We use Stirling's formula to approximate the probability of drawing $n$ red and $n$ blue balls from an urn with $2n$ red and $2n$ blue balls, assuming we draw $2n$ balls without replacement.
The percent error is quite small (less than 2%) even for modest values of $n$, like $n=10$.
n = 100 truevalue = convert(Float64,binomial(big(2n),n)^2 / binomial(big(4n),2n)) approximation = 2/(sqrt(2π*n)) truevalue, approximation, 100*(approximation/truevalue-1)
function b(n,p,k) # Computes (n choose k) * p^k * (1-p)^(n-k) result = 1.0 for j = 0:k-1 result *= (n-j)/((j+1))*p end result *= (1-p)^(n-k) return result end PowerCurve(n,p,m) = sum([b(n,p,k) for k=m:n])
PowerCurve (generic function with 1 method)
plot(x=[50:100],y=[PowerCurve(100,0.6,k) for k=50:100],Geom.line)