So I did make a math mistake, but I think we’re in broad agreement. Let me be explicit for a race with total expected votes N=400 (e.g. seat on a city council for one district of a small town)
With N=400 sigma = sqrt(400 * 0.5 * 0.5) = 10 a 6-point lead means expected votes would be: A : 212 B : 188 This corresponds to a win probability for A of cdf(12/10) ~88%
Changing one’s vote from A to B changes the expected counts to: A : 211 B : 189 This corresponds to a win probability for A of cdf(11/10) ~86% So yes, it’s only 2% change vs my earlier assertion of 8%, my mistake.
But I think we agree that sigma matters! And my point is that in small local elections, sigma is small, and your vote counts for a lot!
I agree that if you only care about federal policy, this doesn’t apply (I’d missed that in the initial post). But if you care about libraries, or how aggressive the police are, those are local issues where someone can have a strong influence in policy.
So I did make a math mistake, but I think we’re in broad agreement. Let me be explicit for a race with total expected votes N=400 (e.g. seat on a city council for one district of a small town)
With N=400
sigma = sqrt(400 * 0.5 * 0.5) = 10
a 6-point lead means expected votes would be:
A : 212
B : 188
This corresponds to a win probability for A of cdf(12/10) ~88%
Changing one’s vote from A to B changes the expected counts to:
A : 211
B : 189
This corresponds to a win probability for A of cdf(11/10) ~86%
So yes, it’s only 2% change vs my earlier assertion of 8%, my mistake.
But I think we agree that sigma matters! And my point is that in small local elections, sigma is small, and your vote counts for a lot!
I agree that if you only care about federal policy, this doesn’t apply (I’d missed that in the initial post). But if you care about libraries, or how aggressive the police are, those are local issues where someone can have a strong influence in policy.