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)

Monty Hall Problem

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…

  1. # Monty Hall problem simulation
  2.  
  3. print("Monty Hall problem simulation")
  4. print("-----------------------------")
  5.  
  6. # define variables
  7. win.switch.count <- 0
  8. win.noswitch.count <- 0
  9. sim.count <- 50000
  10. switch.count <- 0
  11. noswitch.count <- 0
  12.  
  13. print(paste("Running", sim.count, "simulations"))
  14.  
  15. # run simulation for x times
  16. for (i in 1:sim.count) {
  17.     # create doors, a vector 1 to 3, for door 1, door 2, and door 3
  18.     doors <- c(1, 2, 3)
  19.  
  20.     # set price behind a random door
  21.     price.door <- sample(1:3, 1)
  22.  
  23.     # the guest picks up a random door
  24.     guest.door <- sample(1:3, 1)
  25.  
  26.     # monty needs to open a door where there is a goat
  27.  
  28.     # monty door is not price and is not guest
  29.     doors[price.door] <- 0
  30.     doors[guest.door] <- 0
  31.  
  32.     # monty choose the max() door from the resultant vector
  33.     monty.door <- max(doors)
  34.  
  35.     # re-insert the price door back in list!!
  36.     doors[price.door] <- price.door
  37.  
  38.     # pick a number from 1 to 10 and if it is 5 or less, guest will switch door
  39.     if (sample(1:10, 1) < 6) {
  40.         # guest switch choice
  41.         doors[monty.door] <- 0
  42.         # since we are neutralising the vector and leaving only the only door not 
  43.         # chosen by the guest or monty
  44.         # the sum of doors will be the door left
  45.         guest.door <- sum(doors)
  46.         switch.count <- switch.count + 1
  47.         if (price.door == guest.door) {
  48.             win.switch.count <- win.switch.count + 1
  49.         }
  50.     } else {
  51.         # no switch
  52.         noswitch.count <- noswitch.count + 1
  53.         if (price.door == guest.door) {
  54.             win.noswitch.count <- win.noswitch.count + 1
  55.         }
  56.     }
  57. }
  58.  
  59. # display output
  60. probability.switch <- win.switch.count / switch.count
  61. probability.noswitch <- win.noswitch.count / noswitch.count
  62. print(paste("P(switch door) =", probability.switch))
  63. print(paste("P(no switch) =", probability.noswitch))
  64. 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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