# When does a single vote count?

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understandinguncertainty.org was produced by the Winton programme for the public understanding of risk based in the Statistical Laboratory in the University of Cambridge. The aim was to help improve the way that uncertainty and risk are discussed in society, and show how probability and statistics can be both useful and entertaining.

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1,362 Cambridge academics recently voted on ‘no confidence’ in the universities minister David Willetts, and this resulted in an exact draw with 681 voting each way: by the rules it meant the motion, or ‘Grace’, was not passed. A natural question to ask is: what was the chance of this happening?

It is always tricky to say something about the chance of events that have happened. What does this mean? One interpretation is the odds would have been given in advance for an exact draw, given what was known at the time. This would involve extensive judgement. Perhaps a more intuitive interpretation is the odds on a draw given what we know now about the circumstances of the vote, for example that it was very even matched and around 30% of the electorate voted.

A starting point is to work out the chance of an exact draw, given 1,362 people voted and each vote had a 50% chance of being for or against the motion – this could also be roughly interpreted as being the chance of a draw given the electorate were equally split and the voters were a random sample (this is an approximate interpretation as the voting pool is finite). This chance is around 1 in 46 (see below for mathematical details).

Suppose we relax the assumption that the voting preference is exactly 50% and instead assume there is a chance $p$ that any particular voter is in favour of the motion, where $p$ is unknown. If we assumed that $p$ was equally likely to be anything between 0 and 1, then the chance of specific number of votes each way is $1/(N+1)$, and so this would be the chance of a draw. But we are interested in voting preferences ‘near’ 0.5. One interpretation of ‘near’ uses the data from the vote itself, from which we can conclude that a 95% uncertainty interval for $p$ is between 0.47 and 0.53. If we assume $p$ is a random quantity with a distribution centred on 0.5 and with this 95% interval, then the probability of an exact draw with 1,362 votes drops to around 1 in 65.

Finally, it could be argued that we should not condition on exactly 1,362 votes. The first issue is that the number of votes must be even, and this condition halves the probability to around 1 in 130. This assessment is barely changed by assuming an even number of votes near 1,362, and so our final conclusion is that, given the benefit of some hindsight, there was around a 1 in 130 chance of an exact draw.

#### Some maths!

Let $R$ be the number of votes for the motion, when the voter preference is $p$ and there are $N$ voters, where $N$ is even. Then $R$ is Binomially distributed with mean $Np$ and variance $Np(1-p)$ - since we are assuming $p \approx 0.5$ then we can approximate this variance by $N/4$. Then approximating the Binomial distribution by a Normal with the same mean and variance leads to a probability of exact draw, ie $R = N/2$, of

\begin{eqnarray*}

P(R = N/2) &\approx& \frac{1}{\sqrt{2\pi}\sqrt{N/4}} \exp\left[ -\frac{1}{2}\frac{(N/2 - Np)^2}{N/4}\right] \\

&\approx & \sqrt{\frac{2}{N\pi}} \exp \left[ -2N(\frac{1}{2}-p)^2 \right]

\end{eqnarray*}

If we assume $p=0.5$, we obtain $P(R=N/2) = \sqrt{\frac{2}{N\pi}}$ and so if $N = 1362$, the probability of a draw is around 1/46.

Taking a Bayesian approach to estimating $p$, based on the observed equal vote and a uniform prior for $p$ gives a posterior distribution $p \sim {\rm Normal}(0.5, \frac{1}{4N})$. A 95% interval for $p$ is therefore $0.5 \pm 2 / \sqrt{4N}$, or $0.5 \pm 1/\sqrt{N}$. For $N=1362$ this goes from 0.47 to 0.53. Using the observed data to estimate the chance of a draw looks like cheating, but this interval is appropriate if we consider a draw a plausible outcome with this many votes.

The probability of a draw would therefore be

\begin{eqnarray*}

P(R = N/2) &=& \int P(R = N/2| p)\,\, p(p|{\rm data}) dp \\

& = & \sqrt{\frac{2}{N\pi}} \int \exp \left[ -2N(\frac{1}{2}-p)^2 \right] \,\, \frac{1}{\sqrt{2\pi/(4N)} } \exp \left[ -2N(\frac{1}{2}-p)^2\right] dp \\

& = & \sqrt{\frac{1}{N\pi}} \int \frac{1}{\sqrt{2\pi/(8N)} } \exp \left[ -8N(\frac{1}{2}-p)^2/2\right] dp \\

& = & \sqrt{\frac{1}{N\pi}}

\end{eqnarray*}

which for $N=1362$ comes to 1/65.

Finally we consider $N$ as unknown but ‘around’ 1362. Since the chance varies slowly with $\sqrt{N}$ the 1/65 will be insensitive to $N$ and high and low values will cancel out. However $N$ must be even, and so the final assessment of the chance of a draw is around 1 in 130.

*I am grateful to Allan McRobie and other correspondents for helpful discussion, although the final conclusion is my responsibility!*

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