The Monty Hall problem is a famous brain teaser, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975. It was also mentioned in the hollywood movie 21.

The Problem can be described as follows-

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. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The straight forward answer is, 'You switch your choice'. Switching your choice gives you a higher probability of winning. This answer though counter intuitive is based on simple probability concepts.

There are three doors in the beginning, one with a car behind it and other two with a goat. When you select a door randomly, the probability of selecting the door with the car behind it is 1/3, while probability of selecting the door with a goat behind it is 2/3.

After you have selected a door, the host eliminates a door out of the other two. Now you have two options, you can either stick with your earlier choice, or you can switch your choice. If the door you had originally selected has a car behind it, sticking with that door will win you the car. But if the door you had originally selected has a goat behind it, then switching your choice will win you the car.

As shown earlier, probability of selecting a door with the car behind it is 1/3, while probability of selecting a door with a goat behind it is 2/3. Hence probability of winning when you stick with your earlier choice is 1/3, while probability of winning when you switch your choice is 2/3.

Let's try running a simulation and verify the answer.

The following Python code sets up a simulation of the game. In this simulation,

- Three doors are set up.
- A car is randomly assigned to a door, and other two are kept empty.
- A door is selected randomly as contestant's choice.
- Out of the other two doors, a door not containing the car is eliminated.
- Contestant's choice is switched.
- Door corresponding to the contestant's final choice is opened. If the door contains the car, contestant wins.

In [1]:

```
from random import randint
def setupDoors() :
#Set up three doors, assign car randomly to one of the doors, car is represented by *
doors = dict.fromkeys([1, 2, 3])
doors[randint(1,3)] = "*"
return doors
def chooseDoor():
#Select a random door out of the three doors
return randint(1,3)
def eliminateDoor(doors, selectedDoor):
#Eliminate a door which isn't selected by the player and doesn't have a car behind it
eligibleDoors = []
for key, value in doors.items():
if key != selectedDoor and value != "*":
eligibleDoors.append(key)
if len(eligibleDoors) == 1:
doors.pop(eligibleDoors[0])
else:
doors.pop(eligibleDoors[randint(0,1)])
return doors
def switchDoor(doors, selectedDoor):
#Switch the player's choice
for key in doors.iteritems():
if key[0] != selectedDoor:
return key[0]
def openDoor(doors, selectedDoor):
#Open the player's final selection
if doors[selectedDoor] == "*":
return 1
else:
return 0
def playGame(output):
doors = setupDoors()
if(output):
print "Initial doors = "
print doors
doorChoice = chooseDoor()
if (output):
print "\nPlayer's choice"
print doorChoice
newDoors = eliminateDoor(doors, doorChoice)
if(output):
print "\nDoors after eliminating one door = "
print newDoors
doorChoice = switchDoor(doors, doorChoice)
if(output):
print "\nFinal choice"
print doorChoice
result = openDoor(newDoors, doorChoice)
if(output):
if(result == 1):
print "\nPlayer Won"
return
else:
print "\nPlayer Lost"
return
return result
```

Let's run one iteration of the game and check the output.

In [3]:

```
playGame(output = True)
```

In [11]:

```
import matplotlib.pyplot as plt
import scipy.stats as stats
import numpy as np
def runHundredGames():
result = []
for i in xrange(100):
result.append(playGame(output = False))
return sum(result)
def runSimulation():
finalResult = []
avgResult = []
for k in xrange(1000):
finalResult.append(runHundredGames())
avgResult.append((sum(finalResult)/float(len(finalResult))))
plt.plot(range(1000), avgResult)
plt.grid(True)
plt.title("Plot 1 - Simulations vs Avg games won in 100 games \n")
plt.ylabel("Average games won in 100 games")
plt.xlabel("Number of Simulations")
plt.show()
#Fitting a normal distribution
sFinalResult = sorted(finalResult)
normalDistFit = stats.norm.pdf(sFinalResult, np.mean(sFinalResult), np.std(sFinalResult))
plt.plot(sFinalResult ,normalDistFit)
plt.hist(finalResult,normed=True)
plt.grid(True)
plt.title("Plot 2 - Histogram of Number of games won in 100 games \n")
plt.ylabel("Frequency")
plt.xlabel("Number of games won in 100 games")
plt.show()
return finalResult
result = runSimulation()
```

Plot 1 shows the relation between 'Avg number of games won in 100' and 'Number of simulations'. As number of simulations increase the average number of games won stabilizes near 66.66.

Plot 2 is a histogram of Number of won in 100 games, the blue curve is a Normal distribution with Mean = Simulation Mean and Standard Deviation = Simulation Standard Deviation. The histogram is centered around 66.