Joho the Blog
An Entry from the Archives

It's dead simple. If you have a 2-to-1 chance of getting a donkey with your first choice, then by the same 2-to-1 odds you'd be right to switch out of it.

What confuses people is the seeming binary nature of the choice: donkey or new car. That's why a lot of people answer that it makes no difference whether you switch or not. They forget that the odds of turning up one or the other are not the same. It's like you're flipping a weighted coin: one side is a donkey, the other the new car, but as there are two donkeys and only one car, one side has a 2/3 chance versus only 1/3 for the other.

In other words, you can't divorce your second choice (the one Monty gives you after he's opened another door) from your first. You're not starting from a clean slate; rather, the chances are 2-to-1 that you are sitting on a donkey. That remains true, no matter what other information Monty chooses to share with you.

Clear?

Totally not.

David, the first of your links says in part, “If the contestant picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game.” If that is how the game works, you know you have picked the wrong door when the host reveals the second. If you had picked the door with the car behind it, there would presumably have been a “Ta-da!” from the band, the host would have said something fatuous, and everyone would be applauding. But the second door has been opened to reveal the donkey, and that leaves only the third, and the Ferrari. How that squares with the link’s remaining discussion of statistics, I’m not sure.

This problem is the single reason why I've never fully grokked statistics. Teachers always base the entire class on students getting this, and if you don't, it's hard for someone to whom it's obvious to explain it.

It's clear to me that initially, when you pick your door, the odds are 2 to 1 that there's a donkey behind it. But after Monty has revealed the other donkey, there are only two doors that are mysteries. One of them has a donkey, and one a car. You're sitting on one of 'em. I can understand, but not *believe*, that the initial odds for picking your door still hold. The situation feels like it *has* changed. You're picking from two mystery doors, one with a prize behind it, and it's hard to believe the odds aren't 50/50.

I do get the 1000 door example though. It's easier to see that your odds of getting the car become incrementally larger every time you switch, once one of 999 donkey doors has been eliminated. I just have a hard time believing that it really holds true at the three door level, the smallest number of doors possible for the experiment.

Part of the problem, I think, is that you only really start to be able to notice the statistical odds at play in large numbers of games. I played the applet demo game you linked above. First I played 300 games (it's fast!) where I didn't switch doors. For the first 20 games, I was winning about half the time. It was only as the total number of games got larger and larger that my wins started to look like 1/3, rather than 1/2. When I played 100 games where I switched (not as fast!), my odds of winning initially looked a tad better than 50%, but it took a while for them to start looking like 2/3.

My difficulty is in believing that the odds that hold for large numbers of repetitions also hold for just one time, just one game, the one chance in front of Monty Hall on TV with money at stake. I can accept that technically it's true, but...

I'm *still* seeking a truly good explanation of this, tailored to the statistically-irreverent such as myself. If you find one let me know!

David, here's my attempt to explain:

The key point to remember is that... they will never show you the door with the car behind it!

So that means they are skewing the picture. They are not opening a random door. Instead, half of the time when they choose that door they are forced to do so, because otherwise they would reveal the car. (It's the half of the time when you haven't won the car, and so the car is out there. The other half of the time you have won the car, and so they can show you either door they please.)

To calculate your chances you just have to realize that your door's chances didn't change - they remain, as before, 1 out of 3. But the other chances have been sorted out, so they must now all be resting in that other door, and so that's 2 out of 3.

It probably can never make "intuitive sense" but if you realize:
A) something is up - that door they opened ain't completely random
B) meanwhile, my door's chances didn't change
C) so that third door has got to be bearing those missing chances
Then at least it hangs together.

Peace, Andrius

----

Here's another attempt:
Which would you prefer to have:
1) To choose one door out of three?
2) Or to choose two doors out of three and have one of the loser doors thrown away?

The second option is better. In fact, it's twice as good.

Here's what I hope is a clear view of why switching is right:

Monty's offer can be phrased as "Would you like to take *one* door, or *two* doors"? Put your choices - one door, or two doors - into different sealed envelopes."

If you take *one* door, your probability of winning is 1/3.

If you take *two* doors, your probability of winning is 2/3.

Clear? Obvious?

Now Monty says "Here's the catch - if you take two doors, I, who know where the prize is, will open one wrong door of the two doors you picked, show you it is wrong, and ask you if you'd like to switch, and go back to the envelope with just one door in it".

Does anyone doubt that. phrased this way, inversely, which way you're better off?

The key is to understand that Monty gives you NON-RANDOM, DEPENDENT information. That's what the 1,000 box version is about.

Like David, I've been hung on the seeming binary nature of resultant following the opening of the first wrong door. Like David, I've been able to keep explanations of why this is wrong-headed in my own poor oxygen starved skull for between 3 and seven seconds.

Following Seth's inversion I think that now I get it.

Monty says... You get two doors, I get one.

I say... Can I have door number one?

Monty says... Knock yourself out. Oh, and incidentally, check this out -- door number three is a donkey door. Want to switch to number two?

I say... Yeah, I really do, because my odds that door number one, my door, has a donkey are still 2 in 3. But the odds that door number two has a donkey are now only one in two.

By the way, do you get to keep the donkey?

"the odds that door number two has a donkey are now only one in two"

Actually, 2/3.

You pick one. There's a 2/3 chance it's in "the other two". Monty shows you one of "the other two" is wrong. There's still a 2/3 chance it's in "the other two", and just one viable door is in that set, so it has the whole 2/3.

This is the meaning of NON-RANDOM, DEPEDENT.

IF there are 1,000 boxes, you pick one, Monty opens 998 wrong boxes- the remaining box is not 50/50, but 999/1000.

This is an elaborate version of the old joke: "The probability of any event happening is 50/50, because either it will happen, or it won't happen".

That demo site made me understand the problem. Thanks!

I have to admit that I had never heard of this until is was introduced/shown to me as an example for the IMS QTI-specification. After reading your post though, I'm not sure whether it is a good thing we're able to express this problem in QTI or that it makes people hate the spec just because of it. ;-)

So what are the chances that Schrodinger's cat will be behind the door you choose?

Erica's reasoning suggests the source of problem - equally likely event and conditional probability. Erica says: "But after Monty has revealed the other donkey, there are only two doors that are mysteries. One of them has a donkey, and one a car. You're sitting on one of 'em. I can understand, but not *believe*, that the initial odds for picking your door still hold. The situation feels like it *has* changed. You're picking from two mystery doors, one with a prize behind it, and it's hard to believe the odds aren't 50/50."

Since the objects are not moved around after the revelation, the two doors are not equally likely events AFTER the revelation. That means, the probability you have a donkey is 2/3. Since there is only one other possible door, the conditional probability that the other door has the donkey, given the revealed door also has a donkey is the total probability that you have the car, which is 1/3.

Are you serious about damning?

Seth, I suppose you would also try to convince me that the odds of two heads in flipping a pair of fair coins is one in four. Don't bother. I will NEVER let facts and/or logic stand in my way of my intuition or my understanding of TRUTH.

{g}

Aswath: Thank you! I think your way of wording it is helping me wrap my brain around this in a way that might actually stick.

So basically, I have two doors that *feel* like mystery doors, but they are *not* in actual fact mystery doors. The extent of their mysteriousness is dependent on the revealedness of the door Monty opens to show a donkey. OK. I think I can get that.

What it's hard to grok is that conditional probability could remain a fact about an object even after the initial event that feels relevant to its probability (but isn't actually) has passed. The probability of there being a donkey behind the door I've chosen becomes a permanent fact of my door for the duration of this round of the problem, no matter what else happens, or what door is opened. It still feels totally bizarre, but I'm willing to understand it that way.

Erica:

Your last question can be explained this way: Since the contestant selects a door before Monty reveals a door, we could say that this event and the conditioning event are independent. That means the probability of the selected door having a car remains the same even after Monty opens a door with a donkey.

I am explaining further at the risk of potentially confusing is because somebody commented about non-intutive nature of probablity and could stump new students. If it is true it is because most of the time the problem is poorly worded due to linguistic limitations are explanations are given verbally (as was done by Ms. Marilyn vos Savant), which might be stated confusingly or be misinterpreted. But there is no confusion if probablistic statements are stated algebraically. This is my opinion.

THE EASIEST ANSWER:

Assume from the begining you are going to use the switching strategy, if you start on donkey, you will win. (Think about it, the other donkey is shown and you swith to the car). If you start on a car, you will loose. (One donkey will be shown, and you take the other). Therefore if you assume the switching stratgy you have a 2/3 chance of winning

Believe it nor not, but Marilyn may have been wrong! Was is stated that Monty ALWAYS gave the contestants a second chance? As far as I know about the problem it said "YOU are on a show... and so on". It doesn't say anything about how long this show has been running, how many times you've seen it on TV before being a contestant yourself... and most of all, it was NOT said that giving you a second change was part of the rules.
Now, image you're in this show and things happen as stated. You make your choice, Monthy opens another door and asks you if you want to change your mind. What do you do?
Here are the alternatives:
- Monty always offers to change : you should switch and you win 2 times out of 3
- Monty is a nice guy... if you had chosen right, TADA! there goes the music, you win! But you have made the wrong choice, so, he shows you another wrong door. You change and you ALWAYS win..
- Monty is a nasty guy. You chose the wrong door: TADA! he opens the door you've chosen and shows you the donkey... You chose the right door, he offers you to change. Because he knows you're a clever person, he known you'll change your mind. But: you change and you're 100% sure to lose!
- Monty sometimes offers contestants to change because he likes certain persons and dislikes others.
- Monty sometimes offers to change because there is some time left before the end of the show.
- And so many other possibilities we're not told about...
The problem is: we're not told... so we do not know!
Can math solve this problem? YES they can! That's called STRATEGY theory. The demonstration is rather long because one must consider ALL events as possible (with unknown probabilities): Monty being nice or nasty, the probaility a nice/a nasty guy will offer you a second chance depending on your first choice was,etc...
Working it all out, he're the answer: you toss a coin and change if heads (or tails, that's up to you) come out. You then have... a 50% probability of winning the car!!!!!!
It is VERY important that you toss the coin so that Monty can in NO WAY predict what you're going to do. If he knew what you're going to do he might make you win...or lose (but you don't know).

very simple the Monty Hall problem. It really makes no differences to the probabilty of winning the car if you switch or not.

The confusion comes because the probability of PICKING the car first chane is always 1 in 3 whilst the probability of WINNING the car first choice is always zero. The game show host is always going to reveal a goat on the first choice, even if you initially choose a goat. This is illogical as one would expect him to want to keep the car, but really he wants to keep the game show going. they can afford to lose a car.

Similarly, the probability of them losing the car on the first choice is also zero. He will not reveal the car but will reveal a goat. Very simple.

Thus the choice is, and always was, going to be narrowed down to a 50 50 chance. Therefore the odds od WINNING the car are always 50 50. The initial choices, whether from 3 doors or a thousand are irrelevant. The host will always reveal another goat until there is only a goat and a car left.

Making a new choice from a 3 to 1 option to a 2 to 1 option SEEMS advantageous, as indeed it would be if that were the case (which it is not). Your chance of winning the car first go was zero, as was his chance of losing it. Whether you picked the car the first time is irrelevant. You would not be awarded it if you did, nor would you be awarded a goat. Just another new choice. A 50 50 choice.

Staying with the same door is as much of a choice as changing doors. thus the odds remain at 50 50. The probability is unchanged.

Thank you.

Paul Burton
UK