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Birthday Paradox Calculator

Calculate the probability that at least two people in a group share the same birthday. Demonstrates the surprisingly high likelihood of shared birthdays in small groups, a classic result in probability theory.

Reviewed by Chase FloiedUpdated April 16, 2026

This free online birthday paradox calculator provides instant results with no signup required. All calculations run directly in your browser — your data is never sent to a server. Enter your values below and see results update in real time as you type. Perfect for everyday calculations, homework, or professional use.

Number of People

The number of people in the group.

Days in Year

Number of possible birthdays (365 for standard year, or use 366 to include leap day).

How to Use This Calculator

Enter your input values

Fill in all required input fields for the Birthday Paradox Calculator. Most fields include unit selectors so you can work in your preferred unit system — metric or imperial, whichever matches your problem.

Review your inputs

Double-check that all values are correct and that you have selected the right units for each field. Incorrect units are the most common source of calculation errors and can produce results that are off by factors of 2, 10, or more.

Read the results

The Birthday Paradox Calculator instantly computes the output and displays results with units clearly labeled. All calculations happen in your browser — no loading time and no data sent to a server.

Explore parameter sensitivity

Try adjusting individual input values to see how the output changes. This is a quick and effective way to develop intuition about how different parameters influence the result and to identify which inputs have the largest effect.

Formula Reference

Birthday Paradox Calculator Formula

See calculator inputs for the governing equation

Variables: All variables and their units are labeled in the calculator interface above. Input fields accept values in multiple unit systems — select your preferred unit from the dropdown next to each field.

When to Use This Calculator

•Use the Birthday Paradox Calculator when you need accurate results quickly without the risk of manual computation errors or unit conversion mistakes.
•Use it to verify calculations made by hand or in spreadsheets — an independent check can catch errors before they lead to costly decisions.
•Use it to explore how changing input parameters affects the output — a quick way to develop intuition and identify the most influential variables.
•Use it when collaborating with others to ensure everyone is working from the same numbers and applying the same assumptions.

About This Calculator

The Birthday Paradox Calculator is a free, browser-based calculation tool for engineers, students, and technical professionals. Calculate the probability that at least two people in a group share the same birthday. Demonstrates the surprisingly high likelihood of shared birthdays in small groups, a classic result in probability theory. It implements standard formulas and supports both metric (SI) and imperial unit systems with automatic unit conversion. All calculations are performed instantly in your browser with no data sent to a server. Use this calculator as a quick reference and sanity-check tool during design, analysis, and learning. Always verify results against primary engineering references and applicable standards for any safety-critical application.

About Birthday Paradox Calculator

The birthday paradox calculator reveals the surprisingly high probability that at least two people in a relatively small group share the same birthday. With just 23 people, the probability exceeds 50%, and with 70 people it is over 99.9%. This result is called a 'paradox' because it contradicts most people's intuition -- we tend to think about the chance of someone sharing OUR birthday (which requires many more people) rather than ANY two people sharing A birthday. The birthday problem has practical applications in cryptography (hash collision probability), computer science (hashing algorithms), and quality assurance (duplicate detection). This calculator computes both exact and approximate probabilities for any group size.

The Math Behind It

The birthday paradox works because the number of possible pairs in a group grows much faster than the group size itself. In a group of n people, there are C(n,2) = n(n-1)/2 pairs to compare. With 23 people, that is 253 pairs, each with a 1/365 chance of matching. The exact calculation multiplies successive probabilities: the second person avoids matching the first (364/365), the third avoids both (363/365), and so on, giving P(no match) = product of (365-i)/365 for i from 0 to n-1. The complementary probability P(at least one match) = 1 - P(no match) crosses 50% at n = 23. An accurate approximation uses e^(-n(n-1)/(2*365)). In cryptography, this result is critical for understanding hash collision attacks: a hash function with an output space of size N will likely produce a collision after roughly sqrt(N) random inputs, far fewer than N. This is why cryptographic hashes need very large output spaces. The generalized birthday problem extends to finding matches among items drawn from d possible values, with the 50% threshold at approximately 1.2 * sqrt(d). The birthday paradox illustrates why human probability intuition fails for combinatorial problems and serves as a gateway to understanding collision resistance in security systems.

Formula Reference

Birthday Problem (exact)

P(no match) = 365/365 * 364/365 * ... * (365-n+1)/365

Variables: n = number of people; each factor accounts for one additional person having a unique birthday

Birthday Problem (approximation)

P(no match) ≈ e^(-n(n-1) / (2*365))

Variables: Uses the approximation (1-1/365)^k ≈ e^(-k/365) for the product of probabilities

Worked Examples

Example 1: Classic: 23 people in a room

What is the probability that at least 2 of 23 people share a birthday?

Step 1:P(no match) = (365/365)(364/365)(363/365)...(343/365).

Step 2:This product equals approximately 0.4927.

Step 3:P(match) = 1 - 0.4927 = 0.5073.

With 23 people, there is a 50.7% chance that at least two share a birthday, just over half.

Example 2: Office of 50 people

In an office of 50 employees, what is the probability of a shared birthday?

Step 1:Number of pairs = C(50,2) = 1,225.

Step 2:P(no match) ≈ e^(-1225/365) = e^(-3.356) ≈ 0.0350.

Step 3:P(match) = 1 - 0.0350 = 0.9650.

With 50 people, there is a 97% probability of at least one shared birthday pair.

Common Mistakes & Tips

!Confusing 'any two people share a birthday' with 'someone shares MY birthday' -- the latter requires about 253 people for 50% probability.
!Assuming birthdays are uniformly distributed across the year -- actual birth rates vary by season, which slightly increases the collision probability.
!Applying the birthday paradox formula beyond the valid range (n > daysInYear makes the probability exactly 1 by the pigeonhole principle).

Related Concepts

Probability Calculator

Calculate basic probability operations including complement probabilities used in the birthday problem derivation.

Combinations Calculator

Compute C(n,k) values, which determine the number of pairs in the birthday problem and many other counting problems.

Frequently Asked Questions

Why is it called a paradox if the math is correct?

It is called a paradox because the result strongly contradicts most people's intuition, not because there is a logical contradiction. We instuitively think of 365 days and 23 people and feel the probability should be low, but we fail to account for the 253 pairs being compared. It is more accurately a 'veridical paradox' -- a truth that seems absurd.

How many people do you need for a 99% probability?

You need only 57 people for a 99% probability of at least one shared birthday pair. At 70 people, the probability exceeds 99.9%. The curve rises steeply between 20 and 60 people, then flattens as it asymptotically approaches 100%.

How does the birthday paradox relate to cybersecurity?

In cryptography, birthday attacks exploit the same mathematics to find hash collisions. A hash function with N possible outputs will likely produce a collision after about sqrt(N) random inputs. This is why secure hash functions need at least 256-bit outputs: finding a collision requires approximately 2^128 attempts, which is computationally infeasible with current technology.