A vote in my university had 2800 eligible voters, 16.8% turnout, and 54.7% of yays.

Many people on both sides are wondering if the small electorate, low turnout, and narrow margin means the results were misleading.

Is there a formula or a theorem to estimate what is the chance of 1400 or more out of all 2800 being nays when 213 out of 470 are nays?


  1. It was a binary choice.

  2. No one was absolutely indifferent to it. Everyone wanted to show up and vote either yay or nay.

  3. Everyone had to vote either yay or nay; no one - who showed up - was allowed to abstain.

  4. There was undisputed manipulation of the electoral procedure, but this is hard to quantify, so one could ignore it in order to try to arrive at any result at all.

  5. The consensus is that there was a bias in favor of yays showing up (e.g. voter suppression, I think).

  6. Assuming no bias could give a rigorous lower bound.

  7. Therefore, we can assume that the sample is an unbiased representative of the population.

  • You can force people to vote and you can't know what they would have voted for if they would have voted. Commented Apr 11 at 4:58
  • 12
    Everyone wanted to show up and vote either yay or nay. I'd say the 16% turnout utterly disproves that notion, and does so most emphatically. You've missed on that so badly it raises serious doubt about every single subjective interpretation you've posted.
    – Just Me
    Commented Apr 11 at 12:51
  • 4
    @JustMe my thoughts exactly. Regardless of the results, 4 of each people actively trying to vote and being refused would be a major scandal, specially when you consider a university where all of the voters are expected to have a minimum of education. The OP would not need to get guesses about what the result would have been with more participation to claim interference. My guess is that the OP has misled himself into believing the issue is more important to the voters than the voters actually think.
    – SJuan76
    Commented Apr 11 at 12:56
  • 1
    @SJuan76 I can see other, more passive ways to "manipulate the election procedure" to influence the result than turning people away. Scheduling it for an inconvenient time or place, for example. That said, even with something like that in place, given that 17% were able to make it to the vote it's unlikely that the rest cared as much as OP thinks they did, or they'd have gone through the trouble to make it to an inconvenient vote, too.
    – Bobson
    Commented Apr 11 at 13:55
  • 2
    Bias in the turnout doesn't imply voter suppression. Often there's a correlation between the voter's preference and their motivation to vote.
    – Barmar
    Commented Apr 11 at 23:02

6 Answers 6


A vote is not an opinion poll. Instead of extrapolating the behavior of a subset of people to everyone, it measures the behavior of all people.

The conclusion from a vote is that set out by the rules, and no other conclusions must be drawn from the results. Speculating about opinions not voiced, even if backed by scientific reasoning, is inherently undemocratic.

Not showing up for the vote is also a relevant and democratically justified behavior. No side in a controversy might claim non-voters were on their side.

If the rules did not prescribe a minimum turnout (or some variant like "a majority of yays plus a minimum percentage of yays from all eligabe"), or if this rule was met, the outcome of the vote is approval. Supposed those voters that did not show up knew and understood that implication, they knew their behavior left the decision to those that voted – a de facto agreement with whatever outcome happened.

Personal comment: This point seems to be difficult to grasp. As long as not voting is a free choice, it contains a decision: the will not take a side and to leave the decision to others. I oppose speculating about "which side they were on" because their opinion is not unknown. It was voiced, and respect demands that no-one misrepresents that.

If you will, imagine a dialogue of the form: "But in reality, you must be on one side or the other! Everyone is!" – "No, I am not on any side." – "But if you were really, really forced to take one side, which would it be?" – "I do not want to be forced."

In calling speculation undemocratic, I am defending the right of people not to be forced to accept a false dichotomy.

Now all that is left is a judgement about the fairness of the vote.

In your case, 2330 of the eligable voters abstained from voting. There are three possible explanations for their behavior (each of them individually – no collective reason can be implied):

  • he/she did not know about the vote,
  • he/she was actively hindered from voting,
  • he/she chose freely not to vote.

Can you find quantities for each of these?

For the first reason, you might ask, how was the vote publicised? Was every voter informed in person about the vote? If, for example, letters or e-mail were sent, you could estimate a number of non-deliveries as an upper bound. Without individual information, you are limited to a qualitative judgement.

For the second reason, this is mostly a qualitative judgement, unless there were organisatorial measures to hinder individual voters. If voters that showed up were denied, and these were documented, you could name a lower bound. If the opportunity to cast a vote was made technically impossible for some, you might be able to quantify that.

For the third reason, they were not hindered from voting.

You could call the vote unfair if the number of people hindered by the first two reasons is large enough that, had they voted, the outcome could have been different. There were 257-213=44 more yays than nays. If 45 persons wanted to vote, but were hindered, their vote could have potentially changed the outcome.

This is no speculation about how they would have voted, it only names a lower bound on how many more persons were needed to prevent an approval.

Can you prove 45 persons or more were hindered from voting? Then the outcome of the vote should be voided. Otherwise, the democratic conclusion is that it stands.

  • @GabrielAndrade - "Barring political oposition from voting by declaring them criminals" is felony disenfranchisement, which has indeed been criticized as undemocratic, but is independent from compulsory voting (you can have voluntary voting by current and previous prisoners, and you can have compulsory voting that excludes them). Anyway, compulsory voting is fairly unusual in the grand scheme of things. Only about 21 countries have it, and only a handful of those (such as Brazil, Belgium, Peru and Greece) have practically relevant penalties for refusing to do so.
    – Obie 2.0
    Commented Apr 11 at 20:14
  • @GabrielAndrade - Well, Australia, for instance, has both compulsory voting and limited felony disenfranchisement, whereas a lot of countries both lack felony disenfranchisement and have voluntary voting. So, no, the two things may happen to coincide in, say, Brazil, but not universally (and I would argue that it is primarily true because both felony disenfranchisement and compulsory voting are rare at the international level that they rarely coincide, and less because of intrinsic incompatibility).
    – Obie 2.0
    Commented Apr 11 at 20:53
  • 1
    Note that 'poll' is polysemic in English. Wikipedia: "An opinion poll, often simply referred to as a survey or a poll (although strictly a poll is an actual election)". What you mean is that a vote is not a randomly/scientifically sampled poll. Instead the sample is (usually) self-selecting. Which is also why assuming that a random subsample showed up to vote is typically not a good assumption. So, answers here/below which just apply some Gaussian distribution etc. are bad. Commented Apr 12 at 6:33
  • @thegodsfromengineering Thanks for the language advice. As an aside, my critique is more involved than talking about the randomness of samples, it reminds that abstaining from an opinion is an opinion itself. Any form of subsuming non-voters in the opinions of voters is a misrepresentation.
    – ccprog
    Commented Apr 12 at 13:35

If the election results reflect a random, unbiased sample, a 95% confidence interval would say that the measure was supported by 54.7 - 4.11 (50.59%) to 54.7 + 4.11 (58.81%) of the total population. A 99% confidence interval would be 54.7 - 5.4 (49.3%) to 54.7 + 5.4 (60.1%). So if you assume an unbiased sample, it is plausible though very unlikely that the population as a whole opposed the measure.

There are various sample size calculators online where you can plug in the numbers and play around with the confidence interval you want.

Of course, the math behind these calculations assumes an unbiased sample which is unlikely in any sort of election. Even in the absence of electoral manipulation there is almost always an enthusiasm gap. Even if no voter is indifferent, one side's supporters are likely to care more and thus be more likely to prioritize voting over other things they wanted to do that day.

  • People who show up to vote are hardly ever a random subsample of the whole (vote-entitled) population. So the first para isn't terribly useful. Commented Apr 12 at 6:38
  • Also in the 1st para, I'm not seeing why you assume the absentees were opposed, instead of being indifferent. Absent=opposed is a parly reasonable assumption if the measure only passes if certain a quorum is met, in which case the opposition can/does strategically boycott the vote. Most opinion polls that try to predict something other than an election result have a scale of agreement/disagreement, with more than 2 options. Commented Apr 12 at 6:45
  • I'm not assuming that the absentees were opposed. I'm making the assumption (requested in the question) that there was no bias in who showed up so the election results were an unbiased sampling of the population as a whole. If the election required a certain quorum to be valid (which wasn't specified in the question and 16.8% turnout leading to passage would suggest not) then that would be one more reason that the election was unlikely to be an unbiased sample. Commented Apr 12 at 14:03

If no one else would vote for yay: 54.7 * 16.8 = 9.19% is the minimum votes for yay.
If everyone else would vote for yay: 9.19+(100-16.8) = 92.39% is the maximum votes for yay.

That doesn't say much.

There are some things you could do with the assumption that the sample (i.e. who exactly turned out) is a representative of the population.

But that is not the case. Those who turned out cannot be considered as a representative of the population. There are other factors at play into who goes to vote.

Estimates such as opinion polls and exit polls follow a good method of sampling, but even those frequently go wrong. So we can't really say.

  • 4
    @GabrielAndrade No, those assumptions are simply incorrect.
    – whoisit
    Commented Apr 11 at 2:06
  • 1
    @GabrielAndrade If you assume no bias in who turned out, then the final percentages would be exactly the same as they are now, just based on 100% turnout instead of 16.8%. After all, no bias would mean that for every 5.4 extra people who showed up to vote yay, another 4.6 would show up to vote nay, just like with the current numbers. So the question is how much bias was there, and the only way to answer that is to have some other value to compare it to, like the polling that this answer mentions.
    – Bobson
    Commented Apr 11 at 11:18
  • @GabrielAndrade - I see what you're looking for, now. I just posted an answer about margin of error, which is the term for what you're trying to calculate.
    – Bobson
    Commented Apr 11 at 11:51

Is there a formula or a theorem to estimate what is the chance of 1400 or more out of all 2800 being nays when 213 out of 470 are nays?

We can estimate that, of 2800, 2330 did choose not to vote.

Unless you consider a scenario of voter suppression (which you do not mention), you cannot seriously say "hey, those people would have voted the way I did want them to vote." The fact is that such people did have a right to vote and did choose not to do so.

It is like asking if I would have chosen coca cola or pepsi, after I had tell you that I want to drink beer.

If there were an scenario of voter suppression, you would need way more data: how voters were suppressed, and an estimate about how those supprewsed voters would have voted. For example, if group A is 90% nay and, of 1000 voters of group A 1000 were unable to vote despite wishing to, you could estimate 900 more nays and 100 more yays.

The maths are simple, what would be difficult worker be making the assumptions of what those people would have voted, and establishing in the first place of voter suppression has taken place.

  • 4
    This makes little sense. In one side you say everybody was very involved, on the other side less than 1 in 5 voters did vote. Either it is public and notorious that many eligible voters were not allowed to vote (and then you do not need the kind of cabalistic figures you are asking about), or you are misleading yourself about the feelings among the students about the issue. Which is kind of normal, because people tend to get involved themselves with people who share their views...
    – SJuan76
    Commented Apr 11 at 12:52

The consensus is that there was a bias in favor of yays showing up (e.g. voter suppression, I think).

Therefore, we can assume that the sample is an unbiased representative of the population.

These two statements are a self-contradiction in your 'proof'. (Unless by 'sample' you mean the whole population, including those who didn't vote, which is a pretty nonsensical use of the term 'sample' then.)

Whether the 1st one (quoted) is a reasonable assumption depends on how well the vote was advertised etc., and what are the rules for the results to be [legally] valid (e.g. is there a minimum quorum for the result to be valid, and if it's not deemed valid what are the practical implications.)


Based on your comments, it sounds like what you're actually looking for is the "margin of error" in the turnout results. It's usually applied to surveys, where the pollster is trying to extrapolate the opinions of a large group of people from interviewing a much smaller subset of them, which is basically what you're trying to do here.

The formula for margin of error is pretty straightforward, although knowing what to plug in if you're not a math person could be challenging. Fortunately, there's online calculators for it, such as this one. Plugging in your numbers of a population of 2800 and a sample size of 470, and choosing a 95% confidence level (more on that in a moment), we get a margin of error of 4%.

What this means is that you can be 95% confident that the real number if everyone had voted would be between 50.7% and 58.7%.

The confidence level is an arbitrary choice you make to trade off between precision and accuracy. The more precise you get with your result (the smaller the range), the less likely you are to be accurate in that the real number is within that range. 95% is the industry standard for polling and scientific papers, but that still means that one out of 20 times, the real number is outside your range. This particular calculator I used doesn't give fractional margin of error, but from plugging in other confidence levels, you can be 99% certain that the result is between 49.7% and 59.7%, or you can be 80% certain that the result is between 51.7% and 57.7%.

  • It sounds like OP is particularly concerned about the possibility that, had everyone voted, the percentage of yays would have been below 50%, but judging from your data, the odds of that happening were less than 1%.
    – F1Krazy
    Commented Apr 11 at 12:46

You must log in to answer this question.