Gelman et al only got to the conclusion that the probability of a random vote being decisive is about 10^-7 by having a model of how different states’ votes relate to one another.
True, that number may have changed somewhat. It may have decreased somewhat due to the voting population being larger, or increased somewhat due to the election being projected to be somewhat closer than 1992 turned out to be. But I’d expect the probability to mainly just shift between states. So 10^-7 made a pretty good baseline.
Accordingly, they have California as very unlikely to be tied, and really quite likely to be decisive if tied. I repeat: this is not at all the same as saying that it’s at all likely actually to be decisive.
Since you press the issue, I looked up how exactly Nate Silver defines the probabilities of being decisive that he uses on 538. He says:
The most rigorous way to define this is to sort the states in order of the most Democratic to the least Democratic, or most Republican to least Republican. Then count up the number of votes the candidate accumulates as he wins successively more difficult states. The state that provides him with the 270th electoral vote, clinching an Electoral College majority, is the swingiest state — the specific term I use for it is the “tipping point state.”
So you’re right; the probability of a state being decisive is not quite the same as the probability conditional on it being tied. [Edit: actually they are not even close to the same. And they wouldn’t have been even if 538 defined tipping-point state differently.] But (probability of single vote in California being decisive) = (probability of CA being the decisive state) * (probability that CA is tied given CA is the decisive state) = (probability of CA being the decisive state) * (probability of a randomly selected vote being a vote that ties CA given CA is the decisive state) / (probability of a randomly selected vote being cast in CA).
The assumption that I made was that (probability of a random vote tying CA given that CA is the decisive state) is close to (probability of a random vote tying whatever the decisive state happens to be), which seems fairly reasonable. I did NOT assume that (probability that CA is tied given that CA is the decisive state) is close to (probability that CA is tied). ( [Edit: ignore this parenthetical comment] At least I think I didn’t, but I’m tired right now, so it is conceivable that I could have made an error on that front. If I did mess that up, then the actual probability of a vote in CA swinging the election should be greater than 1 in 1 billion, but still probably less than 1 in 100 million.)
The three states’ political fluctuations are completely independent of one another.
That assumption is not even close to correct in US presidential elections. Not that that makes much of a difference (I think).
(2) conditional on Blue being tied, Blue will almost certainly be decisive.
Almost certainly? I thought you said Red was solid and Purple was only slightly bluish, so there should be a significant chance that Red and Purple both vote red, and Blue’s surprise red vote is redundant.
Pr(Blue decisive) is also 0.1%
Sounds like you are using the Gelman et al meaning of a state being decisive. Not what 538 calls being the tipping-point state, which is what I was using. That’s why you get garbage when you plug that into my formula.
The assumption that I made was that (probability of a random vote tying CA given that CA is the decisive state) is close to (probability of a random vote tying whatever the decisive state happens to be), which seems fairly reasonable.
So, first of all, there’s a factor of 2 error in there: your last equality says, in effect, Pr(CA tied | …) = Pr(a random vote ties CA | …, and that vote is in CA) but when CA is tied only half the votes there tie it.
I’m already late for work, so will look harder at the rest of what you’re saying later. (I find myself somewhat convinced both by my toy example and by the more-detailed argument you’re now making, but the two don’t seem consistent with one another. I expect I’m missing something.)
That assumption is not even close to correct in US presidential elections.
I know. That’s one reason why the example is a toy. But no part of your argument appeals to the correlations between states’ results, and the point of the toy example is to show that the calculation you’re doing produces completely wrong results in a toy example, so in the absence of anything in your argument that makes it apply to the real election and not to the toy example something’s got to be wrong with the argument.
Almost certainly?
Sorry, “slightly bluish” was meant to describe vote share rather than win probability. I’m assuming that P is a win for the Blue candidate about 90% of the time, which with simple-but-unrealistic models of voters will happen if it’s reasonably big and, say, 55% Blue.
Sounds like you are using the Gelman et al meaning of a state being decisive. Not what 538 calls being the tipping-point state, which is what I was using.
I was intending to use the 538 meaning. Pr(Blue decisive) is small because in almost all elections the state that gets the winner that crucial third EV—the one whose EV is in the middle when you line them up in order—is Purple. What do you find wrong with this reasoning?
So, first of all, there’s a factor of 2 error in there: your last equality says, in effect, Pr(CA tied | …) = Pr(a random vote ties CA | …, and that vote is in CA) but when CA is tied only half the votes there tie it.
Nope. Half the votes prevent a Romney victory, and the other half prevent an Obama victory.
I’m already late for work, so will look harder at the rest of what you’re saying later.
Your confusion is understandable, especially since I confused myself and started bullshitting you for a while before rederiving what I did in the first place. Sorry about that.
But no part of your argument appeals to the correlations between states’ results
That’s right. Sorry, I shouldn’t have been stressing the high correlations between voting fluctuations in different states.
Sorry, “slightly bluish” was meant to describe vote share rather than win probability. I’m assuming that P is a win for the Blue candidate about 90% of the time
Ah, ok.
I was intending to use the 538 meaning. Pr(Blue decisive) is small because in almost all elections the state that gets the winner that crucial third EV—the one whose EV is in the middle when you line them up in order—is Purple. What do you find wrong with this reasoning?
Numbers from your toy example:
Pr(Blue tied) = 0.1%
Pr(Blue decisive | Blue tied) = 90%
Pr(Blue decisive) = 0.1%
implications:
Pr(Blue decisive and tied) = 0.09%
Pr(Blue decisive and not tied) = 0.01%
This is not plausible. Presumably Pr(Blue decisive and votes blue by 1 vote) is also roughly 0.09%, in which case Pr(Blue decisive and not tied) cannot possibly be less than that. Assuming Red never enters the picture, Blue is decisive whenever it ends up voting more reddish than Purple does. Given how often Blue ties, I would expect this to actually happen fairly frequently.
Apologies for the slow response; I’ve been unreasonably busy. Executive summary of what follows: Yup, you were right.
So I tried generating more realistic numbers with the general structure of my toy example, and my conclusion is: Oops, you’re right and my example is no good. Sorry. And I think I agree with your simple probability-pushing argument that 538′s probability for California being decisive isn’t consistent with the numbers from Gelman et al being applicable in the 2012 election.
So, it seems to me that there are (at least) the following possibilities. (1) Gelman et al had a good model, and it remains reasonably applicable now, and 538 had too low a probability of California being decisive. (2) Gelman et al had a good model, but the political landscape has changed, and now California is less likely to be decisive than their model said it was in 1992. (3) Gelman et al had a screwed-up model, and their probabilities weren’t right even in 1992.
I agree with you that #2 is the least likely of these, and I offer the following statistic which, if cited at the outset, might have saved us a good deal of argument :-). In 1998, California went Democratic by about 51:48. In 2012, California went Democratic by about 59:39.
I accordingly agree with you: Academian’s numbers for his own case, which used the Gelman et al figures for California, likely gave much too high an expected value for his vote in California.
I agree with you that #2 is the least likely of these, and I offer the following statistic which, if cited at the outset, might have saved us a good deal of argument :-). In 1998, California went Democratic by about 51:48. In 2012, California went Democratic by about 59:39.
I assume you meant #2 is most likely? And you’re right; I should have pointed that out initially (even though it was before the election, I could have used 2008 figures).
True, that number may have changed somewhat. It may have decreased somewhat due to the voting population being larger, or increased somewhat due to the election being projected to be somewhat closer than 1992 turned out to be. But I’d expect the probability to mainly just shift between states. So 10^-7 made a pretty good baseline.
Since you press the issue, I looked up how exactly Nate Silver defines the probabilities of being decisive that he uses on 538. He says:
So you’re right; the probability of a state being decisive is not quite the same as the probability conditional on it being tied. [Edit: actually they are not even close to the same. And they wouldn’t have been even if 538 defined tipping-point state differently.] But (probability of single vote in California being decisive) = (probability of CA being the decisive state) * (probability that CA is tied given CA is the decisive state) = (probability of CA being the decisive state) * (probability of a randomly selected vote being a vote that ties CA given CA is the decisive state) / (probability of a randomly selected vote being cast in CA).
The assumption that I made was that (probability of a random vote tying CA given that CA is the decisive state) is close to (probability of a random vote tying whatever the decisive state happens to be), which seems fairly reasonable. I did NOT assume that (probability that CA is tied given that CA is the decisive state) is close to (probability that CA is tied). ( [Edit: ignore this parenthetical comment] At least I think I didn’t, but I’m tired right now, so it is conceivable that I could have made an error on that front. If I did mess that up, then the actual probability of a vote in CA swinging the election should be greater than 1 in 1 billion, but still probably less than 1 in 100 million.)
That assumption is not even close to correct in US presidential elections. Not that that makes much of a difference (I think).
Almost certainly? I thought you said Red was solid and Purple was only slightly bluish, so there should be a significant chance that Red and Purple both vote red, and Blue’s surprise red vote is redundant.
Sounds like you are using the Gelman et al meaning of a state being decisive. Not what 538 calls being the tipping-point state, which is what I was using. That’s why you get garbage when you plug that into my formula.
So, first of all, there’s a factor of 2 error in there: your last equality says, in effect, Pr(CA tied | …) = Pr(a random vote ties CA | …, and that vote is in CA) but when CA is tied only half the votes there tie it.
I’m already late for work, so will look harder at the rest of what you’re saying later. (I find myself somewhat convinced both by my toy example and by the more-detailed argument you’re now making, but the two don’t seem consistent with one another. I expect I’m missing something.)
I know. That’s one reason why the example is a toy. But no part of your argument appeals to the correlations between states’ results, and the point of the toy example is to show that the calculation you’re doing produces completely wrong results in a toy example, so in the absence of anything in your argument that makes it apply to the real election and not to the toy example something’s got to be wrong with the argument.
Sorry, “slightly bluish” was meant to describe vote share rather than win probability. I’m assuming that P is a win for the Blue candidate about 90% of the time, which with simple-but-unrealistic models of voters will happen if it’s reasonably big and, say, 55% Blue.
I was intending to use the 538 meaning. Pr(Blue decisive) is small because in almost all elections the state that gets the winner that crucial third EV—the one whose EV is in the middle when you line them up in order—is Purple. What do you find wrong with this reasoning?
Nope. Half the votes prevent a Romney victory, and the other half prevent an Obama victory.
Your confusion is understandable, especially since I confused myself and started bullshitting you for a while before rederiving what I did in the first place. Sorry about that.
That’s right. Sorry, I shouldn’t have been stressing the high correlations between voting fluctuations in different states.
Ah, ok.
Numbers from your toy example: Pr(Blue tied) = 0.1% Pr(Blue decisive | Blue tied) = 90% Pr(Blue decisive) = 0.1% implications: Pr(Blue decisive and tied) = 0.09% Pr(Blue decisive and not tied) = 0.01% This is not plausible. Presumably Pr(Blue decisive and votes blue by 1 vote) is also roughly 0.09%, in which case Pr(Blue decisive and not tied) cannot possibly be less than that. Assuming Red never enters the picture, Blue is decisive whenever it ends up voting more reddish than Purple does. Given how often Blue ties, I would expect this to actually happen fairly frequently.
Apologies for the slow response; I’ve been unreasonably busy. Executive summary of what follows: Yup, you were right.
So I tried generating more realistic numbers with the general structure of my toy example, and my conclusion is: Oops, you’re right and my example is no good. Sorry. And I think I agree with your simple probability-pushing argument that 538′s probability for California being decisive isn’t consistent with the numbers from Gelman et al being applicable in the 2012 election.
So, it seems to me that there are (at least) the following possibilities. (1) Gelman et al had a good model, and it remains reasonably applicable now, and 538 had too low a probability of California being decisive. (2) Gelman et al had a good model, but the political landscape has changed, and now California is less likely to be decisive than their model said it was in 1992. (3) Gelman et al had a screwed-up model, and their probabilities weren’t right even in 1992.
I agree with you that #2 is the least likely of these, and I offer the following statistic which, if cited at the outset, might have saved us a good deal of argument :-). In 1998, California went Democratic by about 51:48. In 2012, California went Democratic by about 59:39.
I accordingly agree with you: Academian’s numbers for his own case, which used the Gelman et al figures for California, likely gave much too high an expected value for his vote in California.
I assume you meant #2 is most likely? And you’re right; I should have pointed that out initially (even though it was before the election, I could have used 2008 figures).
Yes, of course I meant most likely. Duh. I’ve edited my comment for the benefit of our thousands of future readers.
testing: blah