# An intro to: Birthday Paradox

Fonte: https://vbfelix.github.io/posts/0011-birthday-paradox/index.html

```{r setup,echo=FALSE,message=FALSE,warning=FALSE}
suppressWarnings(library(ggplot2))
suppressWarnings(library(relper))
suppressWarnings(library(dplyr))
suppressWarnings(library(tidyr))
suppressWarnings(library(janitor))
suppressWarnings(library(knitr))
suppressWarnings(library(kableExtra))
suppressWarnings(library(forcats))
```

In this post, we will learn about the likelihood of birthdays colliding.

# Introduction

The birthday paradox is a well-known and somewhat perplexing probability problem that concerns the likelihood of two people in a group sharing their birthday.

The results frequently surprise people due to their initial probability assumptions, which is why it is referred to as a paradox.

What is the probability that two people in a room with 23 random people share the same birthday?

# Twice the wishes

First, we compute the total number of pairs using the combinations with no repeat formula, which is given by:

$$
\frac{n!}{r!(n-r)!},
$$ {#eq-combination}

where:

-   $n$ is the number of observations;

-   $r$ is the the number of observations to be selected.

So we can apply the @eq-combination to our example:

$$
\begin{align}
\frac{n!}{r!(n-r)!}
&= \frac{23!}{2!(23-2)!} \\
&= \frac{23!}{2!\times21!} \\
&= \frac{23\times22\times21!}{2!\times21!} \\
&= \frac{23\times22}{2} \\
&= \frac{506}{2} \\
&= 253. \\
\end{align}
$$ {#eq-pair-23}

As seen in @eq-pair-23, we have a total of 253 pairs.

# When Statistics Blow Out the Candles

Given that a year has 365 days, if the first person is born on a single day, the second person only needs to be born on any other day, so the likelihood of two people having different birthdays is:

$$
\frac{364}{365} \approx 0.9972.
$$ {#eq-prob-diff-birthday}

Taking this probability into account for each pair, we can compute the probability of all pairs having different birthdays:

$$
\left(\frac{364}{365}\right)^{253} \approx 0.4995.
$$ {#eq-prob-pair-diff-birthday} Then, we can quickly compute the probability of a pair matching their birthday by doing the complementary event of @eq-prob-pair-diff-birthday:

$$
1 - \left(\frac{364}{365}\right)^{253} \approx 0.5005.
$$ {#eq-pair-same-birthday} So, in a group of only 23 people, there is a greater than 50% chance that at least two of them have the same birthday.

# The more the merrier?

But what if we want to calculate this probability for a pair in a group of more or fewer people? We can generalize the @eq-pair-23 to:

$$
\begin{align}
\frac{n!}{r!(n-r)!}
&= \frac{n!}{2!(n-2)!}  \\
&= \frac{n\times(n-1)\times(n-2)!}{2(n-2)!} \\
&= \frac{n\times(n-1)}{2}. \\
\end{align}
$$ {#eq-pair-n}

Then, we use the @eq-pair-n in @eq-pair-same-birthday:

$$
1 - \left(\frac{364}{365}\right)^{\frac{n\times(n-1)}{2}}.
$$ {#eq-birthday-paradox}

Finally, with @eq-birthday-paradox we can see the probability behavior as the group size changes.

```{r, echo = FALSE}
same_birth <- function(n = 23){
  p <- 364/365
  power <-  n*(n-1)/2
  1- (p^power)
}

tibble(n = 2:50) %>% 
  mutate(p = same_birth(n)) %>% 
  ggplot(aes(n,p))+
  geom_line()+
  plt_theme_xy(margin = .5)+
  plt_water_mark(vfx_watermark)+
  scale_x_continuous(
    expand = c(0,0),breaks = seq(2,50,4),
    sec.axis = sec_axis(trans = ~.,breaks = 23)
      )+
  plt_scale_y_mirror(breaks = seq(0,1,.1))+
  labs(
    x = "Group size",
    y = "",
    subtitle = "Probability of a pair of people having the same birthday in a group"
  )+
  geom_vline(xintercept = 23, linetype = "dashed",col = "firebrick3")+
  geom_hline(yintercept = .5, linetype = "dashed",col = "firebrick3")+
  annotate(geom = "point",x = 23,y = .5,col = "firebrick3", size = 3)

```