# Intuition, and the Monty Hall Problem

Statistics is hard. It’s one of those fields, where your intuition seems to take you gently and seductively towards safety, and then you wake up in a bathtub full of ice, missing a kidney. Interface design is often the same way (but, for once, that’s not what I’m writing about). When we say that we find something “intuitive” it means that we have seen something like it before—we are saying that it is familiar. The thing with statistics is that similar looking situations can behave very differently.

The first time I had this point driven home was when I was presented the Monty Hall problem. Posit: You are on a game show and three doors stand before you. Behind two of the doors lies nothing, behind one door is a pile of cash. The game begins by you picking one of the doors. The host (Monty Hall) then opens one of the two remaining doors, revealing one that is empty. Then it’s your choice: Do you stick with the door you originally picked, or do you switch to the other door? Does it matter?

Intuition says is that it doesn’t matter. It’s 50%-50%. It’s the same as throwing a coin—the money is in either one door or the other. This kind of thinking will lose you a kidney. You’ve been tricked by “intuition” to not got suckered by the gamblers fallacy, only to get suckered by something new.

It’s best to switch doors. In fact, switching doors will make you win twice as often as staying. While I’ve been able to prove that for a long time—by enumerating all possible outcomes and counting&mash;it never really clicked in my head *why* it’s true. It never became familiar. It never became intuitive. Until a couple days ago. Here’s the explanation the occurred to me (in the shower, as always) that made it all make sense:

The chance that you pick the right door on the first choice is 1/3: there are empties and one prize. Conversely, that means that there is a 2/3 probability that prize is in *one* of the other two doors.

If you could open the other two doors, see which ones contains the money, and take it, you would have clearly do that. That’s twice as good as getting to open only one door! Well, when Monty opens one of those doors that doesn’t contain the prize, he’s effectively giving you that ability: there is still a 2/3 probability that the prize is behind the set of doors you didn’t pick.

It’s obvious, now, that switching will give you a 2/3 chance of winning. By opening a door that doesn’t contain a prize, Monty has effectively concentrated that 2/3rds probability into the one remaining door.

This way of looking at it makes it really easy to calculate similar problems. For example, take the 5 door case: you pick a door, Monty opens 3 empty doors, and you have to choose if you should switch. Your first pick has a 1/5 chance of getting the prize, which means that there is a 4/5 chance that it is in the set of doors you didn’t pick. Monty opens 3 of those 4 doors, which means that 4th unopened door takes that full 4/5 probability. Thus, switching is 4 times more likely to win than is staying.

What happens if Monty only opens two doors, instead of three? Should you switch? It’s easy to figure out. You still have 1/5 chance of getting the door on your first pick. The other doors still have a 4/5 chance of having the prize. He opens two, which means that 4/5 probability is spread equally over the remaining two doors. Thus switching will give you a probability of half of 4/5ths, which is a 4/10ths probability of winning. That’s better than 1/5 chance of winning, so switching is better.

What’s the moral of the story? That intuition is malleable and that statistics is hard, until you find the right way of framing the problem. And that’s a general rule.

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Tags: intuition, monty hall problem, statistic

Dan

Huh? Once I got to your second picture the little mathematician inside me cried out and I just had to post. It’s been a while since I took Probabilities and Statistics but this is simple enough compared to some of the stuff I had to do in that class.

Once Monty opens the second door, you are given new information (that the door does NOT contain the money). Now your probabilities change to 1/2 for each of the still unopened doors… it’s still 50/50. Not 33.3/66.7.

2/3 only works if you didn’t know what was behind the door Monty picked and you could pick to open BOTH at once.

The same applies to your first 5 door example. The chance is still 50/50. because the three opened doors now have 0 probability (since there’s no way the money is there since Monty just showed you it isn’t) which redistributes the full 100% probability equally among the other two doors.

For your last five door example, two doors are opened and their probability drops to 0, as before. Now you have three doors, and since you have no information about which door is more likely to contain the money (assuming the assignment of the money to a door is completely random every time) every door has a 1/3 probability.

Going back to your original reasoning for your method, you claim you enumerated all possible outcomes for the problem and found your 66.7% success rate with your strategy. OK I’m gonna see if I can figure out what you’re thinking here.

We have three doors, times three choices you can make. Depending on the door and the first choice Monty has either one or two empty doors to choose from to show you. Then you have your second choice of one of two doors. Choosing to switch is 50% of this sample, while staying is the other 50%.

I’ll number the doors 1, 2, and 3, and tally the results below in a little table with columns: Money Door, First Choice, Monty Door, Second Choice, and Won

Stay:

1 1 2 1 Y

1 1 3 1 Y

1 2 3 2 N

1 3 2 3 N

2 1 3 1 N

2 2 1 2 Y

2 2 3 2 Y

2 3 1 3 N

3 1 2 1 N

3 2 1 2 N

3 3 1 3 Y

3 3 2 3 Y

Switch:

1 1 2 3 N

1 1 3 2 N

1 2 3 1 Y

1 3 2 1 Y

2 1 3 2 Y

2 2 1 3 N

2 2 3 1 N

2 3 1 2 Y

3 1 2 3 Y

3 2 1 3 Y

3 3 1 2 N

3 3 2 1 N

6/6 and 6/6.. The only way I see you can get a better chance than 50% could be different is if whoever hides the money or Monty uses a pattern instead of complete randomness and you pick up on it, then it gets interesting (if you realize Monty always picks the smallest numbered empty door, for instance, whenever he picks 2 you know for sure the money is behind door #1). But this has nothing to do with your strategy.

I think you have succeeded in showing statistics is hard, just not the way you were trying. :) Sorry. I honestly have no clue how you’re coming up with your probabilities.

Dan

After further thought, the only way I see this working is if you make the first choice win the player money if they get it right… but then if it’s wrong (2/3 chance) monty will either pick your door or the other empty door. If he picks yours you have a 50/50 chance, if the other one your chance goes up to 100% since there’s only one unknown door left. So the chance works out to 1/3 + (2/3 * 1/2) = 2/3

But if that’s it, “staying” would be ludicrous since you already know your first guess was wrong, and there’d be no reason NOT to switch, and whoever is putting the money behind doors is swamped with Statistics students who win all his money and he goes bankrupt. Unless he went to Statistics class too and charges enough money to offset the 2/3 probability (IE at least 2/3 the prize).

Wikipedia mentions the problem you’re talking about here so I guess there must be something to it, but I’m still not seeing it.

Dan

OK, found the full Wiki article, I’m gonna go figure this out now. :)

Mikael

So if there is a hundred doors and he opens 98 of the empty ones, that means you will have a 1% chance that your first choice is right and if you take the other door a 98% chance of getting rich?

Yes but only if your first choice was not one of the two doors still left unopened, if the first choice was one of the unopened there would be a 98% probability on both.

And that means it is still a 50%/50% (98/98) chance on getting the gold.

So switching door will not give you a better chance or probability or anything else for that matter.

Just my two cents.

Axel Hecht

Aza’s right, up to the point that I don’t think this is actually statistics, but combinatorics.

Good old days in school, I don’t yet know whether our teacher didn’t get it himself, but that topic was hell confusing up to the point where I actually discarded what he said and picked up the book over the weekend.

One detail that’s been missing in the comments here is that Monty *never* opens the door you picked.

The real trick here is to model the additional knowledge that you get from Monty opening one door. And the best way to do that is go back to the original set like Aza did. Assuming that you’re standing in front of a problem that you don’t know anything about (i.e. just two doors) is wrong, so even if you have two doors, the probability is really not 50 50.

As you can tell from the chances, if you’re actually in the show and only have one go through the game, you still have a one third chance of getting nothing.

[ICR]

This in conjunction with the Wikipedia explination helped me to understand this (http://en.wikipedia.org/wiki/Monty_hall_problem).

Here’s how I understand it, the same as Aza just phrased slightly differently:

The probability that, in the first instance, you picked the right door is 1/3. Thus the probability that you picked the wrong door is 2/3 – the car is twice as likely to be behind one of the other doors. Even after one of the duff doors has been opened, the probability of having picked the wrong door in the first instance is still 2/3. It’s just now, as Aza says, concentrated on the one remaining other door.

[ICR]

Okay, according to Wikipedia that’s wrong. It works in this instance, but not to more generalised problems. Bugger.

Ron

@Dan, Mikael: Comfy in that bathtub of ice? You’d thnk when someone says “hey, this is hard and doesn’t work the way you’d think”, you wouldn’t be quite so quick to jump up and make total asses of yourselves. And Mikael, you might want to switch from cents to pesos for future posts–it’ll be a more accurate valuation.

[ICR]

Oh no, it does still work (I think) :S I’ll stop posting now.

Dorus

I do not completely agree with this explanation, even when it’s a good way to visualize the problem, it’s not complete. Wikipedia explain it right;

Crucial about the door problem is that:

1. the game master never opens the door you picked.

2. the game master never opens the door with the money.

The first fact is explicit named in the initial question, so no need to assume anything about that. The second fact is a bit more tricky to find. If we do allow the game master to open the door with the money, the game master does not add any new information to the game, he just eliminate a random door, what mean that the remaining 2 doors keep 1/2 change to be correct, but also that the change the game is a fluke is 1/3th.

Sander

For the skeptics, here is a simulation you can run in Python:

from random import randint

games = switch_win = not_switch_win = 0

for games in range(0, 100000):

prize = randint(1, 3)

first_choice = randint(1, 3)

if prize is first_choice:

shown = randint(1, 2)

if shown is first_choice:

shown += 1

else:

shown = 6 – prize – first_choice

if first_choice is prize:

not_switch_win += 1

else:

switch_win += 1

print ‘games: %d’ % (games + 1)

print ‘won by switching: %d’ % switch_win

print ‘won by staying: %d’ % not_switch_win

The results here looked like:

games: 100000

won by switching: 66729

won by staying: 33271

Sander

Hm, wait, logics seem to have escaped me for this morning. The whole ‘shown’ thing isn’t relevant in my simulation. But erm, well, it just proves that you should just switch choices anyway :-)

Abdulkadir Topal

Which brings us to browser: If your first choice has been IE you should switch, it’s very likely you’ll win, qed ;)

Winz

It’s easier to understand when you first list all the possibilities from the first step to the final choice. Do NOT separate the 2 steps.

Then you will see that the whole thing is clear:

Monty KNOWS which door is the good one, so in the 2 cases out of 3 where you did not first pick the good door, he will eliminate the wrong door in the “switching choice” group. As Aza explained, it doesn’t matter if it is the 2nd or the 3rd one.

Daniel Einspanjer

The important thing that finally allowed me to “get” this problem when I sat down to solve it is that Your initial choice is random but Monty’s choice is very much not random. Because his choice is not random, it does not affect the initial 1/3 to 2/3 probability.

The other statement that helped me follows (expanded to the chose one door and Monty shows you 98 losing doors and offers to allow you to switch):

You have a 99% chance of forcing Monty to give you the choice of switching to the winning door (because he always has to open 98 doors). Only on the 1% chance that you originally chose the winning door will Monty be able to open 98 losing doors and offer you the choice of switching to the 99th losing door.

mawrya

As some other posts have noted, this problem is often stated without the two important facts required to understand the answer:

1. Monty never opens the door you initially choose.

2. Monty knows where the money is and always opens a door with nothing behind it.

If you ever mention the door puzzle without also mentioning these two criteria, your blog immediately becomes filled with posts arguing the 50/50 view, and rightly so.

Gijs

Hah. Actually, we treated this topic extensively in my Intensional Logic class. Your explanation is correct, as far as I can tell, save for one subtle point, which is explicit in the python script someone pasted above, and is not amongst the two rules several people have posted here already:

Monty picks randomly between the two other doors, if you already picked the door which holds the money.

I’ll try to show why this is important.

Let’s consider the case where you picked door 1. Let’s imagine that Monty’s strategy is to instead always pick the first unopened non-money door. Now:

If the money is behind door 1, he’ll open door 2. Switching gets you 0, not switching gets you 1.

If the money is behind door 2, he’ll open door 3. Switching gets you 1, not switching gets you 0.

If the money is behind door 3, he’ll open door 2. Switching gets you 1, not switching gets you 0.

Now observe, that if Monty opens door 2 (2 of the cases mentioned above), your probability of winning by switching is 1/2, assuming we believe the a priori chances of the money being behind the doors are distributed evenly. If he opens door 3, you should always switch, and your odds of winning are 1.

Of course, the alternate strategy of always opening the other door gets you the opposite advice, and if we assume equal probability for the random strategy and these two determined strategies, it is better to switch. But in this case we assume we have knowledge about the likelihood the game host will use these strategies, which is not true, strictly speaking. So, the logician/professor who teaches the course claims that you can’t actually decide what is better to do unless you have information about Monty’s strategy. Which just proves you should never become a logician, as far as I’m concerned. :-)

Leaf

Problems like these would make me fail in life X___X I’m generally strong at probability because it tends to be intuitive. But this is one of the those which behave otherwise…

Your pic has finally helped me cognize this problem, although it was clear to me when I started drawing on the paper:

In the three doors, the chosen door is marked with +, the one with treasure is marked with $; the unchosen marked with _.

As the puzzle is true to any order of the doors, there are only three cases:

1. [$+] [_] [_]

2. [$] [+] [_]

3. [$] [_] [+]

As expected, the _ gets opened

Case 1: No-Switch gets money

Case 2: Switch gets money

Case 3: Switch gets money

This way it can be explained with least statistics involved. If need be, we can replicate it with money put under second and third doors with similiar results..

Stacy

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Gregg

This particular problem was covered by a local newspaper columnist a couple years back. Although in that case there was a new car behind one door and goats behind the rest. Many people fell for the trap, including myself. So I ended up making a page to test it. Enjoy :)

Aza Raskin

@Gregg: Very cool!

Chris H

Thank you so much for your elegant game explanation! I always get lost in the statistics-I-only-saw-once. Design to the rescue!

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It’s easier to understand when you first list all the possibilities from the first step to the final choice. Do NOT separate the 2 steps.

Then you will see that the whole thing is clear:

Monty KNOWS which door is the good one, so in the 2 cases out of 3 where you did not first pick the good door, he will eliminate the wrong door in the “switching choice” group. As Aza explained, it doesn’t matter if it is the 2nd or the 3rd one.

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