Tag Archives: beginner

The Monty Hall problem

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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

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InDesign scripting : lesson 36

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throughout this site (and others) you’ll find lots of applescripts which are direct responses to real-world workflows. but, chances are, you’ll need to do some cut-pasting, and additional code writing to create a script which does exactly what you want in your workflow.

this can be a bit intimidating and not a little frustrating as you try to get the bloody thing to work.

so here’s a post (and there may be more to follow) which gets back to basics and addresses a common beginner question :

When I’m cutting and pasting sections of a script, how many ‘end tells’ do I need to include?

new scripters will often be confronted with errors like these when trying to compile or run a script in development :
end error 01
end error 02

one of the main problems with this kind of error, as compared with many other types of errors, is that Applescript Editor may not highlight the exact line where the problem needs to be fixed.

the important thing to remember is that the number of ‘end tells’ will be exactly the same as the number of ‘tells’.

when writing code from scratch, it’s best practice to write the ‘tell’ and ‘end tell’ first — then go back and fill in the rest of the code :

tell application id "com.adobe.InDesign"
-- rest of code here
end tell

you can have tells within tells (or nested tells) — these get closed off in the reverse order that they were opened — but notice, still, the number of ‘end tells’ is the same as the number of ‘tells’ :

tell application id "com.adobe.InDesign"
  -- code here addresses the application
  tell active document
    -- code here addresses the active document
    tell layout window 1
      -- code here addresses the layout window
    end tell
    -- code here addresses the active document
  end tell
  -- code here addresses the application
end tell



other commands
this is also the case with other commands that need an ‘end’ statement. for example, if you have an ‘if’ command within a ‘tell’ command, the ‘if’ needs to end before the ‘tell’ ends :

tell application id "com.adobe.InDesign"
  if x = y then
    -- some more code
  end if
end tell

when writing code you can also close off with just ‘end’ and the editor will work out which end goes with which command when the script is compiled. so you can write it out like this :

tell application id "com.adobe.InDesign"
tell active document
if x = y then
repeat 5 times
-- some more code
end
end
end
end

… and, when compiled, it will become this :

tell application id "com.adobe.InDesign"
  tell active document
    if x = y then
      repeat 5 times
        -- some more code
      end repeat
    end if
  end tell
end tell

but you can see that even here — all the commands get closed down in the reverse order that they were opened. the ‘repeat’ is closed first, then the ‘if’, then the second ‘tell’ and lastly the first ‘tell’.



… and another thing
whenever possible, it’s best to avoid one application ‘telling’ another application :

tell application id "com.adobe.InDesign"
  set mgFolder to file path of active document
  tell application "Finder"
    set mgOLDFolder to (mgJobFolder & "OLD:" as string) as alias
  end tell
end tell

… is better writen as :

tell application id "com.adobe.InDesign"
  set mgFolder to file path of active document
end tell
tell application "Finder"
  set mgOLDFolder to (mgJobFolder & "OLD:" as string) as alias
end tell

… where we close the InDesign tell before opening the Finder tell



are there other scripting issues that keep cropping up for you?

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