I have a PhD in statistics, to bring this into a simple perspective: a binary process like voting between two choices are typically modelled by a binomial model where one choice is '0' and the other choice is '1'.
Even if you choose Biden getting p=0.999 (99.9%) of the votes, on average he would get 891 votes. The estimated value of Binomial(900,0.999) is n*p = 891.
If you want to estimate the odds of Biden getting all votes, even assuming something ridiculous like 99.9% support, run a monte carlo simulation on the Binomial simulations. Then count the occurences of the result summing to 900.
Here are the probabilities of Biden getting 900 votes for each value p (probability of a vote going for Biden)
theta = 0.5 probability = 0
theta = 0.6 probability = 0
theta = 0.7 probability = 0
theta = 0.8 probability = 0
theta = 0.9 probability = 0
theta = 0.95 probability = 0
theta = 0.975 probability = 0
theta = 0.99 probability = 0.000114
theta = 0.995 probability = 0.011067
theta = 0.996 probability = 0.027168
theta = 0.997 probability = 0.067105
theta = 0.998 probability = 0.164542
theta = 0.999 probability = 0.407043
Even with 97.5% of the votes, probability is rounded to zero using 6 digits of significance. With 99% support probability of this happening is 0.01%. Statistical significance is finally broken at 99.7% support (giving a 6.7% probability of occurrence) and with 99.9% of the vote in 40.7% of the cases would this happen. This is statistically impossible.
Feel free to run the rather simple simulation yourself with this R script.
m = 10^6
n = 900
t = c(0.5, 0.6, 0.7, 0.8, 0.9, 0.95, 0.975, 0.99, 0.995, 0.996, 0.997, 0.998, 0.999)
for(p in t)
{
mu = sum(rbinom(m,n,p) == 900)/m
cat("theta = ", p, "\t probability = ", mu, "\n")
}
Hope the Trump lawyers will consult statisticians in this, a lot of what has happened is statistically not just improbable but virtually impossible.
I have a PhD in statistics, to bring this into a simple perspective: a binary process like voting between two choices are typically modelled by a binomial model where one choice is '0' and the other choice is '1'.
Even if you choose Biden getting p=0.999 (99.9%) of the votes, on average he would get 891 votes. The estimated value of Binomial(900,0.999) is n*p = 891.
If you want to estimate the odds of Biden getting all votes, even assuming something ridiculous like 99.9% support, run a monte carlo simulation on the Binomial simulations. Then count the occurences of the result summing to 900.
Here are the probabilities of Biden getting 900 votes for each value p (probability of a vote going for Biden)
theta = 0.5 probability = 0
theta = 0.6 probability = 0
theta = 0.7 probability = 0
theta = 0.8 probability = 0
theta = 0.9 probability = 0
theta = 0.95 probability = 0
theta = 0.975 probability = 0
theta = 0.99 probability = 0.000114
theta = 0.995 probability = 0.011067
theta = 0.996 probability = 0.027168
theta = 0.997 probability = 0.067105
theta = 0.998 probability = 0.164542
theta = 0.999 probability = 0.407043
Even with 97.5% of the votes, probability is rounded to zero using 6 digits of significance. With 99% support probability of this happening is 0.01%. Statistical significance is finally broken at 99.7% support (giving a 6.7% probability of occurrence) and with 99.9% of the vote in 40.7% of the cases would this happen. This is statistically impossible.
Feel free to run the rather simple simulation yourself with this R script.
m = 10^6
n = 900
t = c(0.5, 0.6, 0.7, 0.8, 0.9, 0.95, 0.975, 0.99, 0.995, 0.996, 0.997, 0.998, 0.999)
for(p in t)
{
mu = sum(rbinom(m,n,p) == 900)/m
cat("theta = ", p, "\t probability = ", mu, "\n")
}
Hope the Trump lawyers will consult statisticians in this, a lot of what has happened is statistically not just improbable but virtually impossible.