The Birthday Problem
by aoum, Mar 13, 2025, 11:48 PM
The Birthday Problem: Probability of Shared Birthdays
The Birthday Problem is a fascinating probability puzzle that asks: In a group of randomly chosen people, what is the probability that at least two people share the same birthday? The result is surprisingly counterintuitive—much higher than most people expect.
1. Understanding the Birthday Problem
If you randomly select people and record their birthdays (assuming 365 days in a year and ignoring leap years), the probability that at least two of them share a birthday grows rapidly as the group size increases.
Here are some key probability benchmarks:
These probabilities seem surprising because most people intuitively compare everyone’s birthday to their own. However, the correct approach compares every pair of people, which significantly increases the number of potential matches.
2. Why Does This Happen?
At the heart of the birthday problem lies the concept of combinatorial growth. With each new person added to the group, the number of potential birthday pairs grows rapidly.
In a group of
people, the number of unique pairs is given by the binomial coefficient:
![\[
\binom{n}{2} = \frac{n(n - 1)}{2}.
\]](//latex.artofproblemsolving.com/4/a/8/4a85bfc9cb6e0393d2e212897b7378c0757c45af.png)
For 23 people:
![\[
\binom{23}{2} = \frac{23 \times 22}{2} = 253 \text{ pairs.}
\]](//latex.artofproblemsolving.com/f/2/1/f21e9b09b6d891613ce31b081690e720cbb3cac4.png)
The rapid growth of potential comparisons means the likelihood of a match increases quickly as the group size increases.
3. The Mathematics Behind the Birthday Problem
To calculate the probability that no one shares a birthday, we multiply the probabilities that each successive person has a unique birthday:
For
people, the probability that no two share a birthday is:
![\[
P(\text{No Match}) = \frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{365 - n + 1}{365}.
\]](//latex.artofproblemsolving.com/7/4/9/749c2346ac67a6238dd81f288412dbe67dc63d9e.png)
This can be rewritten using the product notation:
![\[
P(\text{No Match}) = \prod_{k=0}^{n-1} \left( \frac{365 - k}{365} \right).
\]](//latex.artofproblemsolving.com/0/9/4/0946a363afd3d28845f20d09cfd7b9d3198f3d3c.png)
The probability that at least two people share a birthday is the complement:
![\[
P(\text{At Least One Match}) = 1 - P(\text{No Match}).
\]](//latex.artofproblemsolving.com/1/a/c/1ac6048dc22efdb590ba28fee9ac23e1f7b233f8.png)
4. Example Calculation (23 People)
If
:
![\[
P(\text{No Match}) = \frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{343}{365} \approx 0.4927,
\]](//latex.artofproblemsolving.com/e/b/d/ebd30f312b27b16abb134f995e817fe363c1c421.png)
Thus,
![\[
P(\text{At Least One Match}) = 1 - 0.4927 = 0.5073 \approx 50.7\%.
\]](//latex.artofproblemsolving.com/8/c/f/8cf82ea2e1780802f883972f0d956332e4baad91.png)
5. Python Code to Simulate the Birthday Problem
Here’s a Python script to estimate the probability using a Monte Carlo simulation:
6. Approximating with the Poisson Distribution
For larger groups, the product formula becomes computationally intensive. We can approximate the probability using the Poisson distribution:
![\[
P(\text{At Least One Match}) \approx 1 - e^{-\frac{n^2}{2m}},
\]](//latex.artofproblemsolving.com/0/3/c/03cc9bae4716e98f9eb8cdfdc9c341471eec88d8.png)
where:
For
:
![\[
P(\text{At Least One Match}) \approx 1 - e^{-\frac{23^2}{2 \times 365}} \approx 0.507.
\]](//latex.artofproblemsolving.com/3/3/0/330b514ae8a3e890bd10f9e3d53c0fca84df6f42.png)
7. Fun Facts About the Birthday Problem
8. Applications of the Birthday Problem
9. Exploring the Generalized Birthday Problem
The birthday problem extends to other domains:
10. Conclusion
The birthday problem illustrates how quickly probabilities grow when considering combinations. Despite the vast number of possible birthdays, relatively small groups have a high likelihood of sharing one. This seemingly simple puzzle has profound implications across mathematics, computer science, and data security.
References
The Birthday Problem is a fascinating probability puzzle that asks: In a group of randomly chosen people, what is the probability that at least two people share the same birthday? The result is surprisingly counterintuitive—much higher than most people expect.

1. Understanding the Birthday Problem
If you randomly select people and record their birthdays (assuming 365 days in a year and ignoring leap years), the probability that at least two of them share a birthday grows rapidly as the group size increases.
Here are some key probability benchmarks:
- With 23 people, the probability that at least two share a birthday is approximately 50.7%.
- With 50 people, the probability exceeds 97%.
- With 70 people, the probability is 99.9%, meaning it is almost certain that two people share a birthday.
These probabilities seem surprising because most people intuitively compare everyone’s birthday to their own. However, the correct approach compares every pair of people, which significantly increases the number of potential matches.
2. Why Does This Happen?
At the heart of the birthday problem lies the concept of combinatorial growth. With each new person added to the group, the number of potential birthday pairs grows rapidly.
In a group of

![\[
\binom{n}{2} = \frac{n(n - 1)}{2}.
\]](http://latex.artofproblemsolving.com/4/a/8/4a85bfc9cb6e0393d2e212897b7378c0757c45af.png)
For 23 people:
![\[
\binom{23}{2} = \frac{23 \times 22}{2} = 253 \text{ pairs.}
\]](http://latex.artofproblemsolving.com/f/2/1/f21e9b09b6d891613ce31b081690e720cbb3cac4.png)
The rapid growth of potential comparisons means the likelihood of a match increases quickly as the group size increases.
3. The Mathematics Behind the Birthday Problem
To calculate the probability that no one shares a birthday, we multiply the probabilities that each successive person has a unique birthday:
- The first person has any birthday:
.
- The second person must avoid the first person's birthday:
.
- The third person must avoid the first two birthdays:
.
For

![\[
P(\text{No Match}) = \frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{365 - n + 1}{365}.
\]](http://latex.artofproblemsolving.com/7/4/9/749c2346ac67a6238dd81f288412dbe67dc63d9e.png)
This can be rewritten using the product notation:
![\[
P(\text{No Match}) = \prod_{k=0}^{n-1} \left( \frac{365 - k}{365} \right).
\]](http://latex.artofproblemsolving.com/0/9/4/0946a363afd3d28845f20d09cfd7b9d3198f3d3c.png)
The probability that at least two people share a birthday is the complement:
![\[
P(\text{At Least One Match}) = 1 - P(\text{No Match}).
\]](http://latex.artofproblemsolving.com/1/a/c/1ac6048dc22efdb590ba28fee9ac23e1f7b233f8.png)
4. Example Calculation (23 People)
If

![\[
P(\text{No Match}) = \frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{343}{365} \approx 0.4927,
\]](http://latex.artofproblemsolving.com/e/b/d/ebd30f312b27b16abb134f995e817fe363c1c421.png)
Thus,
![\[
P(\text{At Least One Match}) = 1 - 0.4927 = 0.5073 \approx 50.7\%.
\]](http://latex.artofproblemsolving.com/8/c/f/8cf82ea2e1780802f883972f0d956332e4baad91.png)
5. Python Code to Simulate the Birthday Problem
Here’s a Python script to estimate the probability using a Monte Carlo simulation:
import random
def birthday_simulation(num_people, trials=100000):
matches = 0
for _ in range(trials):
birthdays = [random.randint(1, 365) for _ in range(num_people)]
if len(birthdays) != len(set(birthdays)):
matches += 1
return matches / trials
for n in [10, 23, 30, 50, 70]:
probability = birthday_simulation(n)
print(f"For {n} people, the estimated probability of a shared birthday is {probability:.4f}")
6. Approximating with the Poisson Distribution
For larger groups, the product formula becomes computationally intensive. We can approximate the probability using the Poisson distribution:
![\[
P(\text{At Least One Match}) \approx 1 - e^{-\frac{n^2}{2m}},
\]](http://latex.artofproblemsolving.com/0/3/c/03cc9bae4716e98f9eb8cdfdc9c341471eec88d8.png)
where:
= Number of people
= Number of days in a year (365)
For

![\[
P(\text{At Least One Match}) \approx 1 - e^{-\frac{23^2}{2 \times 365}} \approx 0.507.
\]](http://latex.artofproblemsolving.com/3/3/0/330b514ae8a3e890bd10f9e3d53c0fca84df6f42.png)
7. Fun Facts About the Birthday Problem
- The birthday problem generalizes to hash collisions in computer science, where the likelihood of two items mapping to the same hash value is calculated similarly.
- This problem is an example of the pigeonhole principle: If you have more items than categories, at least one category must contain more than one item.
- The distribution of real-world birthdays is not perfectly uniform—certain dates (e.g., September) have higher birth rates.
- This paradox applies to various fields like DNA profiling, cryptographic security, and data science.
8. Applications of the Birthday Problem
- Cryptography: Explains the vulnerability of hash functions to birthday attacks.
- Data Science: Used in large datasets to calculate the likelihood of duplicates.
- Networking: Analyzes packet collisions in communication protocols.
- Probability Theory: Demonstrates how quickly probabilities compound with growing combinations.
9. Exploring the Generalized Birthday Problem
The birthday problem extends to other domains:
- k-Matches Problem: What is the probability that at least k people share a birthday?
- Non-Uniform Distribution: Real-life data, such as hospital records, show non-uniform birthday frequencies, requiring adjusted calculations.
- Higher Dimensions: Similar probabilistic problems occur in multidimensional spaces, such as the collision probability of random vectors.
10. Conclusion
The birthday problem illustrates how quickly probabilities grow when considering combinations. Despite the vast number of possible birthdays, relatively small groups have a high likelihood of sharing one. This seemingly simple puzzle has profound implications across mathematics, computer science, and data security.
References
- Wikipedia: Birthday Problem
- Rosen, K. H. Discrete Mathematics and Its Applications (2018).
- Feller, W. An Introduction to Probability Theory and Its Applications (1968).
- Mitzenmacher, M. Probability and Computing: Randomized Algorithms (2005).
- AoPS: Birthday Problem