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monty hall problem

Lets say if instead of monty showing you the gate , one of the goat ACCIDENTALLY comes out of one of the doors, which is different from the door you have chosen. Will switching still help? I think now it won't. Am i right?

vivek

vivek gupta
Saturday, April 06, 2002

I don't think so. The reason  is that goat is still coming out. I mean after u selected a door, the goat is definitely coming out of one of the doors which is definitely not the one in which money is hidden.

thanks,

****
Monday, April 08, 2002

Yeah basically it depends on whether monty hall knows or not.

Mike Lee
Wednesday, April 10, 2002

No, switching will not increase your chances of winning.  The goat emerging from a non-chosen box eliminates that box from the calcuations.  At that point, the game changes from a one-in-three choice to a one-in-two choice.  So, if you compare your odds before the goat exits and after  it exits, your odds improve.  But the odds on the non-goat, non-chosen box also increases.  They both increase to 50%

J. Hancock
Wednesday, April 10, 2002

The Monty Hall problem lives!

Always keep this in mind: _when you picked the first door_ the chance of that door having a car behind it was 1/3.

Now that you know one of the other doors has a goat behind it, it _doesn't_ change the fact that the door you _originally picked_ had a 1/3 chance of being the "car" door. It's incorrect to say that the probabilities "change" in the way you describe -- rather, they're rebalanced between the two doors that remain closed, but _your door still has a 1/3 chance of having the car behind it_. (Otherwise, the act of Monty opening a "goat" door would magically change the past).

Consequently, if you swap, two times out of three you'll get the car. If you stick, one time out of three you'll get the car.

I find that where the confusion creeps into this problem is that it's seldom explicitly stated whether Monty opens the other door at random (and just happens to pick a goat door), or deliberately chooses a door with a goat behind it (because he knows where the goats and car are). (Mike pointed this out above, "depends on whether monty hall knows or not"). Thus, it's not stated whether, in some cases, he'll open the door, show you a car, and you go home happy. Or, to take Vivek's example, can we consider the case where the car "accidentally wanders out" through its door?

But in any case, your strategy should remain the same (to swap to the remaining door). The events that _lead up_ to the situation DO NOT alter the probabilities at the time of your decision (to swap or stick). In the final analysis:

1. Your "original choice" door has a 1/3 chance of having a car, a 2/3 chance of having a goat.
2. Your choise is now reduced to only two doors. The original door, or the remaining third door.
3. Therefore, the remaining third door is twice as likely to be the "car" door as your original choice.

Adrian

Adrian Gilby
Thursday, April 11, 2002

Lets see it this way. In the original case when monty knew what was behind which door, he made use of his knowledge which led to the concentration of probability onto the the single unchosen door. There, you knew that monty will not choose the gate which you have chosen irrespective of the fact whether you have a goat or money behind your door but he could well have chosen the other closed door .Hence there came asymmetry between the two left over doors because one was in the possible choices of Monty and the other was not.
But in the case where a goat randomly came out of one of the doors (and only by chance it was not your door), no extra knowledge has been made use of. Therefore the two leftover doors are totally symmetrical with respect to the probabilities associated with them. Hence switching won't help.
And how did the probability of your chosen door increased from 1/3 to 1/2 ? It increased because of the very fact that the goat did not come out of your door.

vivek gupta
Friday, April 12, 2002

I'm not sure I see your point.

Let's assume the following:

1. You pick door A.
2. One (and only one) of the goats reveals itself.

The probabilities are as follows:

1. Prize behind door A. Goat B revealed.
2. Prize behind door A. Goat C revealed.
3. Prize behind door B. Goat A revealed.
4. Prize behind door B. Goat C revealed.
5. Prize behind door C. Goat A revealed.
6. Prize behind door C. Goat B revealed.

Now, I'm sure we agree that each possibility has a 1/6 chance of occuring.

And I'm sure we also agree that the probability of (prize behind door A) is 1/3, the probability of (prize behind door B) is 1/3, the probability of (prize behind door C) is 1/3.

Now it gets interesting. Let's say that the goat DOESN'T come out of door A.

This means we can eliminate possibilities 3 and 5:

1. Prize behind door A. Goat B revealed.
2. Prize behind door A. Goat C revealed.
4. Prize behind door B. Goat C revealed.
6. Prize behind door C. Goat B revealed.

But here's the important point: The probability of (prize behind A) is still 1/3. Likewise for (prize behind B) and (prize behind C).

So the probabilites are:

1. Prize behind door A. Goat B revealed.  1/6
2. Prize behind door A. Goat C revealed.  1/6
4. Prize behind door B. Goat C revealed.  1/3
6. Prize behind door C. Goat B revealed.  1/3

All we've done is eliminated the events where the goat came out of door A. We haven't changed the probability distribution of where the prize is.

To me, it still makes sense to switch -- because there's still a 2/3 probability that the prize is behind B or C.

Adrian

Adrian Gilby
Sunday, April 14, 2002

I agree with you till the point that initially all doors have the probability of 1/3.
BUT once the goat comes out , we not only know that it did not came out of door A but we also know which doors out of B & C , it actually came out. Lets say it came out of door B. Now there are only two possibilities left
1.Prize behind door A
2.Prize behind door C.
And no such event has occured which gives more weight to any one of them (because,unlike the original problem, both were candidates  from which the goat could have come out ) , therefore both have equal probability of 1/2. (In the orginal problem, door A was not in the candidate list for the
Monty to choose from, therefore door A did not get any advantage while the fact that out of B & C, Monty chose B increased the probability of door C.)

So in this case, switching won't help

vivek gupta
Monday, April 15, 2002

The trouble with the scenario you describe is that suddenly door A has an extra 33% chance of hiding the prize, even though you, the contestant, didn't do anything.  That's impossible.

Suppose instead of 3 doors, it's 100.  One has the prize; the other 99 have goats.  You pick a door.  Then goats spontaneously come out of 98 other doors.  Should you switch then?

Paul Brinkley
Tuesday, April 16, 2002

This is very much possible that a contestant makes a choice - and later on the probability of that choice increases even if the contestant does not do anything.

For example lets say if there were only two doors A & B , one having prize and the other having goat. Initially the contestant chooses one of the gates lets say A. Clearly the probability of his success is 50%. Now if the goat comes out of the door B, then the probability of prize being behind your door (door A) is now 100%! Thus even when the contestant did not do anything, his chances have increased from 50 % to 100%.

Coming back to more than 2 gates, even if there were 100 doors and 98 goats come out accidentally (and  the door you chose is not one of them), then also switching will not help. Because the only effect of goats coming out is that 98 choices have been removed. The remaining 2 choices still have a proability of 50% each.

vivek gupta
Tuesday, April 16, 2002

I think some of the above posters are correct, it *does* depend on Monty knowing which one has the car, because if he knows, he is giving extra information.  If it is just random, the chance of each remaining door having the prize is equal (in the case of 3 original doors, 50%)

Adrian, think of it this way...

Take a deck of cards, and add a joker.  Shuffle thoroughly, and deal yourself a card face down.  Don't look at it.  The odds of that card being the joker is 1/53.  Start dealing cards face up.  For each card you turn over, the odds of the original card, sitting face up in front of you, being the joker *increase* because you have added knowledge that some number of cards are definitely *not* the joker.  If you continue this so that there is only one card remaining in the deck, the odds that your original card is the joker are 50%, because the other 51 cards are known for sure not to be the joker, and there are only two choices remaining.

It is the difference between looking at the odds that a sequence will happen, and the odds that one particular element will happen.  The odds that you will flip 3 heads in a row are 1/8.  But after flipping 2 heads, the odds that you will flip a 3rd one are 1/2.  IE, the odds that you turned over a joker initially are always 1/53.  But after you see 51 cards that are not the joker, the odds that your particular card is a joker rise to 50%

When the goat randomly walks out of the door, you have added information about the system, so the original probabilities no longer hold.

The reason that the original problem (with Monty revealing one of the doors) works out the way it does is not because of him revealing a goat.  The secret is that he gives out information by *not* revealing a goat in the *other* door.  By him not revealing a goat under that door, the odds that it have the prize are increased.  This does not happen if the goat simply walks out randomly.  If the revealing is random, the odds of each remaining door having the prize are 50%

Mike McNertney
Tuesday, April 16, 2002

Okay, you've all convinced me. I agree that it matters which doors Monty is allowed to open / the goats are allowed to wander out of.

I thought through the fact that there really is a difference between the two scenarios:

Original Monty Hall problem: The _only_ possibility is that one of the doors you _didn't_ pick will be opened (to reveal a goat.)

"New" scenario: _Any_ door might be opened to reveal a goat. (It just so happens that on this particular occasion, the goat came out of one of the two remaining doors that you didn't initially pick.)

Each of these is different to the third scenario which I was also getting confused with, which is: You pick a door. Monty opens _one of the two remaining doors_ at random (sometimes revealing a goat, sometimes a car). (In this particular instance, he revealed a goat.) I think in this scenario you should swap?

Adrian

Adrian Gilby
Tuesday, April 16, 2002

Arrrrrgh...

vivek gupta: "For example lets say if there were only two doors A & B , one having prize and the other having goat. Initially the contestant chooses one of the gates lets say A. Clearly the probability of his success is 50%. Now if the goat comes out of the door B, then the probability of prize being behind your door (door A) is now 100%! Thus even when the contestant did not do anything, his chances have increased from 50 % to 100%."

The contestant's chances were 100% as soon as he picked door A; he just didn't know it.  (The host could ask the contestant if he wants to change his mind, and the chance would still be 50% while the contestant is making up his mind.)  In other words, the chance of winning went from 50% to either 100% or 0% -as- -soon- -as- -the- -contestant- -chose-, and had absolutely nothing to do with any goats.  The contestant very much had to do something to affect his chances.

vivek gupta: "Coming back to more than 2 gates, even if there were 100 doors and 98 goats come out accidentally (and the door you chose is not one of them), then also switching will not help. Because the only effect of goats coming out is that 98 choices have been removed. The remaining 2 choices still have a proability of 50% each."

First of all, the word "accidentally" here is a red herring.  We're only considering the case where 98 doors other than the one chosen are shown to have goats.  Secondly, tell you what:  carry out an experiment.  Get two decks of 52 cards.  Remove all the aces from one of the decks, leaving 100 cards in all.  Shuffle these together.  Have a friend spread the cards out, and ask you to pick one.  We'll say the one remaining Ace of Spades is the "prize"; the other 99 cards are "goats".  Then have your friend look at the remaining 99 cards, and choose any 98 which aren't the Ace of Spades, and flip them over.  Then decide if you want to switch to the remaining face-down card, or stick with the one you have.  Do this 20 times or so and see how often you have the prize.  If you're right, you'd get the ace about half the time by sticking with your choice; if I'm right, you'll be lucky to have the ace at all.

Or more simply, do the original experiment, using three cards.  Again, 20-30 times to get a good sample.

Paul Brinkley
Wednesday, April 17, 2002

Regarding Adrian's post:

The puzzle assumes that Monty always knows everything.  He's the host, so he knows.  (This is supposed "common knowledge" in any game scenario; the host knows how the puzzle is set up, and the only thing host doesn't know is what choices the contestant will ultimately make.)  So yes, Monty will always open a door you didn't pick, and it will always have a goat.

I find it mildly puzzling that people would be concerned whether Monty is revealing a door at random.  First of all, he would never choose to reveal the door the contestant picked, and then ask whether he wants to switch.  That would make no sense.  Secondly, if Monty picks an unchosen door at random, and shows a car, it makes no sense to ask the contestant whether he wants to switch.  Of course he will!  Finally, if the random door hides a goat, we're right back to the original problem.  The -only- case you could make is where the contestant picks the car on the first try, and now Monty has to decide which "goat door" to open.  If Monty flips a hidden coin to decide, then it doesn't matter.  Perhaps if Monty always picked the door farthest to the left, say, then the contestant could use that, but the problem -doesn't say that-.  You have to assume it's random in that case.

Again, even if Monty picked a door at random, sometimes showing a goat, sometimes not, it makes no difference, since you're only asking about the cases where the random event yielded a particular outcome.  It's like saying, "I choose the head side of a coin.  What's on the other side?" and then, later, "I flip a coin.  In the cases where I get heads, what's on the bottom of the coin?"

Paul Brinkley
Wednesday, April 17, 2002

In the variant of the question I mentioned in the beginning of the discssion,  Monty was totally eliminated from the scene. The contestant just chose one of the doors and then out of one of the gates , the goat came out by itself(thats what I meant when I say accidentally). It just happened that that door was not the door the contestant chose.

Coming to the Paul's experiment about two decks. The experiment you are doing is analagous to the orginal monty hall problem and NOT the variant I mentioned. Your friend is choosing 98 cards out of the REMAINING 99 cards- which is exactly what was happening in the orginal problem. And in such cases, switching does help. BUT the correct analogy to my variant will be you choose one of the cards. And then suddenly 98 cards (out of the total 100 cards and NOT 99) got flipped
(BY THEMSELVES) none of them being the desired Ace and it was just by chance that your card was not one amongst them. That is altogether a separate matter that the chances of this event being happening in the first place is very low. But once this happens then the probability of the 2 remaining cards is identical.

vivek

vivek gupta
Wednesday, April 17, 2002

Paul: "I find it mildly puzzling that people would be concerned whether Monty is revealing a door at random. First of all, he would never choose to reveal the door the contestant picked, and then ask whether he wants to switch. That would make no sense. Secondly, if Monty picks an unchosen door at random, and shows a car, it makes no sense to ask the contestant whether he wants to switch. Of course he will! Finally, if the random door hides a goat, we're right back to the original problem."


I agree, if Monty opens a door and reveals a car, the game doesn't make much sense. But it's important to distinguish between a situation where that _could_ have happened but didn't, and a game where revealing a car could _never_ happen.

If we assume that Monty will only ever open one of the two remaining doors, then (contrary to what I originally believed) it _does_ matter whether he picks at random (with some possibility of revealing a car), or is restricted to always showing a goat.

This is the subtle distinction between game types that is a cause of so much confusion.

Game type 1 (originial Monty Hall problem): Monty absolutely must open a goat door, and he can't open the contestant's door. So the only time he can choose randomly between the two doors is when the contestant originally picks the car door. Stick or swap? (Answer: Swap)

Game type 2: Monty randomly picks between the remaining two doors. One time in three, he'll reveal a car, and then we don't know what happens -- perhaps the contestant goes home happy, perhaps they can't have the car, it doesn't matter. But IN THIS PARTICULAR GAME he revealed a goat. Stick or swap? (Answer: Doesn't matter, 50/50 chance).

This is without getting into a discussion of games where Monty can also open the contestant's picked door.

Adrian

Adrian Gilby
Thursday, April 18, 2002

Sorry, one other thing. If Monty was instead told to open one of the doors with a goat, but with _no consideration_ of which door the contestant had picked, then it would be analagous to Vivek's original question.

If these are the game rules, and _given that_ the situation is that he's just opened a door that I didn't pick (or, analagously, the goat just happens to have wandered out from behind one of the two non-picked doors), then I realise that I was wrong to say you should swap. Reasoning: If I picked the "car" door, then the chance of a goat coming out of my door is 0. But if I picked a goat door, the chance of the goat coming out of it is 0.5. Multiplying these probabilities you get:

I picked the car door: 1/3

I picked a goat door, and a goat wandered out of the other goat door: 2/3 * 1/2 = 1/3

I picked a goat door, and the goat wandered out of my door: 2/3 * 1/2 = 1/3, _but this did not occur_.

So we rescale the top two 1/3 probabilities up to 1 again, and get 0.5 for each.

Bayseian probability wraps my head in knots. No matter how hard I try, I can't shake the feeling that my knowledge that something did/didn't occur is altering past events.

Adrian

Adrian Gilby
Thursday, April 18, 2002

Hey, I made a page where you can see the statistical outcomes of playing this game:

http://www.flooble.com/perplexus/boxes.php

(You can chose to either pick a box (door) that definitely doesn't have the prize, or try to have the "goat walk out randomly")

levik
Thursday, April 18, 2002

The scenario where one card is picked out of 100 and 98 unpicked non-aces happen to flip over, and the scenario where one door is picked out of three and a goat happens to trot out of an unpicked door, are both just variations of the scenario I described with the coin.  Another example would be, what's the chance of rolling a 6 on a six-sided die, if a 6 happens to come up on top?

The reason this controlled selection appears to be affecting past events is because it is, sorta.  Consider the three doors.  Here's the numbers.

Pick a random door.  (Actually, the car has to be placed randomly first, but we all seem to agree the probabilities mush together for this step.)
1/3: you picked the car door.
2/3: you picked a goat door.

Random goat jumps out, equal chance for each goat.
1/3: you picked the car, one of the goats trots out.
1/3: you picked a goat, that goat jumped out.
1/3: you picked a goat, the other goat jumped out.

The problem is right here:

We dismiss the cases where our door had a goat.
1/2: you picked the car, one of the goats trots out.
1/2: you picked a goat, the other goat jumped out.

It's like jumping back in time and telling the contestant, "pick a random door, but not that one" and pointing to one of the doors.  Suddenly the contestant only has two doors to choose from at random.  Dismissing cases has the result of backchaining to all previous probabilities, so it's now this:

Pick a random door.  (Actually, the car has to be placed randomly first, but we all seem to agree the probabilities mush together for this step.)
1/2: you picked the car door.
1/2: you picked a goat door.

Random goat jumps out, equal chance for each goat.
1/2: you picked the car, one of the goats trots out.
1/2: you picked a goat, the other goat jumped out.

And things move on from there as you'd expect.  The trouble with this is that -it's not interesting-, at least in my opinion.  It reduces the problem to something trivial.  Again, it's like picking a number from 1 to 10, and asking what the chances are that I picked 2 if we only consider the cases where I picked 2 or 7.

Paul Brinkley
Thursday, April 18, 2002

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