# 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

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