R Simulation – Monty Hall Problem

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? (taken from Wikipedia)

Shall I change doors or stick to my original choice? That is the question! Let’s write an R Script to run a simulation for this problem…

```
# Monty Hall problem simulation
```
```
 
```
```
print("Monty Hall problem simulation")
```
```
print("-----------------------------")
```
```
 
```
```
# define variables
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```
win.switch.count <- 0
```
```
win.noswitch.count <- 0
```
```
sim.count <- 50000
```
```
switch.count <- 0
```
```
noswitch.count <- 0
```
```
 
```

print(paste("Running", sim.count, "simulations"))

```
 
```
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# run simulation for x times
```
```
for (i in 1:sim.count) {
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    # create doors, a vector 1 to 3, for door 1, door 2, and door 3

```
    doors <- c(1, 2, 3)
```
```
 
```
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    # set price behind a random door
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```
    price.door <- sample(1:3, 1)
```
```
 
```
```
    # the guest picks up a random door
```
```
    guest.door <- sample(1:3, 1)
```
```
 
```

    # monty needs to open a door where there is a goat

```
 
```

    # monty door is not price and is not guest

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    doors[price.door] <- 0
```
```
    doors[guest.door] <- 0
```
```
 
```

    # monty choose the max() door from the resultant vector

```
    monty.door <- max(doors)
```
```
 
```

    # re-insert the price door back in list!!

```
    doors[price.door] <- price.door
```
```
 
```

    # pick a number from 1 to 10 and if it is 5 or less, guest will switch door

```
    if (sample(1:10, 1) < 6) {
```
```
        # guest switch choice
```
```
        doors[monty.door] <- 0
```

        # since we are neutralising the vector and leaving only the only door not

```
        # chosen by the guest or monty
```

        # the sum of doors will be the door left

```
        guest.door <- sum(doors)
```

        switch.count <- switch.count + 1

```
        if (price.door == guest.door) {
```

            win.switch.count <- win.switch.count + 1

```
        }
```
```
    } else {
```
```
        # no switch
```

        noswitch.count <- noswitch.count + 1

```
        if (price.door == guest.door) {
```

            win.noswitch.count <- win.noswitch.count + 1

```
        }
```
```
    }
```
```
}
```
```
 
```
```
# display output
```

probability.switch <- win.switch.count / switch.count

probability.noswitch <- win.noswitch.count / noswitch.count

print(paste("P(switch door) =", probability.switch))

print(paste("P(no switch) =", probability.noswitch))

print(paste("Check P(switch) + P(no switch) =", probability.switch + probability.noswitch))

Answer:

Running the script returns something similar to:

[1] "Monty Hall problem simulation"
[1] "-----------------------------"
[1] "Running 50000 simulations"
[1] "P(switch door) = 0.671743823458863"
[1] "P(no switch) = 0.332786360361803"
[1] "Check P(switch) + P(no switch) = 1.00453018382067"
>

Switching increase the chance to win!

For a mathematical explanation we suggest to watch Numberphile’s video at https://www.youtube.com/watch?v=4Lb-6rxZxx0.

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R Simulation – Monty Hall Problem

Answer: