Just for fun and to train R, I tried to proof the Monty Hall Game rule (changing your choice after one gate opened gives you more probability to win), I made this reproducible code (The explanation of every step is within the code):
## First I set the seed
set.seed(4)
## Then I modelize the presence of the prize as a random variable between gates 1,2,3
randomgates <- ceiling(runif(10000, min = 0, max = 3))
## so do I with the random choice.
randomchoice <- ceiling(runif(10000, min = 0, max = 3))
## As the opening of a gate is dependent from the gate you chose (the gate you chose cannot be opened)
## I modelize the opening of the gate as a variable which cannot be equal to the choice.
options <- c(1:3)
randomopen <- rep(1,10000)
for (i in 1:length(randomgates)) {
realoptions <- options[options != randomchoice[i]]
randomopen[i] <- realoptions[ceiling(runif(1,min = 0, max = 2))]
}
##Just to make data more easy to handle, I make a dataset
dataset <- cbind(randomgates, randomchoice, randomopen)
## Then I creat a dataset which only keeps the realization of the games in which we carry on (
## the opened gate wasn't the one with the price within)
steptwo <- dataset[randomopen != randomgates,]
## The next step is just to check if the probability of carry on is 2/3, which indeed is
carryon <- randomopen != randomgates
sum(carryon)/length(randomgates)
## I format the dataset as a data frame
steptwo <- as.data.frame(steptwo)
## Now we check what happens if we hold our initial choice when game carries on
prizesholding <- steptwo$randomgates == steptwo$randomchoice
sum(prizesholding)
## creating a vector of changing option, dependant on the opened gate, in the dataset that
## keeps only the cases in which we carried on playing (the opened gate wasn't the one with the prize)
switchedchoice <- rep(1,length(steptwo$randomgates))
for (i in 1:length(steptwo$randomgates)) {
choice <- options[options != steptwo$randomchoice[i]]
switchedchoice[i] <- choice[ceiling(runif(1,min = 0, max = 2))]
}
## Now we check how many times you guess the prize gate when you switch your initial choice
prizesswitching <- steptwo$randomgates == switchedchoice
sum(prizesswitching)/length(steptwo$randomgates)
When I check the probability without changing my initial choice in the cases in which the game carried on (the gate opening didn't match the one with the prize) I obtain what I exepected (close 1/3 of probability of winning the prize), which refers to the following instruction:
carryon <- randomopen != randomgates
sum(carryon)/length(randomgates)
My problem arises when I check the probability of winning the prize after changing my choice (conditionate, obviously to not having opened the door which holds the prize), instead of getting 1/2 as Monty Hall states, I get 1/4, it refers to the following instruction:
prizesswitching <- steptwo$randomgates == switchedchoice
sum(prizesswitching)/length(steptwo$randomgates)
I know that I am doing something bad because it is already more than proofed that Monty Hall holds, but I am not able to detect the flaw. Does anyone know what it is?
If you don't know what Monty Hall problem is, you can find easy-to-read information at wikipedia: