## Suppose there are two boxes, each containing a mix of red and blue
## balls.  We do not know how many balls of each color are contained
## in each box.
##
## Our goal is to estimate the proportion of red balls in the box that
## has more red balls.
##
## First we do a pilot sample by randomly selecting 5 balls from each
## bucket (with replacement).
##
## Then we sample 10 balls (with replacement) from whichever bucket
## produced more red balls in the pilot sample.  The proportion of red
## balls in this sample of 10 is our estimate.
##
## Use simulation to calculate the bias, variance, and MSE of this
## estimator.
##
## For the simulation, assume the following:
## Box 1 contains 20 red balls and 10 blue balls.
## Box 2 contains 15 red balls and 15 blue balls.


D = NULL

nrep = 1e4

for (k in 1:nrep) {

  ## The numbers of red balls in the two pilot samples.
  P1 = sum(runif(5) < 2/3)
  P2 = sum(runif(5) < 1/2)

  ## The proportion of red balls in the actual sample.
  if (P1 >= P2) {
      D[k] = mean(runif(10) < 2/3)
  } else {
      D[k] = mean(runif(10) < 1/2)
  }
}


bias = mean(D) - 2/3
Var = var(D)
MSE = mean( (D-2/3)^2 )