# PROBLEM DESCRIPTION
#
# Prisoner 1  asks the guard for additional information
#
# If Prisoner 2 is pardoned,  guard names Prisoner 3
#
# If Prisoner 3 is pardoned,  guard names Prisoner 2
#
# If Prisoner 1 is pardoned,  guard names Prisoner 2 
# with prob 1/2 and 3 with prob 1/2

  
# Given that the guard named Prisoner 2, the prob of 
# being pardoned becomes 1/3 for Prisoner 1 and 2/3 
# for Prisoner 3.

###########################################################

N = 10000  # number of simulations
x = 1:N    # pardoned prisoner
y = 1:N    # guard disclosure of non-pardoned prisoner

	for (i in 1:N)
				{
	s = sample(1:3);
	x[i] = s[1];
	if(x[i] == 1){y[i] = sample(2:3)[1]};
	if(x[i] == 2){y[i] = 3};
	if(x[i] == 3){y[i] = 2};
				}

sum(x==1 & y==2)/sum(y==2)   #  P( 1 | guard names 2 )
sum(x==3 & y==2)/sum(y==2)   #  P( 3 | guard names 2 ) 


sum(x==1 & y==3)/sum(y==3)   #  P( 1 | guard names 3 )
sum(x==2 & y==3)/sum(y==3)   #  P( 2 | guard names 3 )