Election predictions

[robve]

(11-05-2020 05:03 PM)David Hayden Wrote:  Example: Trump currently has 49.7% in PA and Biden has 49.1% in Georgia with 95% of the votes counted. What percentage of the remaining votes does Biden need to win?

49.7 [Enter] 49.1 [Enter] 95 XEQ A
Answer: 55.7
So Biden needs 55.7% of the outstanding votes to win Georgia.

Very interesting model and example.

Things have changed quite a bit since this post. I got curious to figure out when a candidate has won a state even though it appears that the candidate has no absolute 100% guarantee to win it, given these numbers.

A 100% guaranteed win requires that all outstanding votes can never contribute to a trailing candidate's win to overtake the current leading candidate. For example, if 1000 votes are outstanding and the difference between the leading and trailing candidate is less than 1000, then the leading candidate has won. But that is too rigid, because the probability (or "odds") of all 1000 remaining uncounted votes to all vote for the trailing candidate is infinitesimal. I suppose we can assume a normal PD for the remaining vote percentages with some given or estimated mean and variance.

Here is my simple model, combined with a statistical estimate for the chance of the leading candidate to win.

L = leading candidate vote percentage
T = trailing candidate vote percentage
V = total votes cast as a percentage
X = minimum vote percentage of the remaining votes for the leading candidate to win

Normalize L, T, V, and X to fractions instead of percentages (just divide by 100), then if

VL + (1-V)X > VT + (1-V)(1-X)

then leading candidate wins, even if 1-X of the votes all go to the trailing candidate. Note that there can be any number of other candidates for this model to work, just think of T to be the closest competitor to L.

Solving for X (the unknown variable), we get:

X > (V(T-L)+R)/(2R)

where R = 1-V.

This is essentially the same as your calculation.

For example, take the state of Nevada. Plugging in the values:
L=.499
T=.479
V=.94
gives
X>.343

Note that there is no 100% guarantee that the leading candidate wins because at least 34.3% of the remaining votes should go the the leading candidate, assuming that in the worst case all other votes go to the competitor. But getting less than 34.3% of the remaining votes seems very unlikely, meaning that winning is likely. But how likely?

I assume the model needs to show at least a 99% probability of winning by the leading candidate for the state to be called, or sufficiently close to 99%.

Assume that the remaining vote distribution is a normal PD with mean .5 (a bit arbitrary but a fair 50% split) and a standard deviation of .1 (10%). The actual values can be determined by sampling the (remaining) votes. In this case we find that p = 0.0116 (e.g. on a calculator plug in these values in the Distribution: Normal PD mode.) The probability for the leading candidate to receive less than the required 34.3% of the remaining votes is sufficiently small, giving about 99% chance of a win. This fits the case for Nevada.

Note that the mean may favor the leading candidate, i.e. a value greater than .5 such as .6 or .7, which means that the variance can be a lot larger than .1 to still reach a 99% certainty of a win.

Disclaimer: this is a fairly simple approach that is not derived from any methods that the statisticians actually apply to make the final call, just a guess. Also, I've made some assumptions that may not be valid. Please do not derive any conclusions and correct me if I'm wrong in your comments and if you have a different method to estimate the probability of an outcome.