first_dd.Rmd

---
title: "Description of the Birthday Problem"
author: "Fernando Hoces de la Guardia"
date: "9/3/2021"
output:
  pdf_document: default
  word_document: default
  html_document: default
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
set.seed(1234)
n.pers = 16
n.sims = 10000
```

# Description of the problem 

What is the probability that at least two people this room share the same birthday?

There are `r n.pers` in this room.

Is it something like  $\frac{`r n.pers`}{365} = `r n.pers/365`$ 

# The mathematical solution  

Define p(n) as the probability that at least one pair has the same birthday, then the 1-p(n) is the probability that all are born in a different day. Which we can compute as:


\begin{align} 
 1 -  p(n) &= 1 \times \left(1-\frac{1}{365}\right) \times
              \left(1-\frac{2}{365}\right) \times \cdots \times
              \left(1-\frac{n-1}{365}\right) \nonumber  \\
           &= \frac{365 \times 364 \times \cdots \times (365-n+1) }{365^n} \nonumber \\
           &= \frac{ 365! }{ 365^n (365-n)!} = \frac{n!\cdot\binom{365}{n}}{365^n}\\
p(n= `r n.pers`) &= `r  round(1 - factorial(n.pers) * 
                          choose(365,n.pers)/ 365^n.pers, 3)`  \nonumber
\end{align}


# The simulated solution 

We will simulate the probability:   

1 - Simulate 10^{4} rooms with `r n.pers` random birthdays, and store the results in matrix where each row represents a room.

2 - For each room (row) compute the number of unique birthdays.

3 - Compute the average number of times a room has `r n.pers` unique birthdays, across 10^{4} simulations, and report the complement.


```{r}
birthday.prob = function(n.pers_var, n.sims_var) {
  # simulate birthdays
  birthdays = matrix(round(runif(n = n.pers_var * n.sims_var, min = 1, max = 365) ),
                     nrow = n.sims_var, ncol = n.pers_var)
  # for each room (row) get unique birthdays
  unique.birthdays = apply(birthdays, 1,
                           function(x)  length( unique(x) ) )
  # Indicator with 1 if all are unique birthdays
  all.different = 1 * (unique.birthdays==n.pers_var)
  # Compute average time all have different birthdays 
  result = 1 - mean(all.different)
return(result)
}
bp_sim = birthday.prob(n.pers_var = 21, n.sims_var = 10000)
print(bp_sim)
```

# Results

- Many people think the solution is $\frac{1}{365} \times n = `r n.pers/365`$  

- The math says: `r  round(1 - factorial(n.pers) * choose(365,n.pers)/ 365^n.pers, 3)`  
- A simulation with `r n.sims` rooms with `r n.pers` people in each room, says: `r bp_sim`