I was chatting with some kids at the weekend, and they were asking each other riddles. However these were more interesting than the usual puns and trick questions. Rather they were logic puzzles - "This statement is false" type of thing. Interestingly the kids were figuring them out faster than some of the adults...

Which got me thinking about the Monty Hall Problem. You may or may not know that Monty Hall was an American game show host, and while he may be pretty famous for being on TV, he is immortalised in the maths puzzle named after him. Mythbusters studied it, and its even been explain in movies.

*You have three doors. Behind 1 is a car and the other two are goats. Pick a door and the prize is yours... but wait... Before you open the door let me show you another door. Behind this other door is a goat. Maybe there's a goat behind your door too! It's not too late to change your mind? Stick with your door, or switch to the final unopened door?*

*Stick you say? Thought you might. Most people do. So lets open then door and see what you won!*

Now of course there's an element of luck in here - you might have picked correctly, or you could be wrong. The interesting question is should you have switched?

Well let's write a program to find out!

Let's start by setting up some variables:

make doorCount number 3

make switch boolean no

make prizes list of strings

make open list of boolean

make pickedDoor number

We've got a couple of lists - prizes which tells us what's behind each door and open which tells us if the door is open yet. PickedDoor is going to be the number of our selection. I've also set up a couple of constants - we have 3 doors, and we're

*not*going to switch.
Now lets bring out some prizes:

when hideCar

.delete all of prizes

.delete all of open

.repeat doorCount

..add "goat" to prizes

..add no to open

.replace item pick random 1 to doorCount of prizes with "car"

when showDoors

.make counter number

.repeat length of prizes using counter

..if item counter of open

...say join "Door" join [counter] join ":" item counter of prizes

..else

...say join "Door" join [counter] join ":" "Closed"

.say ""

hideCar, clears out the lists (just in case), puts a goat behind every door, and sets the door to closed. It then picks a random door and replaces the goat with a car. Next we have showDoors which goes through the doors and either prints out that the door is closed or (if its open) then it prints the prize that was behind it.

Now we can write a start script to begin putting these together.

when start

.broadcast hideCar and wait

.set pickedDoor to pick random 1 to doorCount

.say join "you pick door number " [ pickedDoor ]

.replace item pickedDoor of open with yes

.broadcast showDoors and wait

In this simplified version we hide the car, pick a random door, then open that door and show the results. It's pretty obvious that with three doors we should win about 1/3 of the time.

Now lets start opening some doors.

when openRandom

.make counter number

.forever

..set counter to pick random 1 to doorCount

..if not counter=pickedDoor and not item counter of open

...if item counter of prizes="goat"

....replace item counter of open with yes

....

....say join "Behind door " join [counter] join " is a " item counter of prizes

....stop script

This script tries to open a random door. Except its not totally random. The gameshow host never opens the door you picked, or the door with the car - he always shows an unpicked door with a goat. Therefore, having picked a random door we check that its not the one picked, its not already open, and it does have a goat behind it. Only when all those conditions are true can the door be opened to reveal a goat.

You might note that this code is a little more complex than it needs to be. If we only open one random door then we don't actually need to check that the door is closed (they all are), and when we print out the prize behind the door its

*always*a goat. However its worth writing the code this way as it actually represents what's going on in the real world, and it provides a double check that what we're going is correct.
The final piece of code we need is to switch doors:

when switchDoors

.make counter number

.forever

..set counter to pick random 1 to doorCount

..if not counter = pickedDoor and not item counter of open

...set pickedDoor to counter

...stop script

To switch, we pick a random door, and check that its different to our current pick, and that its currently closed. Again this is more complex that we need, but it accurately represents the puzzle. It means it will keep working if we start adding extra doors or playing with the other parameters.

when start

.say "New Game:"

.broadcast hideCar and wait

.set pickedDoor to pick random 1 to doorCount

.say join "you pick door number " [ pickedDoor ]

.say ""

.broadcast openRandom and wait

.say ""

.broadcast showDoors and wait

.if switch

..broadcast switchDoors and wait

..say join "You switch to door " [pickedDoor]

.else

..say join "You Stick with door " [pickedDoor]

.say join "You won a " item pickedDoor of prizes

Here's the whole game put together. Remember switch is set to no, so we're never switching, and when I run it I get:

you pick door number 2

Behind door 1 is a goat

Door1:goat

Door2:Closed

Door3:Closed

You stick with door 2

You won a car

Yay for me!!! I win a car!

But wait - maybe I was just lucky. What if we played 100 games:

make score number

when start

.repeat 100

..say "New Game:"

..broadcast hideCar and wait

..set pickedDoor to pick random 1 to doorCount

..say join "you pick door number " [ pickedDoor ]

..say ""

..repeat doorCount-2

...broadcast openRandom and wait

..say ""

..broadcast showDoors and wait

..if switch

...broadcast switchDoors and wait

...say join "You switch to door " [pickedDoor]

..else

...say join "You Stick with door " [pickedDoor]

..say join "You won a " item pickedDoor of prizes

..say ""

..say ""

..if item pickedDoor of prizes ="car"

...change score by 1

.say join "You won " join [ score ] " times"

You won 33 times

That's not so good. I won 33 times out of 100, which is exactly(ish) 1/3 of the time. I guess with three doors that's about as good as it gets.

Or maybe not - lets set switch to

*yes.*

You won 65 times

Wow! Thats a lot more interesting. Now we're winning about 2/3 of the time! But how come? We still get to take home one prize from three, and all three doors have an equal change of being a good prize, so why should we be better at picking the wrong door first time and the right door second?

Well actually you just summed it up... On your first pick you're probably going to pick the wrong door because there are three doors - you get the right one 1/3 and wrong one 2/3. If you stick then that's what you end up with - probably wrong, but it you switch then being wrong the first time means you'll definitely be

*right*the second.
The way I like to think about it is to image there are 100 doors, and I open 98 of them (in fact I've already tweaked the code so the number of doors opened is the total-2).

make doorCount number 100

make switch boolean no

What happens when we run this:

New Game:

you pick door number 23

Behind door 30 is a goat

Behind door 73 is a goat

Behind door 42 is a goat

Behind door 74 is a goat

Behind door 47 is a goat

*lots of goats...*

Behind door 21 is a goat

Behind door 19 is a goat

Behind door 10 is a goat

Behind door 4 is a goat

Behind door 69 is a goat

Behind door 72 is a goat

Behind door 50 is a goat

Behind door 44 is a goat

Door1:goat

Door2:goat

Door3:goat

Door4:goat

Door5:goat

Door6:goat

Door7:Closed

Door8:goat

Door9:goat

Door10:goat

Door11:goat

Door12:goat

*lots of goats...*

Door22:goat

Door23:Closed

Door24:goat

*lots of goats...*

Door96:goat

Door97:goat

Door98:goat

Door99:goat

Door100:goat

you stick with door 23

you won a goat

Think about it - what was the chances that the goat was behind door 23? Well 1 in 100, so its pretty obvious we're probably not going to win. Now look at all those doors I opened for you - do you really think I can open 98 doors and accidentally only find goats? every time? I know something you don't - where the car is! and I

*mysteriously*avoided door number 7. As the number of doors goes up it becomes pretty obvious that switching is a good deal.
Yet it turns out most people don't switch. Even mathematicians who should know better fall for it. Unless you've seen the trick before you'll probably stick, because there's doesn't seem to be an obvious reason to switch, and people don't like to change their mind. I picked door 23, and its as good as any other door, so I'll stay with 23... and probably take home a goat.

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