path: root/content/articles/the-birthday-paradox.md

---
slug: the-birthday-paradox
title: "The Birthday Paradox"

# date (optional for articles)
date: "2021-11-11T06:46:00-04:00"

# draft articles are not visible on live site
draft: true

# show on articles page
show: true

# uncomment to show table of contents
toc: true

tables:
  probs:
    cols:
    - id: num_people
      name: Number of People
      tip: Number of people in room.
    - id: probability
      name: Probability
      tip: Probability that at least two people in a room with this many people share a birthday.
    - id: percent
      name: Percent
      tip: Probability represented as a percentage.
      align: right
    rows:
    - num_people: 1
      probability: 0.0
      percent: 0.00%
    - num_people: 2
      probability: 0.00273
      percent: 0.27%
    - num_people: 3
      probability: 0.0082
      percent: 0.82%
    - num_people: 4
      probability: 0.01635
      percent: 1.64%
    - num_people: "..."
    - num_people: 22
      probability: 0.47569
      percent: 47.57%
    - num_people: 23
      probability: 0.50729
      percent: 50.73%
      _css: has-text-weight-bold has-background-info-light
    - num_people: 24
      probability: 0.53834
      percent: 53.83%
      _css: has-text-weight-bold has-background-info-light
    - num_people: 25
      probability: 0.56869
      percent: 56.87%
      _css: has-text-weight-bold has-background-info-light
---

## Introduction

I share my birthday with my mom.  Yesterday while talking her on ~~my~~
her birthday I asked if she had heard of the [Birthday Problem][bp]:

> How many people have to be in a room before the probability that there
> is a shared birthday is greater than 50%?

This is called the [Birthday Problem][bp], and the solution is known as
the *birthday paradox*.  It is an interesting problem because:

* The answer is counterintuitive (hence the name *birthday paradox*).
* The solution relies on elements of [combinatorics][], [set theory][],
  and [probability][].
* The ramifications affect security, particularly [cryptographic
  hash algorithms][hash].

The next section covers the math that you'll need.  If you're already
familiar with [probability][], feel free to skip ahead to [Solving the
Birthday Problem][solving].

## Probability Crash Course

Here is an extremely abbreviated summary of of math that you'll need
from [combinatorics][], [probability][], and [set theory][]:

1. The expression `N!`, where `N` is a non-negative integer, is called a
   [factorial][], and it is means "the product of all the integers from
   `N` down to `1`".  So `5! = 5 × 4 × 3 × 2 × 1 = 120`.  To make our
   lives easier, `0! = 1` by definition.
2. In [probability][], an occurance is called an *event*, and the
   probability of an *event* `A` is denoted `P(A)`.
3. The value of a probability is a real number between `0` and `1`,
   inclusive (written `P(A) ∈ [0, 1]`).  Inclusive means that `0` and
   `1` are valid probabilities, where `0` means "this event is
   impossible" and `1` means "this event will occur with certainty".
4. You can convert a probability to a percentage by multiplying by
   `100` and appending a trailing `%`.  So `P(A) = 0.1` means `10%`,
   `P(A) = 0.2` means `20%`, and so on.
5. The probability of two [independent][] events `A` and `B` both
   occurring is written `P(A ⋂ B)` and read as "the [joint
   probability][] of A and B".  The value is calculated as `P(A) ×
   P(B)`.  This pattern of multiplication continues, so the probability
   of three [independent][] events `A`, `B`, and `C` occurring is `P(A ⋂
   B ⋂ C) = P(A) × P(B) × P(C)`, and so on.  The upside-down U symbols
   represent an [intersection][intersection].
6. The probability of an event `A` **not** occurring is written `P(¬A)`,
   called the *compliment*, and calculated as `1 - P(A)`.  This is
   useful because sometimes it is much easier to calculate the
   probability that an event will not occur, as we'll see below.

Here's what all the chicanery above looks like visually:

{{< figure
  src="/files/articles/birthday-paradox/probability-basics.svg"
  class="image"
  width=492
  height=556
  caption="Visual representation of probability basics."
>}}

To calculate a [discrete][] (countable) probability, you sum up all the
matching events, then divide the sum by the total number of possible
events.

For example, if you put one red marble and three blue marbles in a jar
and then randomly choose a single marble out of the jar, then the
probability that you will choose the red marble is:

```sh
P(R) = probability of choosing a red marble
P(R) = number of red marbles / total number of marbles
P(R) = 1/(1 + 3) = 0.25 = 25%
```
&nbsp;

Here's a more complicated example: Let's say you put 3 red marbles and 5
blue marbles in a jar, pick a marble from the jar at random, and then
roll a fair, [6-sided die][die].  What is the probability that you will
pick a red marble from the jar **and** roll a 5 or a 6 on the [die][]?

```sh
P(R) = probability of choosing a red marble
P(R) = number of red marbles / total number of marbles
P(R) = 3/(3 + 5) = 0.375 = 37.5%

P(D) = probability of rolling a 5 or a 6 on the die
P(D) = number of matching die faces / total number of die faces
P(D) = 2/6 = ~0.333 = ~33.3%

P(R ⋂ D) = P(R) × P(D)
P(R ⋂ D) = 3/8 × 1/3 = 0.125 = 12.5%
```

## Solving the Birthday Problem

Let's get back to the solution.  I'll save you a bit of time and a lot
of headache by telling you in advance that the [birthday problem][bp] is
one of those pesky probability problems where it much easier to
calculate the compliment; that is, the [probability][] that everyone
has a *unique* birthday.

Now that that piece of bookkeeping is out of the way, let's get started.

Ff there is one person in the room, then the probability that everyone
has a unique birthday is `1.0` (`100%`).

```sh
P1 = P(unique first birthday) = 365/365
P1 = 1.0 = 100%
```
&nbsp;

If there are two people in the room, the probability that everyone has a
unique birthday can be calculated by multiplying `P1` -- the probability
from the previous step -- by the probability that the second person's
birthday is not the same as the first person's birthday.  Algebraically:

```sh
P2 = P(unique second birthday)
P2 = P1 × 364/365
P2 = 365/365 × 364/365 = (365 × 364)/(365^2) = 364/365
P2 = ~0.9972 = 99.7%
```
&nbsp;

Add a third person, and a pattern begins to emerge:

```r
P3 = P(unique second birthday ⋂ unique third birthday)
P3 = P2 × 363/365
P3 = 365/365 × 364/365 × 363/365 = (365 × 364 × 363)/(365^3)
P3 = ~0.9917 = ~99.2%
```
&nbsp;

Do you see the pattern?

The probability that everyone in a group of `N` people has a unique
birthday is the product of the numbers between `365` and `365 - (N - 1)`
(inclusive), divided by `365{{< sup "N" >}}`.

To calculate the product of all the numbers between `365` and `365 - (N
- 1)` in the numerator, we can use a [factorial][] trick to cancel out
the terms that we don't need:

```sh
Π(X, Y) = product of numbers between X and Y (inclusive) where X < Y
Π(X, Y) = Y!/(X - 1))!

Π((365 - (N - 1)), 365) = 365!/((365 - (N - 1) - 1))!
Π((365 - (N - 1)), 365) = 365!/((365 + -N + 1 + -1))!
Π((365 - (N - 1)), 365) = 365!/(365 - N)!
```
&nbsp;

Combined with the previous equation:

```sh
U(N) = P(N unique birthdays)
U(N) = 365!/((365 - N)! × 365^N)
```
&nbsp;

Then we calculate the compliment of the probability of N unique
birthdays to determine the probability of shared birthdays in a group of
N people:

```sh
S(N) = P(shared birthdays among N people)
S(N) = P(¬U(N)) = 1 - U(N)
S(N) = 1 - 365!/((365 - N)! × 365^N)
```
&nbsp;

At this point you can set `S(N) = 0.5` and solve for `N` to find an
exact solution, but often for discrete probabilities it's easier to
write a script and let a computer do the work for you.

So I wrote a couple of scripts which use the derived algorithm to do the
following:

1. Calculate the probability of shared birthdays in a room containing
   `N` people for `N ∈ [1, 50]`.
2. Exclude intermediate rows in the table to keep it short.
3. Highlight the relevant results (values of `N` where `S(N) > 0.5`).
4. Render the bar chart and table shown below.

The full script is available in the [companion GitHub
repository][github].  Here is the [Ruby][] code for our derived
equation:

```ruby
# memoize 365!
F365 = 365.factorial

#
# Calculate the probability of at least one shared birthday in a group
# of N people.
#
def shared_birthdays(v)
  1 - F365 / (365 - v).factorial * (365 ** -v)
end
```
&nbsp;

Here are the results:

{{< figure
  src="/files/articles/birthday-paradox/people-vs-probability.svg"
  class="image"
  width=614
  height=461
  caption="Number of People versus Probability of Shared Birthday."
>}}

{{< table "probs" >}}

So the answer to the birthday problem is:

**If there are 23 people in the room, then the probability of at least
one shared birthday is greater than 50%**.

## Implications

This result is called the *birthday paradox* is because it doesn't feel
intuitive.  When asked, most people will guess that a much higher
number of people (50 or 130) are needed before the probability is
greater than 50%.

*TODO:* segue into hash algorithms.

## Further Reading

* [GitHub repository for the scripts used in this article][github]
* [The Birthday Problem (Wikipedia)][bp]

[bp]: https://en.wikipedia.org/wiki/Birthday_problem
  "The birthday problem."
[combinatorics]: https://en.wikipedia.org/wiki/Combinatorics
  "The mathematics of counting."
[set theory]: https://en.wikipedia.org/wiki/Set_theory
  "The mathematics of sets."
[probability]: https://en.wikipedia.org/wiki/Probability
  "The mathematics of determining how likely an event is to occur."
[hash]: https://en.wikipedia.org/wiki/Cryptographic_hash_function
  "Cryptographic hash algorithm."
[intersection]: https://en.wikipedia.org/wiki/Intersection_(set_theory)
  "Intersection."
[independent]: https://en.wikipedia.org/wiki/Probability#Independent_events
  "Independent events in probability theory."
[discrete]: https://en.wikipedia.org/wiki/Random_variable#Discrete_random_variable
  "A random value with a countable number of possible values."
[joint probability]: https://en.wikipedia.org/wiki/Joint_probability_distribution#Coin_flips
  "Joint probability."
[die]: https://en.wikipedia.org/wiki/Dice
  "Dice."
[factorial]: https://en.wikipedia.org/wiki/Factorial
  "Factorial unary operator."
[ruby]: https://ruby-lang.org/
  "Ruby programming language."
[csv]: https://en.wikipedia.org/wiki/Comma-separated_values
  "Comma-separated values."
[html]: https://en.wikipedia.org/wiki/HTML
  "HyperText Markup Language"
[yaml]: https://en.wikipedia.org/wiki/YAML
  "YAML Ain't a Markup Language"
[hugo-table-shortcode]: https://github.com/pablotron/hugo-shortcode-table
  "Table shortcode for Hugo."
[github]: https://github.com/pablotron/birthday-paradox
  "GitHub repository for scripts used in this article."
[solving]: #solving-the-birthday-problem
  "Solving the Birthday Problem"