The Monty Hall problem

[Ziggurat]

This can't be true however as, after opening one of the doors, he then has to give you a choice to switch and you have to have a chance of winning the car.

In the usual phrasing of the problem, he must open a door that you didn't pick which doesn't have the car, and he has to give you a chance to switch. And in that scenario, switching is indeed the best odds.

But in the real game, he doesn't have to reveal another door at all. He can just go with what you picked the first time. So if he does open a door and give you a chance to switch, that could be because he knows you picked the wrong door and wants you to win, or it could be because he knows you picked the right door and wants you to lose. Or it could just be to draw the show out for another minute. The choice is no longer obvious.

 

[Rolfe]

This can't be true however as, after opening one of the doors, he then has to give you a choice to switch and you have to have a chance of winning the car. If he reveals the car, then he eliminates it from the game, meaning you lose out on the prize not by your choice, but by something the host did.

No, he has to open a non-winning door.

It depends on how the puzzle is worded. Not all versions are entirely clear. Most versions simply say what Monty does, on this occasion - not the rules he is or isn't working to.

If you stipulate that he always has to open another door irrespective of what you picked, and that this door must not reveal the prize, then switching will double your chances of winning, for sure.

If he simply has to open another door, at random, then switching confers no benefit, but no disadvantage either.

If you don't know which of these two scenarios is the one in operation (that is, you know he had to open a second door, but not whether he was deliberately avoiding the prize), then you should still switch because it wouldn't harm your chances and might benefit.

If you don't know whether he was obliged to open a second door at all, you do not have sufficient information to assess the correct response. He might intend only to offer the choice if you had already selected the prize.

Scrutinise the wording of the puzzle carefully, and if it's unclear, specify in your answer what assumptions you are making.

Rolfe.

 

[ZirconBlue]

My reaction as well.

Why not consider the Monty Hall card game? Monty spreads a deck of 52 standard-deck playing cards on the table, and asks you to pick out the King of Clubs. After you make your selection (and before you verify whether or not you picked correctly), Monty turns over all of the unselected cards except one. The faces of 50 cards are now showing, but the King of Clubs is not one of them. (The cards are marked in such a way that Monty can always be guaranteed to turn over 50 cards that are something other than the King of Clubs.) If Monty gives you a chance to change your mind and take the other face-down card, do you:

a. Take the other card because your odds of winning are REALLY good.
b. Flip a coin because your odds of winning are only 50-50, so you might as well let a coin flip decide.
c. Tell Monty to go to hell, because you feel confident you picked the King of Clubs the first time.
d. Kick Monty in the crotch for playing these stupid games, and ask him by the way why he isn't dead yet.

First a, then d.

Then drive off in the car before the cops get there.

 

[GlennB]

It's a well known issue in the (very excellent) game of Bridge, and is referred to as the principle of restricted choice^WP.

Monty *must* show you a door that contains a goat. He has no choice. This gives you information. The o/p scenario gives no additional information. It's 50/50.

 

[Daald]

The monty hall problem I know of is that he opens up one of the remaining 2 doors which does not contain the prize.

What finally made this click in my head is looking at it backwards.

I chose the strategy that I will always switch.

I pick 2 doors in my head (1 & 2).

I tell monty I choose door 3. (I am switching anyway so I think this is the least likely).

Monty than gives me the right choice between my original 2 doors.

I win (2/3 of the time )

-------

Another way to look at it is to imagine more doors. Imagine there are 1000 doors, you pick one, and then monty opens up 998 wrong ones. Would you switch then?

 

[Squeegee Beckenheim]

Scrutinise the wording of the puzzle carefully, and if it's unclear, specify in your answer what assumptions you are making.

You make good points, as does Ziggurat.

I would say, though, that the assumption that I was working on and, I suspect, the assumption most people are working on, is that it's a fair system. Perhaps that's not a warranted assumption in a game show, but for a mathematical puzzle? Probably not unreasonable.

 

[hgc]

We had a thread on this about 8 years ago. I can't find it now, but I remember it being lively.

 

[GlennB]

You make good points, as does Ziggurat.

I would say, though, that the assumption that I was working on and, I suspect, the assumption most people are working on, is that it's a fair system. Perhaps that's not a warranted assumption in a game show, but for a mathematical puzzle? Probably not unreasonable.

The 'proper' description of the Monty Hall puzzle is that Monty *must* show you a goat door, no matter what (as indicated in the o/p). He only tosses a coin, figuratively speaking, if he has two goats to choose from, in which case it makes no difference anyway.

 

[Rolfe]

Delete. I misunderstood Glenn's post.

Glenn, you can't even say it's 50/50 on the wording of the OP. You have insufficient information. You have to stipulate what rules Monty is working on before you can say anything.

Rolfe.

 

[Rolfe]

You make good points, as does Ziggurat.

I would say, though, that the assumption that I was working on and, I suspect, the assumption most people are working on, is that it's a fair system. Perhaps that's not a warranted assumption in a game show, but for a mathematical puzzle? Probably not unreasonable.

It's not a question of fair, not really. If you say it's fair, that simply means Monty won't deliberately confine his kind offer to the times you've picked the prize.

If you know he has to open a door no matter what you picked, you could still call it fair - and as I said, if that is all you know, switch, because in that case you can't make your chances worse.

However, you also have to know that he is obliged to avoid the prize when he opens the second door, to get the "correct" answer of 2/3rds to 1/3rd. If the second door is chosen at random, the answer is 50/50.

Rolfe.

 

[Squeegee Beckenheim]

It's not a question of fair, not really. If you say it's fair, that simply means Monty won't deliberately confine his kind offer to the times you've picked the prize.

It would be unfair if he did open a door for some people and not for others because he would be giving some people odds of 1/3 of choosing a car, and other people greater odds.

If you know he has to open a door no matter what you picked, you could still call it fair - and as I said, if that is all you know, switch, because in that case you can't make your chances worse.

However, you also have to know that he is obliged to avoid the prize when he opens the second door, to get the "correct" answer of 2/3rds to 1/3rd. If the second door is chosen at random, the answer is 50/50.

If the second door is chosen at random, then the answer is either 50/50 or 0. If he opens the door with the car behind it, then you cannot pick that door and therefore have zero chance of winning the prize.

As I've said, though, I can't see any time that the problem as you describe it would ever be presented as a mathematical puzzle. What you seem to be describing is a scenario where someone would set out a version of the problem in the hope that someone, recognising the problem would say the same things that people in this thread have been saying in order for the person who set the problem to be able to leap out from behind a bush and shout "gotcha!" at them.

 

[69dodge]

I guess what I am trying to find an eloquent way of saying is that since you know ahead of time that you were going to see a goat before you had to make your final choice, you never had 3 choices to begin with, you always had 2 and you just needed to wait for Monty to show you which two doors you really had to choose between.

Yes, after you pick a door and Monty opens another, you have two closed doors to choose from, but who says they're equally likely to hide the car?

The situation is not symmetric between those two closed doors: One door was picked by you, and you don't know where the car is; the other door was picked by Monty, and he does know where the car is. Your closed door was picked entirely at random from all three doors; Monty's closed door was picked from the two remaining doors in such a way that it hides the car if at all possible. (And it is possible if you didn't pick the car initially, which you probably didn't.)

Monty is essentially telling you where the car is, if he can, and he probably can.

 

[Rolfe]

It would be unfair if he did open a door for some people and not for others because he would be giving some people odds of 1/3 of choosing a car, and other people greater odds.

Yes, that's why I said that for a fair game you assume he always opens another door. It's not stated in the problem though.

If the second door is chosen at random, then the answer is either 50/50 or 0. If he opens the door with the car behind it, then you cannot pick that door and therefore have zero chance of winning the prize.

Indeed, but in the iteration described in the puzzle it is 50/50, because in that case he has not revealed the prize.

As I've said, though, I can't see any time that the problem as you describe it would ever be presented as a mathematical puzzle. What you seem to be describing is a scenario where someone would set out a version of the problem in the hope that someone, recognising the problem would say the same things that people in this thread have been saying in order for the person who set the problem to be able to leap out from behind a bush and shout "gotcha!" at them.

Indeed. But it's where you find the conversation heading if you argue on forums like this, without making the ground rules clear. Some smart-Aleck will come along and point out that the problem doesn't specify what rules Monty is working to, or that he has to work to the same rule each time, and that indeed the real Monty Hall didn't work to the same rule each time.

So yah-boo-sucks, your clever explanation of how the answer is 2/3rds vs. 1/3rd is wrong.

So to avoid these smart-Alecks, I insist that the ground rules be clarified, and point out that if these are different from what we assume, the answer is indeed different.

Rolfe.

 

[Rasmus]

The monty hall problem I know of is that he opens up one of the remaining 2 doors which does not contain the prize.

What finally made this click in my head is looking at it backwards.

I chose the strategy that I will always switch.

I pick 2 doors in my head (1 & 2).

I tell monty I choose door 3. (I am switching anyway so I think this is the least likely).

Monty than gives me the right choice between my original 2 doors.

I win (2/3 of the time )

Brilliant!

 

[Archangel]

Alan Davies has the Monty Hall problem explained to him here, it also gives a demonstration of why you should change.

http://www.youtube.com/watch?v=o_djTy3G0pg

 

[Beerina]

It's a well known issue in the (very excellent) game of Bridge, and is referred to as the principle of restricted choice^WP.

Monty *must* show you a door that contains a goat. He has no choice. This gives you information. The o/p scenario gives no additional information. It's 50/50.

This is the key. Monty knows which door has the goat behind it, so he can open the other one, and indeed must, or else the trick doesn't work.


Imagine 100 doors instead of 3. You pick one. Now Monty opens 98 other doors.

Now the goat is behind either the one you picked, or the other door. As there were 99 other doors, the door you picked has just a 1 in 100 chance, the other 99 doors have a 99 of 100 chance that one of those has the goat.

But...Monty opened all the doors, knowing which ones didn't have the goat. So the one remaining door he didn't open (and you didn't pick) therefore inherits the entire 99% chance of having the goat. For any 99 doors, you can always open 98 of them without showing a goat. Always.


It seems odd with 3 doors (in which case your door has 1/3 chance and the other two 2/3, of which the one he doesn't open inherits the full 2/3 chance) which is why the 99/100 helps to understanding.


Technically speaking, you could also think of it as Monty giving you a chance to keep the door you picked, or open all 99 of the other doors. In that sense, it's a no brainer to pick (or avoid) the goat as the case may be.

 

[Loss Leader]

The question: Mathematically, how would the answer be different if I refused to make my initial selection of door until after the first donkey door had already been revealed? (meaning there is no "switch" but now you have to pick one of the two remaining doors.)

You'd be choosing between TWO doors instead of THREE. So, mathematically, the answer would be different because there are fewer doors. If you refused to make your choice until Monty had opened up both donkey doors, your choice would be different mathematically as well. You'd be choosing one door out of one door.

 

[pgwenthold]

This can't be true however as, after opening one of the doors, he then has to give you a choice to switch and you have to have a chance of winning the car. If he reveals the car, then he eliminates it from the game, meaning you lose out on the prize not by your choice, but by something the host did.

No, he has to open a non-winning door.

Depending on how it is worded, he may not have to open a door at all. He just does.

Unless specified, you don't know why. Maybe he does it everytime. Maybe he doesn't do it everytime. If he doesn't do it everytime, he could do it on random occasions. Or maybe not randomly.

If this is stated explicitly in the problem, fine. If not, anything that is done is a guess.

 

[pgwenthold]

You make good points, as does Ziggurat.

I would say, though, that the assumption that I was working on and, I suspect, the assumption most people are working on, is that it's a fair system. Perhaps that's not a warranted assumption in a game show, but for a mathematical puzzle? Probably not unreasonable.

Oh, for a mathematical puzzle, it is most certainly unreasonable to assume information that is not provided.

You are making the same mistake that got Marilyn Savant ripped to high heaven. Logicians tore her to shreds over this.

They pointed out, absolutely correctly, that the problem, as worded, was not solvable.

In mathematics, if you don't have the information, you can't just make it up and claim that is the answer.

 

[theprestige]

There's two goats and one car. I have 2/3 chance of picking a goat. Monty shows me the other goat. So now I'm pretty sure I've found both goats. The one remaining door probably has the car. That's about as far as my math goes, or needs to go, to solve this problem.

The rest is up to Monty.