Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Birthday Paradox in Python
The birthday paradox is a very famous problem in the section of probability.
Problem Statement − There are several people at a birthday party, some are having the same birthday collision. We need to find the approximate number of people at a birthday party on the basis of having the same birthday.
In the probability, we know that the chance of getting ahead is 1/2, same as if we have some coins, the chance of getting 10 heads is 1/100 or 0.001.
Let us understand the concept.
The chance of two people having the different birthday is $$\frac{364}{365}$$ which is $$\lgroup1-\frac{1}{365}\rgroup$$ in a Non-leap year.
Thus, we can say that the first person having the probability of a specific birthday is ‘1’ and for others, it would be different which is,
P(different) = $$1\times\lgroup1-\frac{1}{365}\rgroup\times\lgroup1-\frac{2}{365}\rgroup\times\lgroup1-\frac{3}{365}\rgroup\times\lgroup1-\frac{4}{365}\rgroup...$$
Thus,
P(same) = 1 − P(different)
For example, the number of people having the same birthday for which probability is 0.70.
N = √2 × 365 × log(1-1/p)
N = √2 × 365 × log(1-1/0.70) = 30
Thus, the total approximate no. of people having the same birthday is 30.
Example
import math def findPeople(p): return math.ceil(math.sqrt(2*365*math.log(1/(1-p)))) print(findPeople(0.70))
Output
Running the above code will generate the output as,
30
1K+ Views
