# Really a knife's edge?

Something nice and topical. I just read these two items on the news:

• Obama is in the lead by 50.4% to 48%, with 61% of votes counted. (Ohio)

• With 86% of the vote counted, Virginia is still sitting on a knife edge. Romney is hanging on to a lead of 49.9%, but Obama is snapping at his heels on 48.7%.

Intuitively they sound like "dead certs" not "knife edge", because the populations involved must be huge. Then I realized I didn't know exactly what to do to prove my thesis.

Formalizing my thesis: I'm 99% sure that the current leader won't change once all votes are counted. But the question I have is: how many voters do there have to be in each state to be able to claim this?

NOTE: to be clear on definition, if there are N voters in Ohio, and Obama is leading 50.4:48 based on counting 0.61 x N votes, what is N to have 99% confidence that the Obama ratio will be >=50?

UPDATE: The comments have explained to me that the 86% was of precincts, not voters, with precincts supposed to represent around 2500 voters. But more importantly there may be an urban bias in the precincts that get counted last. If you wish to post an answer please assume my original assumptions: that it is 86% of actual voters, and that there is no bias in the votes counted so far. P.S. Apparently, Virginia ended up with Obama 1,868,191 (50.57%) to Romney 1,767,692 (47.85%).

• Look for the Gambler's Ruin problem. It is a standard result in probability theory that in independent tosses of a fair coin, the lead changes very infrequently. Nov 7, 2012 at 4:50
• As you've no doubt noted by now, Virginia changed leaders. The key issue is different from those mentioned in the question: it's whether the current sample is representative of the population. Nov 7, 2012 at 7:19
• @gung No, I'd not realized about Virginia, fascinating! Are you saying that 61% is not a random sample? Does it mean 61% of geographic sub-areas, rather than 61% of voters? In that case I take it all back!! (Sorry, as you can see I'm ignorant of the election mechanics.) Nov 7, 2012 at 8:05
• The "61% of votes counted" are not 61% of votes throughout the state. They are 61% of precincts, and the order in which they report is probably not random. For example, many polling places stayed open beyond the official time in urban areas where lines were extremely long, so rural areas would be more likely to be in the 61%. Nov 7, 2012 at 13:18
• (+1) to @Wayne's comment. To take it one step further, it should be noted that precincts and districts are purposefully constructed to be as unrepresentative as possible in (very) many cases. So, it makes a huge difference which precincts have reported when trying to assess who may have taken a particular state. Nov 7, 2012 at 14:17

## 1 Answer

This is an old question, but we now have the advantage of knowing the final result:

$$\text{Obama: } 51.16 \text{%} \quad \quad \quad \text{Romney: } 47.28 \text{%}.$$

As you can see, Obama ended up winning this one with a handy lead of a few percentage points, which is quite far away from the preliminary count you mention in your question. The issue with these kinds of problems is that the counted votes are not a random sample of all the votes and so they are not necessarily a "representative" sample of the total. For this reason, you will get falsely confident results if you apply standard sampling methods that assume simple random sampling.

For a number of reasons, there can be systematic differences in the sample of counted votes and the entire set of votes. For example, votes might be counted in roughly the order they were cast (e.g., boxes of votes counted in the order that they were filled), and this could be correlated with the voting direction. (For example, it might be the case that Obama voters are more likely to vote near the end of the day, and Romney voters are more likely to vote earlier in the day.) There may also be differences in when the different precincts are counted, so that the preliminary count is weighted in favour of certain precincts, which are correlated with the voting direction.

For these reasons, prediction models for elections are quite complicated, and they generally try to operate at a level that takes account of the particular polling booths where votes are counted, the changes in voter composition over the day, differences between postal votes and booth votes, and any other factors where there is past data that can give statistical information. There is more uncertainty in these predictions than would be the case for standard statistical methods that are built on simple random sampling.

Good thing you didn't bet money on this one!