# Tag Archives: probability of forecasting the EV exactly

## Probability of exactly forecasting the electoral vote in the last three elections

Richard Charnin
Feb. 11, 2017

I was asked to calculate the probability of my exact forecast of the Electoral Vote in the last three elections (365,332,306). It was a combination of experience and luck. I do not expect to exactly forecast the EV in 2020.

Note that the following calculation is just an approximation.

Assume the following:
1) the probability of Obama winning in 2008 was 0.95; it was also 0.95 in 2012. The probability of Trump winning in 2016 was 0.05.
Therefore the probability of forecasting all three winners correctly is
P1 = 0.045 =.95*.95*.05

2) the winning EV is in the 270-370 range.
The probability of exactly forecasting the EV in a given election is 0.01. The probability of exactly forecasting the EV in all 3 elections is 1 in a million:
P2 =.000001 = 0.01*0.01*0.01

Therefore, the probability of forecasting the winner and the EV in the three elections is
P3 = P1*P2 = .045* 0.000001 or 1 in 22 million.

To put it another way, forecasting the electoral vote exactly in three successive elections would be expected to occur just once in 22 million elections (88 million years).