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.
[code language=“python”] #!/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() [/code]
Run the code, you will always get the probabilty close to 0.6667 if you always switch the door.
[code] Always switch the door. Total: 10000, car: 6625. P = 0.6625 No switch the door. Total: 10000, car: 3309. P = 0.3309 [/code]
Frog riddle
Recently I just watched a Youtube about frog riddle.
[youtube src=“https://www.youtube.com/watch?v=cpwSGsb-rTs"]
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.
[code language=“python”] #!/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() [/code]
Running the script, you will get
[code] 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 [/code]
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,
$latex P(\text{female frog}) = 0.5$
$latex P(\text{at least one male frog}) = 0.75$
$latex P(\text{female frog} | \text{at least one male frog}) = \frac{0.5}{0.75} = 0.6667$
Then, based on the simulation and calculation, you will get the 0.6667.