# The Birthday Problem

## Problem

What is the chance that at least two people in a group of k people will have the same birthday?

#### Note:

By same birthday we mean same day and month but not necessarily the same year. We assume further that the birthdays of the k people are unrelated and each equally likely to be in any of the 365 days of the year.

## Solution

Define the proposition:

### E = "at least 2 equal birthdays"

Let S denote the statement of the problem above. Then, the probability of E given S is,

But

### (not E) = "all different birthdays"

So the problem reduces to the computation of the number of ways in which k people can have different birthdays. There are three cases. First, if k > 365 then necessarily at least 2 people will have the same birthday. Second, if k < 2 then there is no problem. Third, when 1 Hence, the total number of ways is,

### 365*364*...*(365-k+1).

The total number of ways in which k people can have their birthdays (with out constraints) is,

### 365k

The probability we are looking for is then,

> answer := 1 - a!/((a-k)!*a^k);

```                           365!
k
(365 - k)! 365```
 This is a function of k that can be easily computed with

> p := k -> evalf((1 - 365!/((365-k)!*365^k)),4):
> p10 := p(10); p20:=p(20); p23 := p(23); p40:=p(40); p50:=p(50);

```                                 p10 := .1169

p20 := .4114

p23 := .5073

p40 := .8912

p50 := .9704```

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>