Programming a Conditional Probability Problem

Hi all. I have been working my way through the Probability course, specifically the section on Conditional Probability, when I fell upon this linked problem.

By solving the linked problem recursively, you find that you have a 13\frac{1}{3} probability of winning the game.

I thought this would be a fun problem to program, so I wrote a program in Python 3 to solve this using Monte Carlo methods. The program and output of the program can be seen below.

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from random import choice
trials = 10000000                   #number of times the game is played (10,000,000)
lose = 0
win = 0
for game in range(1, trials):
    options = [1,2,3,4,5,6]         #can roll any of these numbers
    chips = 1                       #starting point for each game
    while chips > 0 and chips < 4:  #continue this loop for as long as you haven't lost or won
        roll = choice(options)      #picking random number out of options, like rolling a die
        if roll == 1 or roll == 2 or roll == 3:
            chips -= 1
        elif roll == 4 or roll == 5:
            chips += 1
        elif roll == 6:
            chips += 2
    if chips <= 0:
        lose += 1
    elif chips >= 4:
        win += 1

print(win/lose)                     #probability of winning

Output:

0.5001471644515492

Assuming the program is correct, the probability of winning the game is approximately 12\frac{1}{2}, as opposed to the original solution that states you have a 13\frac{1}{3} chance of winning the game.

Can anyone explain what is going on here (it is very possible that my code has something wrong with it, or I am misunderstanding some fundamental concept. If you catch anything please point it out :)?

#Combinatorics

Note by Sam Everett
3 years, 10 months ago

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Comments

Excellent simulation. Your code is mostly correct, except for a small typo in line 21. Can you figure it out?

Agnishom Chattopadhyay - 3 years, 10 months ago
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