If I have an 11 sided coin where p(1) up to p(10) has probability 0.01. And p(11) has probability 0.9.
How do I calculate after how many throws p(1) up to p(10) each occurred at least a 100 times.
Also how do I create a binominal distribution of this, to determine the minimum and maximum 95 or 99% amount?
I am seeking for a general formula to determine this, it is assumed that p(1)-p(n) of the unbiased side all have the same probability though.
You could approximate the problem by each side being independent and then the probability for all ten sides being above 100 is the probability for one side being above 100 to the power ten.
Below is a code example that compares this approximation with a simulation.
d = 100 ### cutoff-level
m = 20 ### number of bins/dice-sides
n = 5000 ### number of trials for histogram
### fucntion that repeats sampling untill all states are above d
sim <- function() {
states = c(rep(FALSE,m),TRUE)
cases = rep(0,m+1)
count = 0
while (prod(states) == 0) {
count = count + 1
s = sample(c(1:(m+1)),
size = 1,
prob = c(rep(0.1/m,m),0.9))
cases[s] = cases[s] + 1
if (cases[s] == d) {
states[s] = TRUE
}
}
return(count)
}
### do the simulation n times
set.seed(1)
sims <- replicate(n,sim())
### plot and compute histogram
h <- hist(sims, freq = 1, main = "histogram of simulation with curve of approximation",
breaks = seq(20000,30000,400))
### compute approximate counts distribution function
ns = 1:max(h$breaks)
psingle <- 1-pbinom(d-1,ns,p=0.1/m)
pmult <- psingle^m
### add to historgram
counts = sapply( 1:length(h$mids), FUN = function(x) {
pmult[h$breaks[x+1]]-pmult[h$breaks[x]]
})
lines(h$mids,counts*n)
points(h$mids,counts*n, pch = 21, col = 1, bg = 0)
You have to throw it at least 1000 times to have a positive probability of 100 for each. The probability is always less than 1 no matter how many time you throw the die.
Suppose I throw the die $N \ge 1000$ times and let $X_i$ denote the number of times side $i$ is observed. $(X_1,...X_{11})$ has a multinomial distribution.
$$P[X_1 \ge 100, X_2 \ge 100, ...X_{10} \ge 100]$$ $$=\sum_{x_1=100}^{N-900}\sum_{x_2=100}^{N-800-x_1}\sum_{x_3=100}^{N-700-x_1-x_2}...\sum_{x_{10}=100}^{N-\sum_{j=1}^9x_j}P[X_1 =x_1, X_2 =x_2, ...X_{10} =x_{10}]$$ $$=\sum_{x_1=100}^{N-900}\sum_{x_2=100}^{N-800-x_1}\sum_{x_3=100}^{N-700-x_1-x_2}...\sum_{x_{10}=100}^{N-\sum_{j=1}^9x_j}\frac{N!}{x_1!x_2!...x_{10}!(N-\sum_{k=1}^{10}x_k)!}p_1^{x_1}p_2^{x_2}...p_{10}^{x_{10}}p_{11}^{N-\sum_{k=1}^{10}x_k}$$
You definitely do not want to try calculating that unless N is close to 1000.
For large $N$, $(X_1,...X_{10})$ is approximately multivariate normal with mean $(0.01 N, ..., 0.01 N)$ and variance matrix that has $0.01\times N\times(1-0.01)$ on the diagonal and $-0.01^2N$ off diagonal. see here
With $N=11600$, there is about 50% chance of having observed 100 of each face from 1-10.
R program:
library(mvtnorm)
N=11600
sigma=matrix(N*sqrt(0.01*0.01)*(0-sqrt(0.01*0.01)),ncol=10,nrow=10)
diag(sigma)=N*sqrt(0.01*0.01)*(1-sqrt(0.01*0.01))
as.numeric(pmvnorm(lower=rep(100,10),upper=rep(Inf,10),mean=rep(0.01*N,10),sigma=sigma))
For 95% chance, use $N=12900$
For 99% chance, use $N=13600$
