Sure! I’m modeling the election as being nVoters coin flips: if there are more Heads than Tails, then candidate H wins, else candidate T wins.
If you flip n coins, each coin coming up Heads with probability p, then the number of Heads is binomially distributed with standard deviation √(nVoters)(p)(1−p), which I lazily rounded to √nVoters.
Could you explain where P(I swing election) = 1/sqrt(nVoters) is coming from?
Sure! I’m modeling the election as being nVoters coin flips: if there are more Heads than Tails, then candidate H wins, else candidate T wins.
If you flip n coins, each coin coming up Heads with probability p, then the number of Heads is binomially distributed with standard deviation √(nVoters)(p)(1−p), which I lazily rounded to √nVoters.
The probability of being at a particular value near the peak of that distribution is approximately 1 / [that standard deviation]. (“Proof”: numerical simulation of flipping 500k coins 1M times, getting 250k Heads about 1⁄800 of the time.)