seats for each of the three major parties. However, the model in [1] does seem to offer a much better quantification of the inherent uncertainty involved.

As the above projections show there are currently 650 seats in the UK Parliament which means that 326 seats are needed to secure an overall majority. It thus seems that the spectre of another coalition government looms large. Of particular interest is that it seems far from certain that either the Labour Party or the Conservative Party would have enough seats to form a coalition government with the Liberal Democrat Party without support from other minor parties. This may have very important implications for the UK as a whole given the predictions of future SNP successes in [1] and an increasingly bitter divide as highlighted by the recent Scottish Independence Referendum [5].

How can we precisely quantify the likelihood of a hung parliament? At the time of writing the model in [1] suggests that the probability of a hung parliament was 0.92. How can we estimate this probability by using the available data on bookmakers’ odds?

The comparison website [3] also quotes odds from 20 major bookmakers on the possibility of a hung parliament. As an illustrative example, one bookmaker gave the odds of a hung parliament as 4/11. This suggests that the probability of a hung parliament can be estimated by setting (1-p)/p=4/11 which can be solved to give p=11/15. The average probability calculated in this way across the 20 bookmakers was 0.726. Results in [3] suggest that the bookmakers spread will be around 7.7%. This suggests that a reasonable estimate of the probability of a hung parliament is 0.726 plus or minus 3.85%, i.e. (0.698-0.754).

In conclusion, any prediction of the 2015 general election is subject to considerable uncertainty – most notably from a late burst of support for the incumbent government. This uncertainty is difficult to quantify but there appears to be particular uncertainty regarding the number of parliamentary seats that the Conservative Party will win. Our simple approach suggests that the model in [1] slightly underestimates the number of parliamentary seats that the three major parties are likely to win. However, the differences involved are slight and the model in [1] has significant advantages in terms of its quantification of uncertainty. Amid low levels of support for the three main parties a future coalition government appears highly likely though our approach suggests that the model in [1] slightly overestimates the probability of a hung parliament (0.92 versus 0.698-0.754).

References
[1] 2015 UK Parliamentary Election Forecast www.electionforecast.co.uk/
[2] Fry, J. M. (2014a) A statistical prediction of the Scottish referendum. Statistics Views
www.statisticsviews.com/details/feature/6552581/A-statistical-prediction-of-the-Scottish-referendum.html
[3] www.oddschecker.com/politics/british-politics/next-uk-general-election/
[4] Neave, H. R. (1978) Statistics Tables: for Mathematicians, Engineers, Economists and the Behavioural and Management Sciences. Routledge, London.
[5] Fry, J. M. (2014b) A statistical reaction to the Scottish referendum. Statistics Views
www.statisticsviews.com/details/feature/6764961/A-statistical-reaction-to-the-Scottish-Referendum.html

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