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Monty Hall Puzzle

  • There are 3 doors to be chosen from for a definete prize behind one of the doors
  • After choosing the first time, the host reveals the door and provides an opportunity to switch before the prize is revealed
  • If the door revealed at the first time is not the prize door, then host provides a choice to switch or stay

Problem Statement

  • Task is to identify the probability of winning in the case of :
  1. staying with Player's choice
  2. switching to Other choice
  3. switching to Host's choice

Assumptions

  • The Host always lies! :D
# import libraries
import numpy as np
# function to define doors with prizes and no prizes
def define_doors():
    """
    Return :
    --------
    Randomly define door numbers for player's choice, other doors, prize & host choice
    """
    # list of doors
    doors = ["Door 1", "Door 2", "Door 3"]
    
    # random doors for each option
    player_choice = np.random.choice(doors)
    prize_door = np.random.choice(doors)
    host_choice = np.random.choice(np.setdiff1d(doors, [player_choice, prize_door], assume_unique=True))
    other_choice = np.random.choice(np.setdiff1d(doors, [player_choice, host_choice], assume_unique=True))
    
    return player_choice, prize_door, host_choice, other_choice

player_choice, prize_door, host_choice, other_choice = define_doors()
#print(f"Player-{player_choice}\nPrize-{prize_door}\nHost-{host_choice}\nOther-{other_choice}")    
# function to suggest if switch is a good choice or not
def oracle(choice : str)-> str:
    """
    determines if player's choice to switch, stay or listen to host is correct or not
    
    Args:
    ----
    switch - string, either switch, stay or host
    
    Return:
    ------
    boolean - if switch was a right choice or not
    """
    # execute define_doors() for choices
    player_choice, prize_door, host_choice, other_choice = define_doors()
    
    # if staying with player_choice
    if choice == "stay":
        return(player_choice == prize_door)
    # if switch to other_choice
    elif choice == "switch":
        return(other_choice == prize_door)

#oracle(choice='stay')
    
    
# simulate to determine if player_choice to STAY is winning strategy
choices = ['stay', 'switch']
for choice in choices :
    result_set = [oracle(choice=choice) for i in range(100000)]
    wins_prop = np.mean(result_set)
    print(f"Winning percentage for {choice} : {wins_prop}")

Conclusion

  • The winning percentages for 1 million iterations :

    • STAY on Player's choice = 33.411 %
    • SWITCH to Other choice = 66.851 %

  • Hence, the takeaways are as follows :

    • SWITCH to Other choice has a greater chance of winning the prize.