Columbia has a day off for Election Day. So, instead of teaching my class, I’ll do a little lesson here on the use of math and practical reasoning to combine information from multiple sources.

New Jersey governor:

Elliott Morris has the latest polls at Sherrill (D) 50.4%, Ciatterelli (R) 44.7%, that’s a two-party vote share for Sherrill of 0.504/(0.504 + 0.447) = 0.53. Reading the uncertainty off the graph, the 95% interval for Sherrill’s support has a width of about 3 percentage points, that’s +/- 2 se’s, so the poll-based standard error for either candidate’s support is about 0.75%.

If we (naively) convert this directly into a probability forecast, we’d forecast Sherrill’s two-party vote share as normal(0.53, 0.0075), which would give her a win probability of pnorm((0.53 – 0.5) / 0.0075) = 0.9999 . . . that ain’t right!

The problem here is that the uncertainty at that website doesn’t account for nonsampling error.

Ironically, we can see this more clearly using the old-fashioned approach of looking at polls: the last few have Sherrill at 56%, 51%, 51%, 54%, 55%, and 55% of the two-party vote. These six different poll results have a mean of 53% and standard deviation of 2.2%. If we take that to represent our uncertainty, Sherrill’s win probability comes to pnorm((0.53 – 0.5) / 0.022) = 91%.

This is better but it still understates our uncertainty because it doesn’t account for the very real possibility that the polls are off in a common direction. If we give the average polling error for the state election a prior with mean 0 and standard deviation 2 percentage points, the simple poll-based forecast is then normal with mean 53% and standard deviation sqrt(2^2 + 2.2^2) = 3.0%, and Sherrill’s win probability becomes pnorm((0.53 – 0.5) / 0.030) = 84%.

Here’s another source of data, at the New York Times, where Sherrill’s 2-party vote preference shares in the last six polls are 51%, 51%, 52%, 54% 55%, and 54%, which have a mean of 53% and a standard deviation of 1.8%, so pretty much the same story–no surprise, given that these are mostly the same polls! Another site is Real Clear Politics, where the last 6 polls give Sherrill an average of 52% of the two-party vote with a standard deviation of 1.3%. Converting this to a win probability gives pnorm((0.52 – 0.5) / sqrt(0.02^2 + 0.013^2)) = 80%.

Then there are the betting odds. Kalshi has Sherrill priced at 82% and Ciatterelli at 19%, which I’ll normalize to give Sherrill a forecast win probability of 0.082/(0.082 + 0.019) = 81%. I don’t think this is quite right–it’s my impression that these betting markets overprice rare events–but it’s close enough, consistent with the polling information, which makes sense given that bettors are pretty much relying on the polls! Polymarket gives identical odds, which is no surprise given that bettors can choose which market to play.

Virginia governor:

We can do the same thing. Morris’s site lists 5 recent polls of Spanberger (D) vs. Sears (R), with Spanberger’s two-party vote preference at 54%, 55%, 56%, 54%, and 56%; these numbers have a mean of 55% and standard deviation of 0.09%. At Real Clear Politics, the most recent six polls give Spanberger a mean of 56% of two-party support with standard deviation 2.4%. I’ll split the difference and add some uncertainty and put her at 55.5% with sd 2.5%. Using the same rules as before, her poll-estimates two-party vote share is 55.5% with standard deviation sqrt(2^2 + 2.5^2) = 3.2%, which maps to a win probability of pnorm((0.555 – 0.5) / 0.032) = 96%.

Now let’s check out the betting odds: they have Spanberger priced at 97% and Sears at 4%, which after normalizing gives Spanberger a 96% chance of winning! But given the overpricing of rare events, I’d give her a slightly higher chance based on these odds–maybe 97% or 98%?

New York City mayor:

OK, this is the tough one because there are three candidates. Real Clear Polling gives a polling average of Mamdani (D) at 46.1%, Cuomo (I) at 31.8%, and Sliwa (R) at 16.3%. It’s hard to see Sliwa winning, so the relevant number would seem to be Mamdani’s share of the vote that goes to him or Cuomo, which in that average is 0.461/(0.461 + 0.318) = 59%. We can also check out the 6 most recent polls at the New York Times site, which give Mamdani’s two-candidate vote share at 53%, 55%, 60%, 60%, 57%, and 66%, which have a mean of 58% and standard deviation 4.8%.

Following the above rules, we could average these estimates to get a poll-forecast two-candidate vote share for Mamdani of 58.5% with standard deviation sqrt(2^2 + 4.8^2) = 5.2%, with a corresponding win probability of pnorm((0.585 – 0.5) / 0.052) = 95%.

But that’s too high! And I say this before even looking at the betting odds.

Why do I think the 95% win probability estimate for Mamdani is too high? Three reasons:

1. Unlike the New Jersey and Virginia elections, the New York mayor’s race is not a straight party-line contest. So individual votes are less predictable. You see this in the jumping around of the polls.

2. Relatedly, a higher level of between-poll variation could go with a higher level of aggregate polling error, so that +/- 2% is probably an understatement.

3. In a three-candidate election, we’ll often see the third-place candidate fall off at the finish line, because of all the people who don’t want to waste their votes. So Republican candidate Sliwa could well get a bit less than the 16% he’s polling at. And it makes sense to expect most of the Republicans switching off Sliwa for tactical reasons to move to Cuomo, the who’s recently been endorsed by the national Republican establishment. Just for example, if Sliwa’s vote drops by 6%, of which 5% goes to Cuomo and 1% goes to Mamdani, this would put Mamdani, Cuomo, and Sliwa at approximately 47%, 37%, and 10%, with the front-runner at 56% of the vote for the two major candidates.

If we put Mamdani at expected vote 56% with a standard deviation of 5.5% (bumping up from 5.2% for the reasons given above), this corresponds to a win probability of pnorm((0.56 – 0.5) / 0.055) = 86%.

Ok, now the prediction markets. They currently price Mamdani at 90% and Cuomo at 11%, which after renormalizing gives the Democrat a win probability of 89%. Pretty close to my calculation, which again is no surprise given that prediction markets are based on a combination of the polls, polling uncertainty, and, in this case, the possibility of some Republican voters tactically moving away from the Republican candidate to the independent candidate who was endorsed by national Republican leaders.