Donald 'Paddy' McCarthy
has a nice Awk solution to the Monty Hall Problem, which he describes as follow:
The contestant in in front of three doors that he cannot see behind..
The three doors conceal one prize and the rest being booby prizes, arranged randomly.
The Host asks the contestant to choose a door.
The host then goes behind the doors where only he can see what is concealed, then always opens one door, out of the other s not chosen by the contestant, that must reveal a booby prize to the contestant.
The host then asks the contestant if he would like either to stick with his previous choice, or switch and choose the other remaining closed door.
It turns out that if the contestant follows a strategy of always switching when asked, then he will maximise his chances of winning.
Donald's simulator shows that:
A strategy of never switching wins 1/3rd of the time.
A strategy of randomly switching wins 1/2 of the time.
A strategy of always switching wins 2/3rds of the time.
Code
BEGIN {
srand()
doors = 3
iterations = 10000
# Behind a door:
EMPTY = "empty"; PRIZE = "prize"
# Algorithm used
KEEP = "keep"; SWITCH="switch"; RAND="random";
}
function monty_hall( choice, algorithm ) { # Set up doors
for ( i=0; i<doors; i++ ) {
door[i] = EMPTY
}
door[int(rand()*doors)] = PRIZE # One door with prize
chosen = door[choice]
del door[choice]
#if you didn't choose the prize first time around then
# that will be the alternative
alternative = (chosen == PRIZE) ? EMPTY : PRIZE
if( algorithm == KEEP) {
return chosen
}
if( algorithm == SWITCH) {
return alternative
}
return rand() <0.5 ? chosen : alternative
}
function simulate(algo){
prizecount = 0
for(j=0; j< iterations; j++){
if( monty_hall( int(rand()*doors), algo) == PRIZE) {
prizecount ++
}
}
printf " Algorithm %7s: prize count = %i, = %6.2f%%\n", \
algo, prizecount,prizecount*100/iterations
}
BEGIN {
print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"
simulate(KEEP)
simulate(SWITCH)
simulate(RAND)
}
