**The Monty Hall problem** is based on a TV game show where you, as the contestant, are presented with three doors. Behind one door is an awesome prize. You just have to guess which door has the prize behind it.

After you make your choice (C), the host reveals what is behind one of the other doors (it’s never the awesome prize). Then you are given the option to stay with your original decision, or change to the other remaining door.

**The problem is this:** Should you change your mind and choose the other door, or should you stay with your original choice?

NOW, there’s a mathematical proof that you should definitely change doors. And I wanted to set up an Applescript to demonstrate that this is so.

You COULD create a script which let’s you emulate the game. The below script let’s you choose a door, and then change your mind (or not) and gives you a readout of your progress (wins vs losses).

And that can be a bit of fun … for about 2 minutes … tops.

set TheDoors to {"Door 1", "Door 2", "Door 3"} set TheWin to 0 set TheLoss to 0 my mgPlay(TheDoors, TheWin, TheLoss) on mgPlay(TheDoors, TheWin, TheLoss) set WinningDoor to some item in TheDoors set ChosenDoor to (choose from list TheDoors with title "Monty Hall" with prompt "Which door do you choose?") if result is false then error number -128 end if set ClosedDoors to {WinningDoor, (item 1 of ChosenDoor)} set OpenDoors to {} repeat with ThisDoor in TheDoors if ThisDoor is not in ClosedDoors then set end of OpenDoors to (ThisDoor as string) end if end repeat set OpenDoor to some item in OpenDoors set NewDoors to {} repeat with ThisDoor in TheDoors if (ThisDoor as string) is not OpenDoor then set end of NewDoors to (ThisDoor as string) end if end repeat set TheString to "You chose " & ChosenDoor & return & return & "Now Monty has opened " & OpenDoor & return & return & "You may now choose to change your mind or stay with " & ChosenDoor & return & return set FinalDoor to (choose from list NewDoors with title "Monty Hall – Second Chance" with prompt TheString) set FinalDoor to item 1 of FinalDoor if FinalDoor = (ChosenDoor as string) then set TheString to "You chose to stay with " & ChosenDoor & return & return else set TheString to "You chose to change from " & ChosenDoor & " to " & FinalDoor & return & return end if if FinalDoor = WinningDoor then set TheWin to TheWin + 1 set TheString to TheString & "Congratulations. " & FinalDoor & " is correct" & return & "You won this round" & return & return & "You have won " & TheWin & " and lost " & TheLoss else set TheLoss to TheLoss + 1 set TheString to TheString & "Sorry. " & FinalDoor & " is not correct" & return & "THE WINNING DOOR WAS " & WinningDoor & return & "You didn't win this round" & return & return & "You have won " & TheWin & " and lost " & TheLoss end if set ThisChoice to display dialog TheString buttons {"STOP PLAYING", "PLAY AGAIN"} set ThisButton to button returned of ThisChoice if ThisButton is "PLAY AGAIN" then my mgPlay(TheDoors, TheWin, TheLoss) else display dialog "Thanks for playing" buttons {"OK"} end if end mgPlay

But to see a definite pattern in the results, we need to run hundreds of games. And the process above is just too tedious. If we could automate the entire process, we could run 10,000 games in about a second.

First (see below) we set up four counters to keep track of our success rate. The next two lines are two lists to represent the contestant’s possible choices. The next two lines access the first list to randomly select a *Winning Door* and the contestant’s *Chosen Door*. This is what * some item* means. It means choose a random entry from any given list. The last line puts those results into a list of doors which must remain closed for the second part of the game. This list might contain only one door if the

*Winning Door*is the same as the

*Chosen Door*.

set NoChangeWin to 0 set NoChangeLose to 0 set YesChangeWin to 0 set YesChangeLose to 0 set TheDoors to {"Door 1", "Door 2", "Door 3"} set TheChoices to {"Change Choice", "Don't Change Choice"} set WinningDoor to some item in TheDoors set ChosenDoor to some item in TheDoors set ClosedDoors to {WinningDoor, ChosenDoor}

The next section chooses a door to open. Remember, this cannot be the *Winning Door*, nor can it be the contestant’s *Chosen Door*. So first we put all the possible doors to open into a list (this list could contain either one or two doors), then choose a random door from that list to open.

Then the next section puts the two remaining closed doors into a list for the contestant to choose from again.

set OpenDoors to {} repeat with ThisDoor in TheDoors if ThisDoor is not in ClosedDoors then set end of OpenDoors to (ThisDoor as string) end if end repeat set OpenDoor to some item in OpenDoors set NewDoors to {} repeat with ThisDoor in TheDoors if (ThisDoor as string) is not OpenDoor then set end of NewDoors to (ThisDoor as string) end if end repeat

The next section is just a little more complicated. First we get a random choice from that second list from the start of the script. This represents whether or not the contestant changes their mind about which door they want. Then we use some nested ** if** statements to determine the results.

Firstly, if the choice is to NOT change, then we set the *Final Door* to be the same as the original *Chosen Door*. Then we check to see if the *Final Door* is the same as the *Winning Door* randomly selected at the start of the script. If it is, we add 1 to the NoChangeWin counter. If it’s not the same, we add 1 to the NoChangeLose counter.

The second part below – the ** else** statement – is for when the contestant DOES choose to change doors. We use a quick repeat loop to select the other door from the list of two closed doors to be our

*Final Door*. Then run the same analysis as above. If the

*Final Door*is the same as the

*Winning Door*, we add 1 to the YesChangeWin counter. If it’s not the same, we add 1 to the YesChangeLose counter.

set TheChoice to some item in TheChoices if TheChoice is "Don't Change Door Choice" then set FinalDoor to ChosenDoor if WinningDoor = FinalDoor then set NoChangeWin to NoChangeWin + 1 else set NoChangeLose to NoChangeLose + 1 end if else repeat with ThisDoor in NewDoors if (ThisDoor as string) is not ChosenDoor then set FinalDoor to (ThisDoor as string) end if end repeat if WinningDoor = FinalDoor then set YesChangeWin to YesChangeWin + 1 else set YesChangeLose to YesChangeLose + 1 end if end if

The final part is our reporting. This simply creates four lines of text with the various outcomes (win after not changing, lose after not changing, win after changing, lose after changing)

return "Contestant wins after NOT changing : " & NoChangeWin & return & "Contestant loses after NOT changing : " & NoChangeLose & return & return & "Contestant wins after changing : " & YesChangeWin & return & "Contestant loses after changing : " & YesChangeLose

Now all we need to do is put the whole lot together and include a repeat loop to generate as many games as we please. I found that 1,000,000 games takes about 27 seconds to run. But that’s a bit ridiculous because you don’t really need more than 10,000 games to show the pattern.

Here is the complete script with commenting included. Plus, below that, a sample of the output for 10,000 games.

--these are counters to track wins/losses set NoChangeWin to 0 set NoChangeLose to 0 set YesChangeWin to 0 set YesChangeLose to 0 -- this is the list of possible doors to choose at the start set TheDoors to {"Door 1", "Door 2", "Door 3"} -- this is the list of possible contestant choices at the end set TheChoices to {"Change Choice", "Don't Change Choice"} repeat 10000 times -- randomly select a winning door set WinningDoor to some item in TheDoors -- randomly select the contestant's chosen door set ChosenDoor to some item in TheDoors -- put these into a list (they might be the same door) set ClosedDoors to {WinningDoor, ChosenDoor} -- randomly select a door to open -- it can't be the winning door, nor the contestant's chosen door set OpenDoors to {} repeat with ThisDoor in TheDoors if ThisDoor is not in ClosedDoors then set end of OpenDoors to (ThisDoor as string) end if end repeat set OpenDoor to some item in OpenDoors -- set the NEW list of possible doors to choose -- this is a list of two doors (now that one door is open) set NewDoors to {} repeat with ThisDoor in TheDoors if (ThisDoor as string) is not OpenDoor then set end of NewDoors to (ThisDoor as string) end if end repeat -- randomly select if contestant will change their choice or not change -- and count how often contestant wins vs loses set TheChoice to some item in TheChoices if TheChoice is "Don't Change Choice" then set FinalDoor to ChosenDoor if WinningDoor = FinalDoor then set NoChangeWin to NoChangeWin + 1 else set NoChangeLose to NoChangeLose + 1 end if else repeat with ThisDoor in NewDoors if (ThisDoor as string) is not ChosenDoor then set FinalDoor to (ThisDoor as string) end if end repeat if WinningDoor = FinalDoor then set YesChangeWin to YesChangeWin + 1 else set YesChangeLose to YesChangeLose + 1 end if end if end repeat return "Contestant wins after NOT changing : " & NoChangeWin & return & "Contestant loses after NOT changing : " & NoChangeLose & return & return & "Contestant wins after changing : " & YesChangeWin & return & "Contestant loses after changing : " & YesChangeLose

Contestant wins after NOT changing : 1663 Contestant loses after NOT changing : 3353 Contestant wins after changing : 3330 Contestant loses after changing : 1654

If you run this script for 10,000 games over and over again you’ll see very similar results. That is, you will win the Monty Hall game approximately twice as often if you change your choice when presented with the two doors.

Now this may seem counter-intuitive (it did for me) because, *surely*, if you’re presented with only two doors, you’ve got a fifty-fifty chance of picking the right door. I mean, there’s just two doors, so it shouldn’t really matter which one you choose … right?

Well, yes, that *would* be true if you had not already been presented with the three doors.

**Here’s how it works …**

You’re presented with three doors and you choose one. Let’s just say you choose the first door.

If you choose the first door, all the possible scenarios for the winning door look like this (below). And the same is the case if you choose the middle door or the last door – no matter which door you choose, there are three possibilities for the winning door.

Then Monty Hall opens one of the doors. In the first scenario, there’s a choice of two doors to open. In the other two, there’s only one door that could be opened.

But the kicker is this … you don’t know which of the three scenarios is in play. But looking at this diagram it should be obvious that out of the three possible scenarios, only one sees you winning if you keep the same door. Whereas there are two possible scenarios where you will win if you change.

Of course, this does not guarantee you a win if you change doors. But, on average, you’ll win twice as often if you change.

Keep grunting