Likelihood_Homework_2_Rosskopf_Golombek.Rmd

---
title: "Likelihood"
author: "Nina Golombek und Martina Rosskopf"
date: "19 11 2018"
output: pdf_document
---
## Task:

Tsunami scientist Trevor rolls some dice and tells you that the added result is 17. 
Think of an example that applies this problem to natural hazards!

The exercise for this week was to write some R code to find out how many times someone would have to roll a dice to get 17 as
a result of al dices.

First we discussed the different possibilties, where the conclusion was, that three is the minimum amount of dices you to do
for a sum of 17 (6+6+5). The maximum amount would be 17 times (each dice roll with 1).

Now to figure out the likelihood distribution of the dice sum 17, we simulated an experiment, where we use the possible 
amounts of dice rolls between 3 and 17.  


```{r}

sum(sample(1:6, 3, replace = TRUE))   # The sample function includes only whole numbers 
sum(sample(1:6, 4, replace = TRUE))   # Here: between 1-6 for a set number of dices (between 3-17)
sum(sample(1:6, 5, replace = TRUE))
sum(sample(1:6, 6, replace = TRUE))
sum(sample(1:6, 7, replace = TRUE))
sum(sample(1:6, 8, replace = TRUE))
sum(sample(1:6, 9, replace = TRUE))
sum(sample(1:6, 10, replace = TRUE))
sum(sample(1:6, 11, replace = TRUE))
sum(sample(1:6, 12, replace = TRUE))
sum(sample(1:6, 13, replace = TRUE))
sum(sample(1:6, 14, replace = TRUE))
sum(sample(1:6, 15, replace = TRUE))
sum(sample(1:6, 16, replace = TRUE))
sum(sample(1:6, 17, replace = TRUE))

```

### Note:

The statistical expectation for each dice is 3.5
Thus, in order to reach ~17, five dices are most likely (5* 3.5 = 17.5 --> closest to 17)

## Possible solution

For the following solution we set a number n, which sets the amount of experiments we do for each possible dice roll (from 
3 to 17).
In the resolution vector *resvec* we write the solution of how many times the sum was 17 for each roll. In the row *0* we 
have the result for *dice roll=3* and in row *15* the result for *dice roll=15*.

```{r}
n <- 10000
resvec <- rep(0,15)

```

In the following loops we go through the possiblies of dice rolls. First we create a vector with n times i rows, where i 
is the amount of dices. The random possible values of the vector are between 1 and 6. This vector we then write into a 
matrix which has i columns and n rows. 
We then build the sums over each row of the matrix. If the sum is then 17 we add one to the specific row of the *resvec*. 
```{r}
# loop to go through possible numbers of dices
for (i in 3:17) {
  # getting a vector of random numbers between 1 to 6
  vec <- sample(1:6, n * i, replace = TRUE)
  # transform vector to a matrix, amount of coloums equal to number of dices i
  mat <- matrix(vec, ncol = i)
  # loop to go through all rows of matrix
  for (j in 1:n){
    # if sum of a row is 17 then put the counter +1
    if (sum(mat[j,]) == 17){
      resvec[i-2] <- resvec[i-2]+1
    }
  }
}
```
The last loop can also be written in one line. With this line we first get a vector with the sums of each row, where we 
then ask if the sums of the rows are 17. The vector will then consist of TRUE and FALSE values. Calculating the sum over 
this vector gives us how many times the sum was 17, because TRUE is 1 and FALSE is 0. This result will then be written 
into our *resvec*.
```{r}
# last loop can also be calculated by: 
  resvec[i-2] <- sum(rowSums(mat) == 17)
```
Last we are plotting a graph out of our results. Here x gives us the number of dices (between 3 to 17) and y how many 
runs we had, where the sum was 17. 
```{r}
## Plot
x <- 3:17      # set x values
plot(x,resvec, type = "h", col="chocolate", lwd=7,
     xlim=c(3,17),
     main = "Likelihood of dices",                        # Diagram Title
     xlab = "Number of dice rolls",                       # X-Axis Title
     ylab = 'Amount of runs with sum = 17',               # Y-Axis Title
     cex.lab = 1.5, cex.main = 1.5)                       # Font size of Titles

```

With this Plot we can see that it is most likely to get the eye sum 17 if you roll the dice 5 times. This fits to the 
known value, that for one dice the assumed eyes of the dice are 3.5. Therefore 5 dice rolls would give us 17.5 for the 
sum of the dices. 

## Example that applies this problem to natural hazards

Probability of a 1000-year Event in 6 different countries 

1: Germany
2: USA
3: Indonesia
4: Japan
5: Argentina
6: New Zealand

--> Probability for one of these events is always 1/1000 for each country.