Birthday paradox python incorrect probability output

The birthday paradox is a famous problem in probability theory that states that in a group of just 23 people, there is a 50% chance that at least two people have the same birthday. This may seem counterintuitive at first, but it is a result of the pigeonhole principle and the fact that there are only 365 possible birthdays.

In this article, we will explore different ways to solve the birthday paradox problem in Python and discuss their pros and cons. We will start by defining the problem and then present three different solutions.

Solution 1: Brute Force

The first solution is a brute force approach that involves generating random birthdays for a given number of people and checking if there are any duplicates. This solution is straightforward but not very efficient.

import random

def birthday_paradox(n):
    birthdays = [random.randint(1, 365) for _ in range(n)]
    if len(birthdays) != len(set(birthdays)):
        return True
    return False

n = 23
probability = sum(birthday_paradox(n) for _ in range(10000)) / 10000
print(f"The probability of a birthday collision with {n} people is approximately {probability:.2f}")

This solution generates random birthdays for a given number of people and checks if there are any duplicates. It repeats this process multiple times to estimate the probability of a birthday collision. However, since it relies on random number generation, the result may vary each time the code is executed.

Solution 2: Analytical Formula

The second solution involves using an analytical formula to calculate the probability of a birthday collision directly. This approach is more efficient than the brute force method and provides an exact solution.

def birthday_paradox(n):
    p = 1
    for i in range(1, n):
        p *= (365 - i) / 365
    return 1 - p

n = 23
probability = birthday_paradox(n)
print(f"The probability of a birthday collision with {n} people is approximately {probability:.2f}")

This solution calculates the probability of a birthday collision using the formula 1 - (365/365) * ((365-1)/365) * ... * ((365-n+1)/365). It iterates over the range of 1 to n and multiplies the probabilities together. This approach provides an exact solution and is much faster than the brute force method.

Solution 3: Monte Carlo Simulation

The third solution involves using a Monte Carlo simulation to estimate the probability of a birthday collision. This approach is similar to the brute force method but uses a larger number of iterations to obtain a more accurate result.

def birthday_paradox(n, iterations):
    collisions = 0
    for _ in range(iterations):
        birthdays = [random.randint(1, 365) for _ in range(n)]
        if len(birthdays) != len(set(birthdays)):
            collisions += 1
    return collisions / iterations

n = 23
iterations = 100000
probability = birthday_paradox(n, iterations)
print(f"The probability of a birthday collision with {n} people is approximately {probability:.2f}")

This solution generates random birthdays for a given number of people and checks if there are any duplicates. It repeats this process a large number of times to obtain a more accurate estimate of the probability. The result may vary slightly each time the code is executed due to the random nature of the simulation.

After comparing the three solutions, the analytical formula (Solution 2) is the best option. It provides an exact solution and is much faster than the brute force method. The Monte Carlo simulation (Solution 3) can also be used for more accurate results, but it requires a larger number of iterations, making it slower than the analytical formula.

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5 Responses

  1. Who wouldve thought that birthdays could be so paradoxical? Im sold on Solution 3 – Monte Carlo Simulation, sounds fancy!

    1. Monte Carlo Simulation may sound fancy, but its not all its cracked up to be. Solution 2 – The Grid Method is much more reliable. Trust me, Ive been there, done that.

  2. Wow, that article on the Birthday Paradox in Python blew my mind! Who knew probability could be so trippy? 🤯

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