The Monty Hall Problem

The Monty Hall problem is a probabilistic puzzle that confounds lots of people including a Nobel laurate in physics apparently!

The Monty Hall problem is based on a TV quiz show and is stated in this way: A player is presented with three doors and behind one of the doors is a prize. The aim of the game is for the player to select the door with the prize behind. Of course, the player does not know which of the doors the prize is behind.  The player guesses which of the three doors the prize is behind. At that point one of the doors which the prize is not behind and which has not been selected by the player is opened. So now there are two closed doors. The player is then asked whether they wish to stick with the door that they selected or switch to the other remaining closed door.

Intuitively we think of the problem like this.  Originally there is a one in three chance of us selecting the correct door.  Then one of the doors is opened and then it appears the problem becomes a fifty-fifty choice between the two remaining closed doors so it does not matter whether we switch or not. But this thinking is incorrect.  The answer is the player should always switch because they have twice the chance of winning the prize than if they stick to the door they chose originally.

There are lots of mathematical solutions that demonstrate this on the web. Here we show a solution using a simple program.  IF you run the code below you will find that two thirds of the time you will win the prize if you switch, but if you stick you will win the prize on one in three occasions. Running the code reveals to us what intuitively we cannot see.

Further Reading

https://en.wikipedia.org/wiki/Monty_Hall_problem

Code

import random

correct_swap=0

correct_stick=0
SAMPLES=1000000

# run lots of times so we have a statistically valid result 

for i in range(SAMPLES):

# select a random door for the prize to be behind

 door=random.randint(1,3)

# player makes a selection; 

 player=random.randint(1,3)

# Open a door that neither has the prize behind

# and has not been selected by the player

 open_door=random.randint(1,3)

 while open_door==door or open_door==player:

  open_door=random.randint(1,3)

# Choose whether to stick or switch

 option=random.choice(['stick','swap'])

# stick with the same door

 if option=="stick":
  if door==player:
   correct_stick=correct_stick+1
  else:
   correct_swap=correct_swap+1

# swap door

 elif option=="swap":

  # switch to the other unopened door

  if open_door==1 and player==2: player=3

  elif open_door==1 and player==3: player=2

  elif open_door==2 and player==1: player=3

  elif open_door==2 and player==3: player=1

  elif open_door==3 and player==2: player=1

  elif open_door==3 and player==1: player=2

 if door==player:

  correct_swap=correct_swap+1

 else:

  correct_stick=correct_stick+1

print("Stick: ", correct_stick/SAMPLES*100)

print("Swap:", correct_swap/SAMPLES*100)