first_dd_test.Rmd

---
title: "Ther birthday problem"
author: "Fernando Hoces de la Guardia"
date: "9/13/2019"
output: html_document
---


# A Description of the birthday ploblem  
sdfsdf
 sdfsddf
 
 sdfsdsd

```{r}
n.pers = 20  
set.seed(1234)
```

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


## Analytical solution  

But actually when we compute the math. We get  an  surprising result: 

\begin{align}   
 1 - \bar 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}



## Simulations  

1 - Simulate 10,000 rooms with n = `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 10,000 simulations, and report the complement.  

  
```{r}
birthday.prob = function(n.pers, n.sims) {
  # simulate birthdays
  birthdays = matrix(round(runif(n.pers * n.sims, 1, 
                                 365)), nrow = n.sims, 
                           ncol = n.pers)
  # for each room (row) get unique birthdays
  unique.birthdays = apply(birthdays, 1, unique)
  # Indicator with 1 if all are unique birthdays 
  all.different = (lapply(unique.birthdays, length) == n.pers) # Compute average time all have different birthdays 
  result = 1 - mean(all.different)
return(result)
}
n.pers.param = n.pers
n.sims.param = 1e4 
result_num = birthday.prob(n.pers.param,n.sims.param)
result_num
```

Ok, I am convinced that the probability of atl least two people in this room share the same birthday is `r round( result_num, 2 )`