Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you: 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?

Yes

Isn't it a 50/50 chance, therefore it's impossible to know whether it is to one's advantage or not???

I'd be better off with a goat btw as I canna drive. :(

HiccupPercy is correct

[BK: assume that you want to win the car and not the goat. You could always sell the car and buy a horse]

Depends on the type of car and how many goats we're talking about. ;)

Yes it is, and for anybody doubting it, sit down and work out the probabilities.

What do I win?

NO! It's a 50/50 chance you FOOLS!!!! :unsure:

HiccupPercy will now explain why the answer is Yes.

Can someone explain this to me, please, preferably using finger puppets?

No it's not.

This problem is illustrative of the way the human brain has evolved. We are extremely good when it comes to working with simple numbers on a linear basis but relatively poor when it comes to probabilities. Advertising makes use of this same biological impasse by constantly referring to percentages and probabilities.

Is it because goats are conditioned to stand either side of a car?

No, replace goats with "nothing" and it is still true.

Is anyone going to explain this? Percy? GLPL?

You have a 1 in 3 chance of picking the car. After being shown a goat the only door left for you to change to is another goat (changing = LOSE)

You have a 2 in 3 chance of picking a goat. The host shows you the other goat. If you now change you will definitely get the car (changing = WIN)

Changing doors is a winning strategy 2/3rds of the time.

Hmm...

So on your first guess, you have a 1/3 shot.

But after the one wrong answer is revealed, you are improving your chances to 2/3 by switching.

This idiot American version is brought to you by BLY


**edited for bad math**

Perfecto, you win the whisky.

To make it seem more intuitive imagine a situation in which there were a hundred doors, 99 with goats behind them and 1 with a car. If the host opened 98 of the doors to reveal goats and then said "do you want to switch?" you obviously would. This is the same thing with fewer doors.

:confused:

jeezlouise, no wonder you got the other quiz too....I think I'll go into a very dark room with a sparkler for a couple of hours

Oh, I see someone's already posted the answer.

A similar issue with humans and probability is known as the prosecutor's fallacy, and a great example of this is from a few years ago when a woman was falsely imprisoned for murder when both her babies died of cot death, (she was later released).

The argument was that there is say, a 1 in 100,000 chance of cot death, so to have two cot deaths would be 1 in 100,000 * 1 in 100,000 i.e. 1 in 10,000,000,000 chance that she was innocent and that there actually were two cot deaths. Because of these odds she was imprisoned. (these are not the actual figures but they demonstrate the point)

However this is bollocks. What you should be doing is comparing the odds of two cot deaths against the chances of double murder, itself a spectacularly rare event. When you think carefully about the probabilities, you are comparing the likelyhood of one very rare event with another very rare event, and once you do so, suddenly things look more favourable for the woman.

Again, this demostrates the problem the human brain has with probability. We are just not hard wired for it, whereas we are for simple numbers.

I remember when I first heard this, the "Monty Hall Dilemma", it seemed like a 50/50. Fair did my head in, so it did.

There is a site somewhere they computerised the problem, and yes it the switch option.

What if you wanted to win a goat? :unsure:

But it doesn't matter how many doors the host has opened, that doesn't alter what is behind the door I've chosen, nor does it alter what's behind the other door. It's still a 50/50 chance. :confused:

The host knows what's behing each door so he obviously opens 98 doors that he knows have a goat behind them, leaving two doors: one with a goat and one with a car. None of what has just occurred alters the fact that one door has a goat and one a car. It's a 50/50 chance!!! :banghead:

No BK, think of it mathematically. The host's knowledge has nothing to do with it.


When you pick a door in the 100 door example, your odds of being right are 1 out of 100. Not very good.

Let's say that 98 incorrect doors are then opened. You can stay with your pick, but it's still 1/100 odds.

If you switch doors, it's 1/2 odds, because when you made the selection there were far fewer choices.

It's the same with the 3 door problem. You can't look at how many doors are left, you have to look at how many doors there were when you first selected :)

^ I understand it mathematically and yes I agree there, but in REAL LIFE, in that situation, you CANNOT say to the person you're better off switching doors, because that mathematical probability is useless once you get down to the final two doors. It becomes a 50/50 chance, despite all the fancy mathematical reasoning. You have two doors in front of you, one has a car and one has a goat. The chances of you having the car is the same as the chances of you switching to the car.

:cry2:

no no no no :)

Okay, what if there were A MILLION doors. You pick one of them, and then Monty opens up 999,998 doors. Would you switch then? Either you picked the right door out of a million or the one door not opened is the right one! It's not 50/50, the odds are hugely more likely that you picked wrong, and the other door is correct.


http://montyhallproblem.com/ <--- here is a website that breaks it down in more detail.

Ah, yes I see now!!! :D

Blimey. I'm going to have a lie down. Thanks! :beer:

I wonder if this captures the evolutionary story. People's brains aren't inherently wired to do long division, play guitar, write poetry etc without extensive training and practice. And I think there's some anthropological discovery about untutored mathematics - the number systems of hunter-gatherer societies running "1, 2, several, many" or something like that. If we were taught probability along with the multiplication tables, maybe we'd all solve this problem without difficulty.

I just want to know if anyone ever asked Monty if they could keep the goat.

"You see, Mr. Hall, in my country, they are considered most flavorful..."

I look at how many doors are left.
Once one door was opened: it ceased to exist for me....It was as though it was never created.
The host is not only irrelevant, I don't even like him.
Two doors - one guess...I may not even guess; pfui...but I probably will because I have a 50 - 50 chance of getting the car :lol:
and, I'm afraid, an insurance policy because the salesmen know a mark when they see one. :(

BK is right.

Stop trying to meddle with our reality. It's all that some of us have to hold on to.

I'm having a glimpse into a possible future of what it's like to have Alzheimer's reading this - sort of fuzzy, cloudy, going "mmmm" a lot but really rather bewildered and wondering about my tea.

But surely this would be true even if you make a second choice by keeping the first choice? The probability has changed to 50/50 in this case as well.

Edit: I have read the linked website and still think the argument is slightly flawed. The car really is behind one of the doors; it's not a Schroedinger car, that before you know where it is exists with 1/3rd behind each door. Think of a situation where you have first picked the other of the two remaining doors; if the original statement is true, then it would be equally better to change in this case as well - which means that the probability is equal that it is behind door one or door two. Because 100% of the car is behind only one door. A real car doesn't change door just because the probability changes.

Somehow, this reminds me of another problem; let's say you throw a dice 99 times and don't get a 5 in any of the cases. Is the probability for getting a 5 the 100th time higher because of this? No, it's still 1/6th. Statistically, the more times you throw the dice, the more it is likely that each number will have come up 1/6th of the times - but this doesn't change the probability for each throw. (Not really related, I know.)

No, it doesn't. Because Monty will never show you if the door you chose first is wrong, that's one of the conditions of the problem. So your original choice is still based on 1/100 probability.

Read my edit above. :)

The second choice is based on new conditions. And, since the same would be equally true also if you initially had picked the other remaining door, that tells me that the odds indeed are 50/50.

Actually, the more I think of it, the dice problem is slightly related.

If you do the Monty thing several times, it might show that it really is better to change to the other door. But this doesn't change that for each case it's still 50/50.

Think of it with a million doors. You pick a door, Monty shows you 999,998 goats, and you can keep your door or you can switch. Do you really think it's 50/50 that you'll win? You picked that door out of a MILLION. Now you have an option to try another one out of 2. That second door inherits all the probabilities of the 999,998 that held goats.

And the dice example is totally apples and oranges.

Maybe I should point out that switching doors does not gaurantee you success, it simply ups your mathematical odds of winning.

God, I am so almost getting this and yet so not actually getting it.

I can understand in theory how the probability changes when you gain more knowledge, but I can't make the leap to understanding how, in practice, you can do better than 50-50 whether you change your choice or stick with your original choice.

Is this exercise just completely theoretical or does it have real world applications?

You're not convincing me by simply repeating your explanation. ;)

Apples and oranges? Sure, as I wrote, I think they are slightly related. Both are considered fruits. But what I was hinting at is that what is statistically more probable is not more probable for each single choice.