1. Simulate $n$ rolls of a fair dice. Output the relative 
frequency of "6".



 dice <- function(n)
{
        freq.prob.list <- NULL
        six.num <- 0
        for(i in 1:n) {
                if(runif(1) < 1/6)
                        six.num <- six.num + 1
                freq.prob.list <- c(freq.prob.list, six.num/i)
        }
        return(freq.prob.list)
}



Example:
>  dice(1000)[1000]
[1] 0.157

2. Robin and Sheriff aim at the target and hit it with probs 0.8 and 0.7,
respectively. If they shut once, independently of each other, what is 
the probability that the target is hit.

>  0.8 + 0.7 - 0.8*0.7
[1] 0.94


archers2 <- function(p, n)
{
        hit <- 0
        prop <- NULL
        for(i in 1:n) {
                Robin <- 0
                Sheriff <- 0
                if(runif(1) < p[1])
                        Robin <- 1
                if(runif(1) < p[2])
                        Sheriff <- 1
                if(Robin + Sheriff > 0)
                        hit <- hit + 1
                prop <- c(prop, hit/i)
        }
        return(prop)
}

> archers2(c(0.8, 0.7), 1000)[1000]
[1] 0.939

3. Out of 100 coins one has heads on both sides. One coin is 
selected at random and flipped two times. 
What is the probability og 2 heads.

Sol: 1/4 99/100 + 1 * 1/100 = 103/400 = 0.2575.

twoheaded <- function(n)
{
        proportion <- NULL
        twoheads <- 0
        for(i in 1:n) {
                coin <- "GOOD"
                if(runif(1) < 0.01)
                        coin <- "BAD"
                if(coin == "GOOD" && runif(1) < 0.25)
                        twoheads <- twoheads + 1
                proportion <- c(proportion, twoheads/i)
        }
        return(proportion)
}

>  twoheaded(1000)[1000]
[1] 0.25


4. In the group of N people, what is the prob that
there is a at least one match in their birthdays.
[no leap years involved]

Exact: 
exabirth_function(n)
 {
 prob_1
 for ( i in 1: n)
  prob_prob * (365-i+1)/365
  1-prob
}
> exabirth(10)
[1] 0.1169482
> exabirth(23)
[1] 0.5072972
> exabirth(100)
[1] 0.9999997

Simulation:

matchbirth <- function(k, people=23)
{ 
  mat_0
  for(j in 1:k){
         a_sample(seq(1,365), people, replace=T)
         aa_sort(a)
         for(i in 1:(length(aa)-1))
                    {if (aa[i]==aa[i+1])  {mat_mat+1
                                              break}
                    }
                }
 cat( mat/k, "\n")
}

> matchbirth(100, 23)
0.46