How likely?

Yesterday, CUP political party held a general assembly to determine whether to support or not Artur Mas's candidacy to President of the Catalonian regional government. The final voting round among 3,030 representatives ended up in an exact 1,515/1,515 tie, leaving the question unsolved for the moment being. Such an unexpected result has prompted a flurry of Internet activity about the mathematical probability of its occurrence.

The question "how likely was this result to happen?" is of course unanswerable without a specification of the context (i.e. the probability space) we choose to frame the event. A plausible formulation is:

If a proportion p of CUP voters are pro-Mas, how likely is it that a random sample based on 3,030 individuals yields a 50/50 tie?

The simple answer (assuming the number of CUP voters is much larger that 3,030) is P_p(1,015 | 3,030), where P_p(n | N) is the binomial distribution of N Bernouilli trials with probability p resulting in exactly n successes.

The figure shows this value for 40% ≤ p ≤ 60%. At p = 50%, which without further information is our best estimation of pro-Mas supporters among CUP voters, the probability of a tie is 1.45%. A deviation in p of ±4% would have made this result virtually impossible.

A slightly more interesting question is the following:

If a proportion p of CUP voters are pro-Mas, how likely is a random sample of 3,030 individuals to misestimate the majority opinion?

When p is in the vicinity of 50%, there is a non-negligible probability that the assembly vote come up with the wrong (i.e. against voters' wishes) result. This probability is

I_p(1,516, 1,515) if p < 50%,
1 − P_p(1,015 | 3,030) if p = 50%,
I_1−p(1,516, 1,515) if p > 50%,

where I_p(a,b) is the regularized beta function. The figure shows the corresponding graph for 3,030 representatives and 40% ≤ p ≤ 60%.

The function shows a discontinuity at the singular (and zero-probability) event p = 50%, in which case the assembly will yield the wrong result always except for the previously studied situation that there is an exact tie (so, the probability of misestimation is 1 − 1.45% = 98.55 %). Other than this, the likelihood of misestimation approaches 49%+ as p tends to 50%. We have learnt that CUP voters are almost evenly divided between pro- and anti-Mas: if the difference between both positions is 0.7% or less, an assembly of 3,030 representatives such as held yesterday will fail to reflect the party's global position in more than 1 out of 5 cases.