Thursday, November 5, 2009
Science / Election Percentages and Vote Counts
Posted by Jonathan Golob on Thu, Nov 5, 2009 at 6:31 PM
Today was a good day for McGinn, right? The difference in vote counts went up from yesterday, from 462 to 515. Fifty-three more votes went to McGinn.
- Counts in the way of McGinn
Wrong.
This is one of the most annoying fallacies of election reporting. While the final result is determined by the absolute counts, as we're using the ongoing vote totals as a sample of that final count we should be paying attention to percentages. By this more accurate measure, the mayoral race has tightened a bit since yesterday, from 0.44% to 0.40% (when excluding write-ins from the total).
- ... but the percentage tightens
Given the (innovative in city elections) use of last-minute get out the vote efforts by the McGinn campaign, I suspect these ongoing totals might not be as reflective of the true final counts as in the past. But for now, today's results are actually looking better for Mallahan, not worse.
(Don't even get me started about the foolishness of expecting machine or hand recounts to resolve an election whose final counts are within the margin of counting error. We might as well just flip a coin, if we're going to rely upon entropy to make our decisions.)
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2
Excellent points, Jonathan. You realize, of course, that to most people, particularly conservatives, there's no such thing as "margin of counting error". You get the same sort of innumeracy with the census.
3
Me too. I want to know what is the margin of error in tallying? You'd have to be pretty sure of what that number is before you can argue that the outcome is random.
4
I'm think'n co-mayors. y'know even days McGinn odd days Mallahan? tunnel gets dug a bit, then buried back up... it's all good.
6
Phil - Margin of error only applies to the range you can be 95% certain the results fall within if you get a purely random sample. Being that this isn't a purely random sample, and rather a trend comparison between two "samples" (or rather the prior cumulative and a new, smaller sample) the term "margin of error" is useless.
...I think.
...I think.
7
your concept of "margin of error" is indeed incorrect here. in this election, there are N votes for one and M votes for another. the experiment is over, figure it out. when you take a stack of bills to the bank, the bank counts 'em. there is no margin of error. there is an answer and there is only 1.
8
#6, he's talking about the margin of error of the vote difference with respect to the error rate of the counting method, ie, in 95% of recounts, the vote difference would fall within those bounds.
Jonathan's point here though, I think, is that using the same error-prone machines/hand count to resolve a vote difference within the margin of error makes no sense because it's just going to produce another random result (which, in 5% of cases, wouldn't even be in that range).
It's kind of like taking a poll, finding the result to be 49-51 with a margin of error of 3 points, then simply doing another identical poll to find the real results.
Jonathan's point here though, I think, is that using the same error-prone machines/hand count to resolve a vote difference within the margin of error makes no sense because it's just going to produce another random result (which, in 5% of cases, wouldn't even be in that range).
It's kind of like taking a poll, finding the result to be 49-51 with a margin of error of 3 points, then simply doing another identical poll to find the real results.
9
w7ngman -- That wouldn't really be a "margin of error," though, at least not in that sense. Machines do have error rates, but there's no formula for it and they're probably contingent on how common the anomaly that produces the error is.
That is, I doubt a perfectly completed ballot will ever be misread. The error rate is probably predominately, certainly disproportionately, people who screwed up voting in particular, creative ways that confuse an optical scanner. And there's no "margin of error" formula possible for that.
Geeky shit.
That is, I doubt a perfectly completed ballot will ever be misread. The error rate is probably predominately, certainly disproportionately, people who screwed up voting in particular, creative ways that confuse an optical scanner. And there's no "margin of error" formula possible for that.
Geeky shit.
10
In terms of predicting what the error-prone machines will tall from remaining ballots:
137,025 have now been counted.
Assuming 210,000 total ballots, the 95 percent confidence interval (margin of error) would be ± 0.16%. So unless there is something materially different in the ballots to be counted versus what's already been counted, the difference between the two candidates is statistically significant.
137,025 have now been counted.
Assuming 210,000 total ballots, the 95 percent confidence interval (margin of error) would be ± 0.16%. So unless there is something materially different in the ballots to be counted versus what's already been counted, the difference between the two candidates is statistically significant.
12
I'm not convinced that this is good news for the Mallahan campaign. Sure, the percentage difference narrowed, but not as fast as they needed it to.
After today, the share of the remaining ballots that Mallahan needs to get in order to win has increased. (He needs to win them by 516 votes instead of 463, and there are about 27,000 fewer ballots left to count.)
Today's ballot drop also suggests that the McGinn campaign could be right that later ballots will favor them. The movement between yesterday's ballots (which favored Mallahan by 2.25%) and today's (which favored McGinn by 0.21%) is certainly good news for them.
After today, the share of the remaining ballots that Mallahan needs to get in order to win has increased. (He needs to win them by 516 votes instead of 463, and there are about 27,000 fewer ballots left to count.)
Today's ballot drop also suggests that the McGinn campaign could be right that later ballots will favor them. The movement between yesterday's ballots (which favored Mallahan by 2.25%) and today's (which favored McGinn by 0.21%) is certainly good news for them.
13
The result will be: McMallaGin, a half-fat hairy corporate douchebag with a JD and a cellular bicycle. No management experience, but smells a hell of a lot like Factoria.
14
The law takes margin of error -- or something like it -- into account, in that it requires a recount if the difference is within (I think) half a percentage point. If the race is that close, the law wants to eliminate the possibility that it was decided by human or machine error. So we recount. But then the bank analogy made by a previous commenter kicks in. We stack 'em all up and take a look at 'em and see how many votes there really were for each candidate. I hardly think that qualifies as "entropy" or, as the media and Republicans liked to call it "chaos." Plus we get to find out who wrote in "Lizard People."
15
This has nothing to do with sampling errors and 95% confidence intervals or any of that. Some of you are just confused. It's more of a quality control question, like how many bad widgets does your assembly line produce?
The error rate of the tabulation machines is the rate that the optical scanner misreads the marks on the page. Due to how the drunk voter filled in the bubble. Or whatever.
From what I can glean, error rates of 1 in 10,000 are pretty good, so your count will vary by 0.01%. But possibly error rates of 1% to 3% exist, at least according to controversies that have happened around the country in the past with various kinds of machines.
The question is what kind of machines does Washington use and what is their expected error rate?
If it is only 0.01% then the it's going to only vary by about 18 or 20 votes, assuming we have 50% turnout of 375,164 ballots. Far smaller than McGinn's current lead. If it is as high as 1%, which is like Florida's crappy old machines, each time you recount the number is plus or minus 2,000 votes, way higher than the real difference, and that makes the result a toss up.
So it matters a lot how well the OCR machines scan. If the error rate is low enough, then you can't call the result random.
And the error rate of a manual recount is some other number I have no idea of.
So once again, just where are you getting your assumptions about the error rate of the machines used here in on our elections? It makes all the difference.
The error rate of the tabulation machines is the rate that the optical scanner misreads the marks on the page. Due to how the drunk voter filled in the bubble. Or whatever.
From what I can glean, error rates of 1 in 10,000 are pretty good, so your count will vary by 0.01%. But possibly error rates of 1% to 3% exist, at least according to controversies that have happened around the country in the past with various kinds of machines.
The question is what kind of machines does Washington use and what is their expected error rate?
If it is only 0.01% then the it's going to only vary by about 18 or 20 votes, assuming we have 50% turnout of 375,164 ballots. Far smaller than McGinn's current lead. If it is as high as 1%, which is like Florida's crappy old machines, each time you recount the number is plus or minus 2,000 votes, way higher than the real difference, and that makes the result a toss up.
So it matters a lot how well the OCR machines scan. If the error rate is low enough, then you can't call the result random.
And the error rate of a manual recount is some other number I have no idea of.
So once again, just where are you getting your assumptions about the error rate of the machines used here in on our elections? It makes all the difference.
16
I don't know if the idea of looking at percentage differences instead of absolute differences really applies here. It's going to be close enough for a recount, no matter who wins.
Yes, Mallahan's percentage of votes already counted increased slightly, so that's negative for McGinn.
But, the remaining votes to be counted shrunk by, what, 25k? And McGinn's margin increased slightly in absolute terms. So, now Mallahan has to make up more votes, and he has fewer uncounted votes to make them up from. Thus, the percentage of remaining, uncounted votes that Mallahan has to get increased slightly as well.
Looked at from that point of view, it's a losing day for Mallahan.
For example, let's say that before the drop today, there were 100k votes left to count, and McGinn had a 462 vote lead. Mallahan would then need 50,232 votes (100k/2 + 463/2 votes), which is 50.232% of the remaining vote.
After the drop today, there were 75k votes left, and McGinn has a 515 lead. So, Mallahan needs 37758 votes (75k/2 + 516/2), which is 50.344% of the remaining vote.
The remaining vote numbers are just examples, so the percentages may vary, but the relationship between the percentages will not - Mallahan's needed percent of remaining votes went up today.
You only have to win by one vote, not by a certain percentage. Good day for McGinn.
Yes, Mallahan's percentage of votes already counted increased slightly, so that's negative for McGinn.
But, the remaining votes to be counted shrunk by, what, 25k? And McGinn's margin increased slightly in absolute terms. So, now Mallahan has to make up more votes, and he has fewer uncounted votes to make them up from. Thus, the percentage of remaining, uncounted votes that Mallahan has to get increased slightly as well.
Looked at from that point of view, it's a losing day for Mallahan.
For example, let's say that before the drop today, there were 100k votes left to count, and McGinn had a 462 vote lead. Mallahan would then need 50,232 votes (100k/2 + 463/2 votes), which is 50.232% of the remaining vote.
After the drop today, there were 75k votes left, and McGinn has a 515 lead. So, Mallahan needs 37758 votes (75k/2 + 516/2), which is 50.344% of the remaining vote.
The remaining vote numbers are just examples, so the percentages may vary, but the relationship between the percentages will not - Mallahan's needed percent of remaining votes went up today.
You only have to win by one vote, not by a certain percentage. Good day for McGinn.
17
@16 - Yes, but Jonathan can counter that the latest count indicates the vote is trending toward Mallahan, thus making it more likely, if the trend continues, that he will make up the needed ground.
19
15, "The question is what kind of machines does Washington use and what is their expected error rate?"
Indeed. I'd be curious to know. When we say "expected error rate", however, are we talking about error due to machine operation anomaly, or error due to drunk voter? It's an important distinction that I think we've both glossed over.
"Drunk voter" error is where intent was not sufficiently communicated via the ballot. They underfilled the bubble. They filled in two bubbles. They spilled ink on the page. A ballot is judged identically from recount to recount. Correcting these errors with a machine recount is hopeless. The level of ambiguity may even be high enough that the error cannot be corrected by a hand recount. See: hanging chad. Even worse, we'll never have a good idea of the real rate of this type of error due to voter anonymity (ie, you can't call up a voter and ask him what he meant in order to verify the error rate).
"Machine error" discussed in #8, and what I thought Jonathan was alluding to here when he talked about flipping a coin, involves ballots where intent was sufficiently communicated. The scanning hardware failed randomly. A fly landed on the mainboard. A bug occurred in the recognition software and it was not dependent on the content of the ballot. It could also be the result of hand recounts (eg 0.5% of the time, someone tallies the wrong column). A given ballot is judged differently from recount to recount. Note that this is the type of error that an ideal hand recount is always capable of correcting.
Machine error very much has to do with 95% confidence intervals. With respect to machine error, the current counting process will produce only one of 2^n possible outcomes, where n is the number of ballots cast. For all 2^n possible outcomes, the difference between the count result and the actual result is, I suspect, a normal distribution. Key point from #8: "In 95% of recounts, the vote difference would fall within [x of the actual difference]". This is what I was, loosely, calling a "margin of error" re the OP and #1.
I think what Jonathan is getting at here is that, when the "final counts are within the margin of counting error", we could do 5 machine recounts and still not have a good idea of exactly where the certified result lies on that normal distribution.
So yeah, I'd be curious to know more about the machine error rate and how it (hopefully) played into the choice of recount thresholds. Another question: what is the intent of the 2000 vote difference part of the rule?
More...
Indeed. I'd be curious to know. When we say "expected error rate", however, are we talking about error due to machine operation anomaly, or error due to drunk voter? It's an important distinction that I think we've both glossed over.
"Drunk voter" error is where intent was not sufficiently communicated via the ballot. They underfilled the bubble. They filled in two bubbles. They spilled ink on the page. A ballot is judged identically from recount to recount. Correcting these errors with a machine recount is hopeless. The level of ambiguity may even be high enough that the error cannot be corrected by a hand recount. See: hanging chad. Even worse, we'll never have a good idea of the real rate of this type of error due to voter anonymity (ie, you can't call up a voter and ask him what he meant in order to verify the error rate).
"Machine error" discussed in #8, and what I thought Jonathan was alluding to here when he talked about flipping a coin, involves ballots where intent was sufficiently communicated. The scanning hardware failed randomly. A fly landed on the mainboard. A bug occurred in the recognition software and it was not dependent on the content of the ballot. It could also be the result of hand recounts (eg 0.5% of the time, someone tallies the wrong column). A given ballot is judged differently from recount to recount. Note that this is the type of error that an ideal hand recount is always capable of correcting.
Machine error very much has to do with 95% confidence intervals. With respect to machine error, the current counting process will produce only one of 2^n possible outcomes, where n is the number of ballots cast. For all 2^n possible outcomes, the difference between the count result and the actual result is, I suspect, a normal distribution. Key point from #8: "In 95% of recounts, the vote difference would fall within [x of the actual difference]". This is what I was, loosely, calling a "margin of error" re the OP and #1.
I think what Jonathan is getting at here is that, when the "final counts are within the margin of counting error", we could do 5 machine recounts and still not have a good idea of exactly where the certified result lies on that normal distribution.
So yeah, I'd be curious to know more about the machine error rate and how it (hopefully) played into the choice of recount thresholds. Another question: what is the intent of the 2000 vote difference part of the rule?
20
It is difficult to talk about margin of error, but you can talk about error rates. You can try to impute the error rate from what people intend to do when they start marking their ballot, vs. what is counted. For the type of ballots used in King County, the error rate is generally less than 3% when the ballot is voted in a polling place. That's good compared to other voting systems. However, when the ballots are mailed in, the error rate is generally higher, perhaps a bit over 3%. Also, less experienced (younger and typically more liberal) voters have a higher error rate, and this effect is exacerbated by mail voting. Thus mail voting tend to skew an election conservative. If you think of Mallahan as the choice of more conservative voters, then he's the beneficiary of mail voting distoritions.
However, part of the votes lost by machine counting mailed ballots are recovered by hand counting them.. (A machine recount doesn't really mean much --it's window dressing.) This is because less experienced voters tend to mark their ballots in ways that are more often misread by machines, but still legal. So, if Mallahan loses the machine recount, he probably can't win a hand recount. But if McGinn narrowly loses a machine recount, it means a hand recount would show he legally won the election.
However, part of the votes lost by machine counting mailed ballots are recovered by hand counting them.. (A machine recount doesn't really mean much --it's window dressing.) This is because less experienced voters tend to mark their ballots in ways that are more often misread by machines, but still legal. So, if Mallahan loses the machine recount, he probably can't win a hand recount. But if McGinn narrowly loses a machine recount, it means a hand recount would show he legally won the election.
21
@17 - if he were to counter that the latest totals shows a trend towards Mallahan, he would be wrong. McGinn picked up votes (in this particular ballot drop) - it does not show a trend towards Mallahan winning, but toward McGinn winning by less than the margin he has now - the total (extrapolating a trend from this one data point, which is a bad idea, but it's what Jonathan seems to be doing) is converging on a smaller margin for McGinn. Not on a margin for Mallahan. No amount of ballot drops where McGinn picks up votes can make Mallahan win.
23
Thanks a lot for those graphs. Which would have rendered EXACTLY THE SAME no matter what numbers you plugged in. A real efficient use of pixels there.
24
I have an easy solution.
McGinn can be mayor for the first 100 years, and then Mallahan can have his turn.
McGinn can be mayor for the first 100 years, and then Mallahan can have his turn.
26
Great comment thread: educational and thought provoking - it almost gave me hope in the Slog community. Then predictably the slog zombie clowns return to shit and wipe their stink around the comments. Thanks Will. Maybe Baconcat can chime in with some more inane sycophant BS too.
27
This is the stupidest argument I have heard in a long time. If McGinn gains 53 votes each of the next few days his percentage will go down but McGinn will win.
28
@22, I'm pretty sure he gained. At least a couple pounds. Bwahaha. (Sorry for being juvenile, @26.)
29
The declining margin of victory for McGinn is relevant in that it makes a manual recount more likely.
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