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The "Three Doors" mathematical problem

De_Bunk

Scourge of the Believer
De_Bunk
Joined
Feb 4, 2002
Messages
5,523
Im sure some of you have come across this before..but i was fascinated by the debate surrounding this.

Somehow, both sides of the discussion seem to make sense....

Heres the question.

You've got three doors..

Behind two of them is a goat, the other a car

You choose one door to open....but you dont open it yet.

Then, someone opens one of the other doors to reveal a goat.

You are left with two closed doors

But before you open your chosen door, you are given the choice to switch to the other unopened door and open that one instead.

Do you stay with your original choice of door or do you switch

Would switching from your original choice increase the odds of you finding the car...

( i know this is probably old maths problem...but it done my head in..)

If you want the link to the simulation...ask..

DB
 
Pgwenthold...

I half guessed it was old..

But it still done me in...

I didnt believe it until i ran the simulator..


:)

Never mind...

DB
 
Here's the best way I've found to think about it:

You select your door, but don't open it.

Monty offers you both the other doors in exchange for your choice. You'd take that deal in a heartbeat, right?

Well, that's what he's done by showing you which of the other two doors wouldn't be worth anything in that deal.

~~ Paul
 
The answer is really simple.

The straight dope guy is wrong. It doesn't matter if Monty knows which door contains the prize or not.

Think of it like this. Separate the doors into two piles. Put the door you picked in one pile, and the other two doors in a separate pile. Now, Monty cannot pick the door you chose, so he is left with the two doors in the other pile to choose from.

Now at this stage, we will both agree that it is more likely the correct door is not the one you chose, but one of the other two (there are two doors to one, so it's more likely).

So, when Monty removes one of the doors in the other pile, the one left is more likely to be the one than the one you chose, so you should always switch.

The way this puzzle tricks your brain is the fact people think Monty is picking any random door from the three available - he isn't (he can't pick the door you chose, only one of the other two).

If Monty doesn't know which door contains the prize, it's just as likely he'll pick the correct door to open as it is that he'll pick the wrong door. When he picks the wrong door, you know for a fact the other door is most likely the right one.

In short, definitely switch doors no matter what!
 
My brain came out of my ears on this one some years ago. In the process I tried the following "reductio ad absurdum".

Suppose there were 100 doors, 99 goats and one car. You choose, but don't open the door. Monty selects a door, and it's a goat. Do you want to switch? But it goes on. Monty keeps selecting doors, and they're always goats. In the end the only doors still shut are the one you chose at random at the beginning and the single other one Monty has not yet opened.

Will you switch now?

I'm not sure if this illuminates much, but please feel free to discuss.

Rolfe.
 
Another really easy way to think about it:

There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!

Let's construct a scenario to test this out and say the prize is behind door #2.

1) What happens if you pick door #1? Monty will open up door #3. If you switch, you win.

2) What happens if you pick door #2? Monty will either open up door #1 or #2. Switching here will make you lose.

3) What happens if you pick door #3? Monty will open up door #1. If you switch, you will again win.

There existed two cases where you won and only one where you lost!
 
Yes, that was my other "reductio ad absurdum".

There are three possibilities.

1. You selected the car, unknowingly. 1 out of 3 probability. Monty perforce must select a goat. And you will lose by switching.

2. You selected a goat, unknowingly. 2 out of 3 probability. So there is now a 50/50 chance Monty will reveal the car. Isn't there?

2 (a). Monty reveals a goat. You win by switching.
2 (b). Monty reveals the car. The game is up.

In this scenario, it's an even break whether you switch or not, because you only have two of the three possibilities still running, and one of these is a win and the other a lose.

But is 2 (b) actually a realistic option? Does it ever happen?

That's the "reductio ad absurdum". You run the trial ten times, and observe whether Monty ever does reveal the car. If he does, then you gain nothing by switching. But if he never reveals the car, then you know that he is deliberately avoiding it. In this scenario, you will win more often than you lose if you always switch.

So, you always switch. If he's not deliberately avoiding the car, then you gain nothing, but you lose nothing either. If he is, then you definitely gain. So you never lose by switching.

I think. Like I said, my brain came out of my ears some time ago.

Rolfe.
 
Batman Jr. said:


There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!
Click to expand...


I GET IT! Thank you!

So since the chances of getting the wrong door are 2/3, so if you DID, and he gets rid of one for sure, and then you switch, you have a 2/3 chance of getting it right! Haha!
 
sorgoth said:
So since the chances of getting the wrong door are 2/3, so if you DID, and he gets rid of one for sure, and then you switch, you have a 2/3 chance of getting it right! Haha!
Click to expand...
Exactly. Your chances of having picked the car door were 1/3. The chances that the car was behind ONE of the other doors was 2/3. Monty's opening one of the other doors doesn't change those odds. All it tells you is that the remaining door is the one that has the 2/3 chance of being the winner.
 
Humphreys said:
The answer is really simple.

The straight dope guy is wrong. It doesn't matter if Monty knows which door contains the prize or not...
The way this puzzle tricks your brain is the fact people think Monty is picking any random door from the three available - he isn't (he can't pick the door you chose, only one of the other two).

If Monty doesn't know which door contains the prize, it's just as likely he'll pick the correct door to open as it is that he'll pick the wrong door. When he picks the wrong door, you know for a fact the other door is most likely the right one.

In short, definitely switch doors no matter what!
Click to expand...
Right conclusion, but dead wrong analysis, Humphreys.

First off is the evidence from the game show, which I have to guess you've never seen. Monty always revealed a goat door. Why? Because there is no drama in doing otherwise. There's nothing left to choose. The contestant knows she's lost. By revealing a goat door, the drama is increased, and the contestant still has a chance and a seemingly difficult choice to make.

Now for the analysis:
(Assume your explanation is right...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty chooses one of these at random.
o The chance he chose the prize is 1/3, but the game is now over, isn't it?
o The chance he didn't choose the prize is also 1/3, but, given that, the chance you should switch to the other door is 1/2.
o No advantage in switching!

(Assume now that Monty knows...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty selects (with knowledge) a goat.
o That selection, being done with certainty, is not part of the probability picture. Effectively, the 2/.3 probability has now jumped to the door the contestant did not originally choose.
o Switch, d***it, switch!
 
The assumption is that the probability of him revealing a false door is independant of which door you chose. The way the puzzle is phrased, we don't know whether he does this everytime or if he intentionally tries to throw people off.

IIRC correctly, on the game show Monty always revealed one. So in that particular case, always switch choices.

Walt
 
I'm sure someone will eventually mix the Monty Hall problem with something in quantum mechanics to create the ultimate puzzle for screwing with people's heads. :D
 
I'll explain the alternate case where Monty Hall doesn't know which door has the car:

In order for switching to work, you will first have to pick a wrong door and Monty will have to choose the other wrong door. There are six cases:

1) Right and Wrong#1

2) Right and Wrong#2

3) Wrong#1 and Right

4) Wrong#1 and Wrong#2

5) Wrong#2 and Wrong#1

6) Wrong#2 and Right

Notice that we now count right-wrong#1 and right-wrong#2 as two cases. This is because Monty can now pick the right door when we first pick the wrong door. In the other scenario, since Monty knew where the car was, the probability of him picking the other wrong door after you having picked the first wrong door was always 1. We had to count right-wrong#1 and right-wrong#2 as one case before because the probability of each happening is (1/3)(1/2). The probability of the other two cases was (1/3)(1)—Monty had a 100% probability of picking the wrong door, remember?—and since each case must have an equal chance of happening (right-wrong#1+right-wrong#2=(1/3)(1/2)+(1/3)(1/2)=(1/3)1) and both right-wrong#1 and right-wrong#2 have the same outcome, we grouped them into one case. But here, all of the cases become (1/3)(1/2) as Monty has only a 50% of getting the other wrong in the case of first picking one of the wrongs.

However, we know Monty at this point in the game revealed the goat! So, we have to eliminate cases 3 and 6 from above. We are left with 1, 2, 4, and 5. Switching works when both Monty and you selected wrong doors. Which instances match this criterion? Why, 4 and 5. That means the probability of a switch working is an unbiased 1/2!
 
Batman Jr. said:
Another really easy way to think about it:

There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!

Let's construct a scenario to test this out and say the prize is behind door #2.

1) What happens if you pick door #1? Monty will open up door #3. If you switch, you win.

2) What happens if you pick door #2? Monty will either open up door #1 or #2. Switching here will make you lose.

3) What happens if you pick door #3? Monty will open up door #1. If you switch, you will again win.

There existed two cases where you won and only one where you lost!
Click to expand...
This explanation did it for me, thank you. I lost hair over this problem a few years ago and never quite "got it." Now I do.
 
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