Monty Hall problem and frog riddle


Monty Hall paradox

Probability topic is the fundamental concept of the statistics. And machine learning is closely related to statistics. That is why, understand the probability very important if you are doing research, statistics, and machine learning.

Monty Hall is a very interesting problem. It says, if you are given 3 doors to choose. One of them contains a car (which you want), the other two are goats (which you don’t want). After you made your choice, before opening the door, the host will open the door that you didn’t choose yet contains the goat (he knows which door has the goat). Now, if you are given an opportunity to change your choice to another door (which you didn’t choose earlier), are you going to change?

In the first glance, you will feel that whatever you choose, the probability is always 1/3. However, the conditional probability tells you that, if you always make the switch after the host opened the door that has a goat, your probability to win the car will increase to 2/3. What??

In order to prove this, I wrote a Python script.

#!/usr/bin/env python
# This is simulating Monty Hall Paradox

import random


def monty(switch):
    # random shuffle
    doors = [0, 0, 1]  # one of the door contains the car
    random.shuffle(doors)

    openDoor = None

    # choose the first door (not open)
    # if the first door is 1, randomly open the other
    if doors[0] == 1:
        # open the door
        openDoor = random.randint(1, 2)
    else:  # open the door that contains goat
        if doors[1] == 1:
            openDoor = 2
        else:
            openDoor = 1

    # now open the last door
    if not switch:
        return doors[0]
    else:
        if openDoor == 2:
            return doors[1]
        else:
            return doors[2]


def main():
    total = 10000
    car = 0
    for i in range(total):
        car += monty(True)

    print("Always switch the door. Total: {}, car: {}. P = {}".format(total, car, car / total))

    car = 0
    for i in range(total):
        car += monty(False)

    print("No switch the door. Total: {}, car: {}. P = {}".format(total, car, car / total))


main()

Run the code, you will always get the probabilty close to 0.6667 if you always switch the door.

Always switch the door. Total: 10000, car: 6625. P = 0.6625
No switch the door. Total: 10000, car: 3309. P = 0.3309

Frog riddle

Recently I just watched a Youtube about frog riddle.

It also mentions about the conditional probability. Interestingly, quite a lot of comments mentioned that the author is wrong.

In order to prove that the author is correct, I wrote another Python script.

#!/usr/bin/env python

import random

# Frog 0 for female, 1 for male


def create_frog():
    return random.randint(0, 1)


def has_croak(pairs):  # also male
    return 1 in pairs


def has_female(frogs):
    return 0 in frogs


def choose_without_croak(choose_two):
    frogs = [create_frog() for i in range(3)]
    # first frog at the right side
    # second and third at the left side

    if choose_two:
        return has_female(frogs[1:])  # choose two frogs

    return has_female(frogs[0:1])


def main():
    total = 10000
    correct = 0
    for i in range(total):
        correct += choose_without_croak(True)
    print('Just choose two frogs. Total: {}, correct: {}. P = {}'.format(total, correct, correct / total))

    correct = 0
    for i in range(total):
        correct += choose_without_croak(False)
    print('Just choose one frog. Total: {}, correct: {}. P = {}'.format(total, correct, correct / total))


# The exact question is,
# "What is the probability of the frogs in the pair has female,
# given that one of them is male?"
def exact_calculation():
    total = 10000
    croak = 0
    correct = 0
    for i in range(total):
        frogs = [create_frog() for i in range(3)]
        if has_croak(frogs[1:]):
            croak += 1
            if has_female(frogs[1:]):
                correct += 1
    print('Total croak: {}, correct: {}. P = {}'.format(croak, correct, correct / croak))


main()
exact_calculation()

Running the script, you will get

Just choose two frogs. Total: 10000, correct: 7498. P = 0.7498
Just choose one frog. Total: 10000, correct: 4974. P = 0.4974
Total croak: 7474, correct: 4998. P = 0.6687182231736687

Based on the result, if you choose two frogs, the probability of survive is close to 0.75. If you choose one frog, the probability is 0.5.

Now, the tricky part is the probability 0.67 mentioned in the video. The question should be “What is the probability of the frogs in the pair has female, given that one of them is male?”

So, based on the question, my similuation needs to get the total count of the male (that has croak), and within these pairs, count the female frogs.

To convert this into mathematical expression,

P(\text{female frog}) = 0.5

P(\text{at least one male frog}) = 0.75

P(\text{female frog} | \text{at least one male frog}) = \frac{0.5}{0.75} = 0.6667

Then, based on the simulation and calculation, you will get the 0.6667.

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