Exit polls showed 96% of black voters in Ohio voted for Obama. If an average precinct records 500 votes, and there are 350 precincts in Cleveland, how many precincts would be expected to go unanimously to Obama? For the sake of argument, assume Cleveland is 100% black and that the Ohio exit poll is exactly representative of the actual vote in Cleveland.
I don’t think that you can assume that the people living in a precinct where everyone, or almost everyone, is black would have the same voting pattern as black people generally. For a start, they are likely to be significantly poorer and less well educated. (Black people who are well off and well educated might want to escape from the “ghetto”, and the same goes for the small minority of Black people who vote Republican.)
This struck me, too. (I posted about it in the Pit…) What I hadn’t realized was that some voting districts can be fairly small – I’m used to larger precincts. Also, I’m from a much more politically diverse neighborhood. Here, the probability of results being unanimous are absolutely zero, but in neighborhoods that are more homogeneous, it could happen more easily.
Someone also pointed out that many districts are gerrymandered in order to reinforce that homogeneity. If it’s known that there are a lot of black, or hispanic, or Jewish residents, the district borders may be re-drawn to include as many of them as possible in a single region. Some precincts are homogeneous by intent and design.
I found an online calculator which (if I did things right) said that if the probability of some event is 4%, the chance of 500 consecutive trials without any case of that event is a bit less than 1 in 400 million.
It will require several assumptions but…
with exactly 500 votes per precinct and 96% chance that a voter will go for Obama then…
[ul]
[li]the chance the first voter votes for Obama is 0.96 = 96%[/li][li]the chance both of the first two voters vote for Obama is (0.96)*(0.96) = 92.16%[/li][li]the chance the first n number of voters vote for Obama is (0.96)^n.[/li][li]so the chance that the first 500 voters all vote for Obama is (0.96)^500 = 0.00000000136652166… a really tiny chance 13.6 in 100 million chance that any particular precinct votes unanimously.[/li]The chances a precinct does not vote unanimously is 1 - 0.00000000136652166 =
0.99999999863 or 99.999999863%
[/ul]
The chance that no precincts vote unanimously is about (0.99999999863)^350 = 0.9999995205 = 99.99995205%
In short, the expected number of unanimous precincts is zero.
All of these calculations disregard the even more remote chance that a precinct in such circumstances might have voted unanimously for Romney.
…
You could also have “zero percent” if it’s less than 1 in 200, and it’s rounded down. And more likely than those urban precincts would be some very small rural precincts-- Consider that town in New Hampshire with 20 voters, that always reports first every year. It would not take a miracle for a place like that to be unanimous.
The town is Dixville Notch and there were only ten votes this year. And when the polls closed after 43 seconds of voting, Obama and Romney each had 5 votes.
As a matter of fact, the first time they held their own midnight election in 1960, they unanimously voted for Richard Nixon, 9-0. Dixville Notch, New Hampshire - Wikipedia
The problem is that precincts are generally very small (a few hundred to a few thousand people), and they tend to have a very non-diverse population. Precincts with people all of the same race and income level are quite common. If you mixed up the population of the U.S. and distributed them randomly over the country, it would be true that (in all likelihood) no precinct would have unanimously voted for Obama or Romney. But they weren’t mixed up and redistributed. Votes in each precinct are very biased to one candidate.
Bumping it up to 99% chance of a voter voting for Obama and keeping the precinct size at 500 voters…
Chance of any one precinct having a unanimous Obama vote is (0.99)^500 = 0.657%
Chance of any particular precinct does not have a unanimous vote is 99.343%
Chance that at least one in 350 precincts has a unanimous vote is 9.953% - not a great chance likely but not absurdly unlikely.
We are not likely to find 350 such precincts all in one major metropolitan area, but perhaps spread out nationwide there might be that many. There were 185,994 voting precincts in the United States as of 2004. Finding a few with a unanimous vote is not entirely unlikely.
Just out of curiosity, has anyone found any rural precincts in the west which went unanimously for Romney?
The calculations you people are doing are based on the assumption that the Obama and the Romney voters are evenly spread around the country. They’re not. They aren’t evenly spread around any state or any city or any country either. People in a given neighborhood tend to be alike in politics.
I did look around for the smallest precinct, and found one with two people and a dog in it:
However, it seems likely that that one either split or went for Obama, as it contains 1 Democrat and 1 Republican. The Republican expressed a dislike for Romney, apparently. The results from the precinct are rolled in with another so the couple’s votes can’t be individually identified.
How are these events dependent on one another?
Per the OP’s stipulation each voter has a 96% chance of voting for Obama - thus they are independent events. The calculations reflect that assumption.
The assumptions reflected in that calculation include other assumptions that fly in the face of reality. One important assumption is that there are essentially an infinite number of potential voters for any given precinct so that each voter going in has the same probability of voting for Obama.
We could change that assumption and suppose there are 1000 eligible voters but only 500 will vote. Suppose that 96% of those 1000 eligible voters will vote for Obama if they show up to vote, and 4% will vote for Romney if they shown up. Then every time someone shows up and votes for Obama the chance the next voter will vote Romney rises. That makes it even less likely to get a unanimous vote at any given precinct.
I wrote:
> . . . any state or any city or any country either . . .
I meant:
> . . . any state or any city or any county either . . .
How is he assuming an infinite number of voters? All he’s assuming is that the voters each vote independently, according to some distribution, and that’s pretty much true in the real world. What you’re doing is assuming, a priori, that a unanimous district is impossible, and then using that assumption to calculate the probability of a unanimous district. The logical flaw in this approach should be obvious.
My calculations were based upon all voters voting independently. Each voter has a 96% chance of voting for Obama. And there is a very tiny, pretty much zero, chance of a unanimous district given the OP’s constraints.
But certain assumptions that go into those calculations do fly in the face of reality.
Think of it like choosing socks from a drawer with the lights out. There are 960 black socks and 40 dark blue socks. They are hard to tell apart without the lights on but you don’t want to wake your spouse so you pull socks at random. You pull out 500 socks. What are the chances you pull all black socks?
The odds that sock #1 is black are 96%. And odds that sock #2 is black are 96% so long as I do not know the result of any prior selection.
But if you know you already pulled one black sock then the odds the second sock is black have changed… it is now 959/999 = 95.996%.
And if you know the first two pulls were black then the odds the third pull is black is 958/998 = 95.992%.
And so on. The odds a particular draw pulls a black sock change if you know the makeup of the sock drawer has changed… unless the sock drawer never changes (ie there are an infinite number and removing one does not change the contents). That is not reality.
The reality is there are a finite number of eligible voters in a precinct. Assuming a poll is correct (OP’s assumption) and 96% of the voters in that precinct prefer Obama then as each one votes for Obama then there are fewer potential Obama voters in the remaining pool of eligible persons and the odds the next voter in line is an Obama voter decrease marginally.
The odds that any given voter choose Obama have not changed… it’s still 96%.
The OP’s question can be restated 1) given 96% of a population of eligible voters will select Obama 2) what are the odds the first 500 voters in a given precinct will all vote for Obama. Inherently the question assumes knowledge and requires measurement of results as you go. This is an application of Bayes Theorem.
We just did a similar thread a few days ago - my search foo is lacking. There were a few Ohio precincts that were 100% Obama. The historical records showed only single digits for McCain in 2008, single digits in 2004 for Bush II, and 0 again in 2000. There were some precincts in Montana (Idaho?) that went all Romney.
Here it is - in the Stupid Republican Idea of the Day, BBQ Pit, starting at post 7582.