## 5.6 Playing the roulette

Consider a European-style roulette which includes numbers between 0 and 36. In roulette there are many different betting alternatives but we consider here the simplest case where we bet on a single number. If we pick the right number then we win 35 times what we bet.

Consider the following scenario. We start playing roulette with a fixed `budget`

. Every sping of the roulette costs one unit of `budget`

. We play until we run out of budget and we always bet on the same number. The following function `roulette`

implements this game.

```
<- function(budget, number){
roulette <- c()
current while(budget > 0){
<- sample(0:36,1)
outcome if(outcome == number) {
<- budget + 35
budget
}else {budget <- budget -1}
<- c(current,budget)
current
}
current }
```

It takes two inputs: the `budget`

and the number we decided to play all the time. It outputs our budget throughout the whole game until it ends.

Let’s play one game with a budget of 15 and betting on the number 8.

```
set.seed(2021)
roulette(15,8)
```

`## [1] 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0`

In 15 spins the number 8 never comes up and therefore our game ends quite quickly.

We can ask ourselves many questions about such a game. For instance:

What is the probability that the game last exactly 15 spins if I start with a budget of 15 and bet on the number 8?

What is the average length of the game starting with a budget of 15 and betting on the number 8?

What is the average maximum wealth I have during a game started with a budget of 15 and betting on the number 8?

We will develop various Monte Carlo experiments to answer all the above questions. In each case, we simply need to modify the function `roulette`

to output some summary about the game.

### 5.6.1 Question 1

- What is the probability that the game last exactly 15 spins if I start with a budget of 15 and bet on the number 8?

If a game where I started with a budget of 15 ends after 15 spins, it means that the number 8 never showed up. We can adapt the function roulette to output `TRUE`

if the length of the vector `wealth`

is exactly equal to `budget`

.

```
<- function(budget, number){
roulette1 <- c()
wealth <- budget
current while(current > 0){
<- sample(0:36,1)
outcome if(outcome == number) {
<- current + 35
current
}else {current <- current -1}
<- c(wealth,current)
wealth
}if(length(wealth) == budget){return(TRUE)} else{return(FALSE)}
}
```

Therefore for the previous example the function should output `TRUE`

.

```
set.seed(2021)
roulette1(15,8)
```

`## [1] TRUE`

Let’s replicate the experiment 1000 times. The proportion of `TRUE`

we observe is our estimate of this probability.

```
set.seed(2021)
<- replicate(1000,roulette1(15,8))
experiment sum(experiment)/1000
```

`## [1] 0.657`

Notice that actually we could have also computed this probability exactly. This is the probability that a Binomial random variable with parameter \(n=15\) (15 spins of the roulette) and \(\theta = 1/37\) (the number eight has probability 1/37 of appearing in a single spin) is equal to zero (no 8 can happen). This is equal to:

`dbinom(0,15,1/37)`

`## [1] 0.6629971`

Therefore the Monte Carlo experiment approximates well the probability.

### 5.6.2 Question 2

- What is the average length of the game starting with a budget of 15 and betting on the number 8?

We can answer this question by adapting the `roulette`

function to output the length of the vector `wealth`

.

```
<- function(budget, number){
roulette2 <- c()
wealth <- budget
current while(current > 0){
<- sample(0:36,1)
outcome if(outcome == number) {
<- current + 35
current
}else {current <- current -1}
<- c(wealth,current)
wealth
}length(wealth)
}
```

In our example, the output should be 15.

```
set.seed(2021)
roulette2(15,8)
```

`## [1] 15`

Let’s replicate the experiment 1000 times and summarize the results with a plot (it may take some time to run the code!).

```
set.seed(2021)
<- replicate(1000,roulette2(15,8))
experiment plot(table(experiment))
```

We can see that the distribution is very skewed, most often the length is 15 spins, but then sometimes the game has a length which is much longer. Therefore, the median of the data is a much better option to summarize the average length of the game. We can compute it, together with a range of plausible values, as

`median(experiment)`

`## [1] 15`

`c(sort(experiment)[25],sort(experiment)[975])`

`## [1] 15 1923`

So we can see that the median is indeed 15: most of the times the number 8 does not appear and therefore the game ends in 15 spins.

### 5.6.3 Question 3

- What is the average maximum wealth I have during a game started with a budget of 15 and betting on the number 8?

We can answer this question by adapting the `roulette`

function to output the maximum of the vector `wealth`

.

```
<- function(budget, number){
roulette3 <- c()
wealth <- budget
current while(current > 0){
<- sample(0:36,1)
outcome if(outcome == number) {
<- current + 35
current
}else {current <- current -1}
<- c(wealth,current)
wealth
}max(wealth)
}
```

Let’s again replicate the experiment and plot the result.

```
set.seed(2021)
<- replicate(1000,roulette3(15,8))
experiment plot(table(experiment))
```

We can then get an estimate of the maximum wealth as well as a range of plausible values as

`median(experiment)`

`## [1] 14`

`c(sort(experiment)[25],sort(experiment)[975])`

`## [1] 14 331`

Since most often the game ends with no 8 appearing, the maximum wealth is most often 14.