The Monty Haul Problem(s): Debunking Distortion

From Wikipedia, the Monty Haul problem is pretty much this:

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?

A lot of people will now say they’ve seen the light on it – they’ll say it IS to your advantage to switch doors. That they were adamant it was 50/50 before, but it’s not!

Here is a revised consideration for that debunking, developed onward from the previous post found under the tilde.

Let’s establish some commitments first – is punching Monty in the face and opening every door until you find the car then driving off in it GTA style a legitimate solution to the Monty Hall problem?

I’m guessing most would say no, since it essentially breaking the rules of the game.

How about this – what if he presented you with two doors, showed you what’s behind them but then covered your eyes and closed and shuffled the doors. BUT you noticed a scratch on the door frame of the door with the car. Monty asks which of the two doors you will choose. He shows you again and dang, does that scratch just stand out to you!  Is that really a solution to the game?

I think people in general might vary on this one. But I think most mathematicians would probably say no, it’s not a solution. It’s a get around. Perhaps even a cheat. It’s breaking the rules of the game.

What about knowledge bleed over? Say it’s proposed to you to play a single player game of memory (trying to finish as quickly as possible), you complete the game then someone proposes a new game of single player memory ‘memory’, but all they do is flip all the cards on the table face down again. For a start, this isn’t how to play memory, it requires shuffling. And secondly you’ve seen all the cards – you are going to have knowledge bleed over from the previous game. You are neither playing memory plus you have bleed over from the previous game.

Is bleed over a legitimate solution? Again people might vary. But I think many mathematicians would probably say no, it’s not a solution. It’s a get around. Perhaps even a cheat.

Further how much are you going to be tempted to use that knowledge if it potentially means winning a new car? How biased towards using it are you going to be?

Indeed, how biased will you be towards saying the second game was a perfectly fine game of memory? Because you could use bleed over to your advantage in winning a car?

So here’s the issue – In the original game Monty asks you to pick a door. Then he kind of ignores what you’d pick as he doesn’t open that and he instead removes another door. He then asks which of the remaining doors you’d pick.

The question is, did Monty just offer you a new game?

Taking it he’s run a new game, it is exactly the same as the single player memory game issue – he’s run one game with you and now you have bleed over from the first game into the second game.

But the original debunking of the 50/50 relies on the knowledge from the first game to be valid for use in the second.

Does the original debunking actually prove there is only one game?

~~~

One of the main arguments I heard for it uses an exaggeration of the problem to highlight the ‘solution’. I think it does the opposite, so I’ll use it myself: Instead of three doors, imagine there are ten! Now there are two choices – the player’s original choice and the second choice Monty provides. If there are ten doors, Monty is going to take away 8 doors, leaving us with two doors.

The thinking goes that the right door is out there, amongst those other 9 doors that the first choice doesn’t cover. And for the second choice Monty has just taken away 8 bad doors. The second option has a 9 in 10 chance of having the winning door! It has to make sense to swap to the second choice!

While some will note that sure, you’re original choice had a 1 in 10 chance of being right. But that’s small odds, right? We can forget about that, right?

Indeed, forget about it so much we do bad math?

Let’s slow down the door removal instead of doing it all at once. Now The first choice has a 1 in 10 chance of being right.

Does that mean the second choice has 9 doors sort of assigned to it?

Well no, the second choice has a 1 in 10 chance by default as well. Monty doesn’t control if you guessed the right door to begin with.

BUT surely that first choice is only a 1 in 10 chance – let’s forget about it like we forget about our chance at winning the state lottery. Dismissible odds (bias).

Here’s part of the problem with the original debunking – there’s a cognitive predilection to assign the second choice a 9 out of 10 chance of having the right answer. An urge to apply a symmetry there. After all the second choice covers 9 of the other doors, right?

If you slow down the removal of the other doors you see that it doesn’t work that way.

If we remove one dud door from the second choice, we now have 9 doors. So does that increase the odds for the second choice?

Remove another and we have 8. So does that increase the odds for the second choice?

The problem here is a fixation on the second choice.

When you remove a door, the odds of the first choice being right go up, from 1 in 10 to 1 in 9.

When you remove another door, the odds of the first choice being right go from 1 in 9 to 1 in 8.

Lets go straight down to 3 doors.

The first choice has a 1 in 3 chance.

The second choice?

We remove a door. Now the first choice has a 1 in 2 chance.

And the second choice a 1 in 2 chance.

50/50

It’s because you’d done bad math and not increase the odds of the first choice as you redid the odds of the second choice. The first choice appeared to be dismissible odds – so they fell out of your work tray when it came to doing the math. You had such low odds of getting the right answer and Monty was removing all the doors but the winner (IF you hadn’t chosen the right choice already) that you automatically ignored looking at your chance of winning a second time. You didn’t increase it as you should because you’d dismissed it from the equation.

Wait, it’s worse than that! You applied a symmetry bias and a permanence bias on top of that – if the first choice has a 1 in 10 chance, you forced the idea of the second choice having a 9 out of 10 chance of succeeding and as if it would just stay that way. Then an icing of confirmation bias; of not going through each door removal one at a time and working out the odds. Or you did, but dismissible odds bias meant you ignored working out the improving odds of the first choice. One or more biases kicked you this way or that.

  • With 10 doors: The first choice has a 1 in 10 chance.
  • With 9 doors: The first choice has a 1 in 9 chance.
  • With 8 doors: The first choice has a 1 in 8 chance.
  • With 7 doors: The first choice has a 1 in 7 chance.
  • With 6 doors: The first choice has a 1 in 6 chance.
  • With 5 doors: The first choice has a 1 in 5 chance.
  • With 4 doors: The first choice has a 1 in 4 chance.
  • With 3 doors: The first choice has a 1 in 3 chance.
  • With 2 doors: The first choice has a 1 in 2 chance. 50% chance.

So how do we express the odds of the second choice without falling to symmetry bias or the other biases?

The trick is there is no second choice, you’ve been fooled (as have I). There is no need to express the odds of ‘the other choice’ because there is no other choice present. The above list shows the odds of winning or losing – that’s it.

Consider if I flipped a coin and asked you heads or tails. You call heads. Then I say ‘Do you maybe want to dance a jig first?’.

You’d think it an utterly redundant question.

The switch question is utterly redundant. It’s the final bias – a ‘red herring bias’, where information given is treated as if it has to matter somehow rather than potentially just be utterly redundant.

Do you want to switch? Do you want to dance a jig?

Neither makes any difference. Only by the (semi white lie) deception that is heavy emphasis does this red herring seem to matter. Are you sure you don’t want to dance a jig? Can you really be certain about your choice of heads on the coin if you don’t dance a jig? Why not work out the odds of how a jig would change the result?

Do you want to switch your understanding of the Monty Haul problem? Ironically if you disagree with this debunk of a debunk, you perhaps wont put much effort into thinking about whether you should switch. It’s a positive outcome bias – thinking about switching because switching might lead to a car seems good because positive outcome. Thinking about switching because it might lead to you being wrong just sucks. Which are you more inclined to put effort into thinking about?

Hopefully the slower, bullet pointed single door by door removal will help, as you would agree the first choice has a 1 in 10 chance when there are 10 doors. And the progression of the odds for that choice being correct will make sense as well. So before you go to argue your logical proofs, please say what is wrong with that bullet lists progression?

The hardest thing is likely you thought you were right about 50/50 to begin with – then you thought you’d eaten humble pie on it, hard earned your new knowledge on the matter, accepted you were wrong and now you proudly know the truth. Sunk cost fallacy.

There’s more pie. That’s a general rule of things in the world. There’s always more pie. Epistemic humility and peace out, ya’ll.

 

3 thoughts on “The Monty Haul Problem(s): Debunking Distortion

  1. You are making quite an extraordinary claim here, given that you disagree with Wikipedia, as well as with almost everyone with mathematical expertise, as well as common belief.

    Luckily, there are two easy ways of verifying the claim.

    The first one is deductive: Simply calculate the probabilities. Don’t use any kind of fancy techniques, or elementary techniques like the formula of Bayes. Simply go through all combinations. First, assume you do not change your guess and go through all combinations of where the car is, what you guess, and what is the outcome. Then do the same when you always change your guess. Calculate the success of both approaches.

    Second one is empirical. Play the game with a friend, or code a computer program to play it with you if you have the skills. Record what happens when you change your guess and when you do not. A number of people have shared such programmes on the internet – this one seems to work, for example, but you can find more with search queries such as “monty hall problem simulation”: http://www.math.ucsd.edu/~crypto/Monty/monty.html . I don’t know if there is an open source program that does this, but if you find, do share.

    I’ve taken the deductive approach, but by listing all the options and by using Bayes’ theorem. The result was that changing the guess is a good idea. I have not done the empirical check.

    I will be genuinely impressed if you carry out the deduction and publicly document it for example on your blog, or alternatively if you somehow rigorously document the empirical approach (write open source code which you share, record the gameplay and describe the experimental set-up so that it can be believed, or something like that).

    Doing this would also be quite an instance of epistemic rigour, and depending on the result, humility. If it turns out that the probability is indeed 1/2 regardless of changing, you can certainly publish your result in an academic journal, at least one such as American mathematical monthly or Journal of recreational mathematics. If you don’t have academic background, then I’ll help you if your demonstration is such that I believe it (if you want the help, of course). This is a promise.

    1. I am revising my position, sadly – I came on today thinking I’d have gotten zero views and could change my post. I’ll explain where I’m coming from.

      When I ran my simulation I made it choose between the doors randomly at the end, giving a 50/50 result. If one were to take the player and turn them away from the doors and shuffle the identical doors then turn the player back, then you’d be back to a 50/50 result. I also played this ( http://www.shodor.org/interactivate/activities/SimpleMontyHall/ ) and may have gotten lucky on my first ten runs as it was coming out at roughly 50% by choosing the same thing.

      I stand by the idea that Monty is removing one prior game by removing a door and he’s presenting a second game. The solution to this is rather like playing a new card game but remembering where the cards are in the deck from the old game (which wasn’t reshuffled). Or playing a second game of ‘memory’, but just flipping over the cards already seen, so you’re not actually playing memory, you’re playing a second game where you got to see all the cards first before they were placed face down. But we act as if we are just playing a second game of memory. This is not the case – that’s not how the game of memory works. This is all a bit like if one were playing cards with Monty and he leaned over so much you could see his cards. Is that really playing a card game when you have that information?

      Thus it is neither a new game nor is it the old game. There’s an incongruence there. I’m considering how to revise my post (I’d probably have just pulled it for the time being if you hadn’t commented) to include this. I think I might go from ‘debunking debunked’ to ‘debunking distortion’. Thanks for your interest, Tommi.

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