Monty Hall problem

Monty Hall problem
You are encouraged to solve this task according to the task description, using any language you may know.

Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.

Problem Statement

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)}}

Note that the player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.

Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.

AWK

#!/bin/gawk -f 

# Monty Hall problem

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
	}
  # One door with prize
	door[int(rand()*doors)] = 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)
		
}

Sample output:

bash$ ./monty_hall.awk

Monty Hall problem simulation:
3 doors, 10000 iterations.

  Algorithm    keep: prize count = 3411, =  34.11%
  Algorithm  switch: prize count = 6655, =  66.55%
  Algorithm  random: prize count = 4991, =  49.91%
bash$

Python

<python> I could understand the explanation of the Monty Hall problem but needed some more evidence

References:

 http://www.bbc.co.uk/dna/h2g2/A1054306
 http://en.wikipedia.org/wiki/Monty_Hall_problem especially:
 http://en.wikipedia.org/wiki/Monty_Hall_problem#Increasing_the_number_of_doors

from random import randrange, shuffle

doors, iterations = 3,100000 # could try 100,1000

def monty_hall(choice, switch=False, doorCount=doors):

 # Set up doors
 door = [False]*doorCount
 # One door with prize
 door[randrange(0, doorCount)] = True

 chosen = door[choice]

 unpicked = door
 del unpicked[choice]

 # Out of those unpicked, the alterantive is either:
 #   the prize door, or
 #   an empty door if the initial choice is actually the prize.
 alternative = True in unpicked

 if switch:
   return alternative
 else:
   return chosen

print "\nMonty Hall problem simulation:" print doors, "doors,", iterations, "iterations.\n"

print "Not switching allows you to win", print [monty_hall(randrange(3), switch=False)

       for x in range(iterations)].count(True),

print "out of", iterations, "times." print "Switching allows you to win", print [monty_hall(randrange(3), switch=True)

       for x in range(iterations)].count(True),

print "out of", iterations, "times.\n" </python> Sample output:

Monty Hall problem simulation:
3 doors, 100000 iterations.

Not switching allows you to win 33337 out of 100000 times.
Switching allows you to win 66529 out of 100000 times.