The Monty Hall problem

[jazzmojo]

I apologize if this has been asked before but my search didn't find what I was looking for. I'm sure many already know what this is, but I'll restate:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
—Whitaker/vos Savant 1990

The answer is that you should switch as it doubles your odds of winning the car from 1/3 to 2/3.

Here is my problem and I follow up with the question. I know from Monty's explanation that behind two curtains are donkeys and behind one is a car. I know my odds are 2/3 donkey and 1/3 car. I know that no matter which door I pick, Monty is going to open one of the other two doors I don't pick to reveal, gasp, a donkey. It makes no difference if I've picked a donkey door or a car door, at least one of the other two doors is a donkey and I'm going to see one of those donkeys no matter what. So, before I choose anything, I know I will be shown a door with a donkey no matter what I do. Since there is no way I can know what's behind any door, it doesn't matter which one I pick since after I am shown the donkey door I will be allowed to choose either of the remaining doors where 1 of the 2 has a donkey and the other 1 of the 2 has a car (the wrong 50/50 chance).

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.)

{In my mind, the choosing door first method does not add any information since you already know before you pick that you will be shown a donkey and then be given the choice of either of the remaining two doors.}

 

[nathan]

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.)

In that case you're choosing from 2 options, each equally likely, because the host has not revealed any information about the doors he didn't open -- you know one of them has a goat and one of them has a car.

So your chance is 50%.

ETA:Neither have you constrained which door the host can choose to open. In the original problem, your first choice constrains what the host can do.

 

[Jack by the hedge]

The offer to change is equivalent to saying "do you want to keep your original door, or would you prefer to have both the other doors?". The odds there are clearly 1/3 vs 2/3. By opening one losing door, the host reduces those two doors to a single choice, but considered as a pair they still retain their 2/3 probability.

If you don't make an initial selection, you don't know which door is paired with that opened door, so your chances are 50:50.

 

[summand]

Jepp. 50:50
In the first example, new information is introduced. If a donkey and a car are behind the doors, the gamemaster has to choose the door with the donkey.

In your new example he can choose between the two doors freely and this introduces no new information.

 

[Guybrush Threepwood]

Here you go, 15 pages of squabbling and nit picking, all the answers are there somewhere.

Monty Hall Problem

 

[GlennB]

It's 50/50.
There are different ways to view why it's different odds from the standard puzzle, but one is that Monty might be showing the contents of the door you would have picked if you had chosen in advance. In the standard method he is obliged to show you a goat from two doors, and sometimes he has no choice - the other conceals a car. Here he always has a choice.

 

[jazzmojo]

Here you go, 15 pages of squabbling and nit picking, all the answers are there somewhere.

Monty Hall Problem

Ah, thank you for that!

Here is a response that seems to follow my thoughts:

Here's the analogy that convinced me.
Say you pick one of the three doors.
Instead of showing you that one of the two doors remaining has a goat behind it (which, to be honest, you already knew; you just didn't know which one), Monty gives you the option to select both remaining doors in exchange for the door you just picked.

What would be the odds then?
They would be exactly the same as in the original problem.

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.

 

[ehcks]

Ah, thank you for that!

Here is a response that seems to follow my thoughts:



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, you do have two choices, but they aren't 50/50. You choose your first door, and it has a 33% chance of being right, meaning that you have a 66% chance of being wrong. One of the other doors is opened with a donkey. You still have a 66% chance of being wrong, or equivalently, you have a 66% chance of being right if you switch.

The best "strategy" then, is to imagine you're trying to choose the wrong door intentionally, so you lose that urge to stick with your first choice.

 

[Rasmus]

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,

But you did!

You get to chose freely from three doors. That that are three choices and not two, regardless of whatever may happen afterwards,

you always had 2 and you just needed to wait for Monty to show you which two doors you really had to choose between.

No.

In the majority of cases, your choice will leave Monty without a choice.

It's very simple. Let us assume you have three doors: Red, Blue and Green.
You initially pick Red, knowing that your chances of holding the car are now 1/3. More often than not you will not have the car at this time.

Now let us assume that Monty reveals the goat behind the Blue door: Since you just figured out that more often than not you will not have picked the car, it follows that it is behind one of the remaining two doors. In most cases, Monty had to show you the door he did, because the car was behind the other door.

I struggled with this forever - here are two things that might help you:

1. Play the game. Use matchboxes and a friend, a computer simulation or whatever else you can. See how often it would have been better for you to change the door.

Simple solution: Throw a die for your initial choice:
1 or 2: Red
3 or 4: Blue
5 or 6: green.

Now throw a die to place the car behind one of the three doors using the same method.

Now play Montys part and reveal a door. If you still have a choice, use the die to dtermine which of the remaining doors to open.

Now keep track of the number of times where changing your initial choice would have been better or worse.


2. Change the puzzle slightly to make things more obvious:
You have a million doors. There is a car behind one of the million doors and goats behind all other doors.

You pick a door, say #15

Now Monty does what he always does: He opens all the doors, except for the door you picked and one other door. He goes from door to door, opening them one by one, revealing goat after goat after goat ... but he skips doors #15 and #584696.

In the end, you are a given a choice: Stick to #15 or change to #584696. Now ask yourself this: Do you really, honestly think that Monty just so happened to skip that one particualr door, or did he skip that door rather than one of the 999,999 doors because it had the car behind it? Would you really think youz were so incredibly lucky when you first picked the door that you actually got the car? Out of a million possible doors?

 

[Paul C. Anagnostopoulos]

I love this problem because the best way of understanding it seems to vary from person to person.

The explanation that Monty is giving you the chance to switch to both of the remaining doors makes it so obvious to me. And yet I explain that to some people and they look at me funny.

Monty says: Would you like both of the remaining doors? You can't actually open both, so I'll open one of them for you. (He has to open one with a goat. If he opened one at random he might open the one with the car and blow the whole deal.)

~~ Paul

 

[Paul C. Anagnostopoulos]

Rasmus, that's the best description of the million-door explanation that I've seen. It becomes similar to the "both of the remaining doors" explanation.

~~ Paul

 

[Rolfe]

Oh God not again.

Switch.

Rolfe.

 

[Brown]

Oh God not again.

Switch.

Rolfe.

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.

 

[Squeegee Beckenheim]

I always think that the best way to look at it if you're having trouble picturing it is to simply write out all the options:

1 2 3
C G G
G C G
G G C
Now pick a door, any door and work through all the options if you switch. Let's say you pick door 1.

Scenario A: You have chosen a door with a car. The host opens one of the two doors with goats. You switch. You now have a goat.
Result: You lose.

Scenario B: You have chosen a door with a goat. The host opens the door with the goat. You switch. You now have the car.
Result: You win.

Scenario C: You have chosen a door with a goat. The host opens the door with the goat. You switch. You now have the car.
Result: You win.

So switching gives you a 2 out of 3 chance to win the car.

 

[Rolfe]

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.

That's roughly how I see it, except your presentation is more elegant.

Rolfe.

 

[Rolfe]

There's an extra little wrinkle of course. Maybe Monty did pick one of the other two doors at random, and it was just chance he didn't reveal the car. So your chances aren't improved if you switch.

You still want to switch, because you don't know. Either switching will double your chance, or it won't make any difference.

The only way you don't want to switch is if Monty will only offer the switch if he already knows you picked the car. The way the puzzle is set up makes it reasonably clear this interpretation is not intended.

However, if you want to be really picky, and include all possible scenarios as that last one, the only answer is "insufficient data".

The thing I like about this puzzle is that the solution involves knowing the intention of the other party, which is counter-intuitive.

Rolfe.

 

[lomiller]

I apologize if this has been asked before but my search didn't find what I was looking for. I'm sure many already know what this is, but I'll restate:



The answer is that you should switch as it doubles your odds of winning the car from 1/3 to 2/3.

Here is my problem and I follow up with the question. I know from Monty's explanation that behind two curtains are donkeys and behind one is a car. I know my odds are 2/3 donkey and 1/3 car. I know that no matter which door I pick, Monty is going to open one of the other two doors I don't pick to reveal, gasp, a donkey. It makes no difference if I've picked a donkey door or a car door, at least one of the other two doors is a donkey and I'm going to see one of those donkeys no matter what. So, before I choose anything, I know I will be shown a door with a donkey no matter what I do. Since there is no way I can know what's behind any door, it doesn't matter which one I pick since after I am shown the donkey door I will be allowed to choose either of the remaining doors where 1 of the 2 has a donkey and the other 1 of the 2 has a car (the wrong 50/50 chance).

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.)

{In my mind, the choosing door first method does not add any information since you already know before you pick that you will be shown a donkey and then be given the choice of either of the remaining two doors.}

Monty is giving you more information if you select a door and then shows you a donkey behind one of the other two then if he simply eliminates one of the donkeys straight off. If you select a door first there is a 2/3 chance he is telling you exactly which door has the car whereas if he simply eliminates a door at the start he is only giving you 50:50 odds. (If he simply eliminated a door at the start he always has 2 to choose from so there is less weight to which one he opens. )

 

[Rasmus]

Rasmus, that's the best description of the million-door explanation that I've seen.

That's how it was explained to me when I finally got it, after it's been explained to me several times over in different ways that just left me confused.

It becomes similar to the "both of the remaining doors" explanation.

It is equivalent, yes. But with just three doors you have that terrible, over-intuitive 50/50 lingering in your head. And the 1/3 you start out with is confusingly close to that as well. More doors made it obvious enough to me to see what it all meant. (Mostly, it is the bit of "open all doors but the one you picked and one other door - how likely is it that the other door was picked randomly?" - much better than one of two remaining doors - which is 50/50 yet again.)

 

[Squeegee Beckenheim]

There's an extra little wrinkle of course. Maybe Monty did pick one of the other two doors at random, and it was just chance he didn't reveal the car. So your chances aren't improved if you switch.

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.

 

[roger]

jazzmojo, before your own thread reaches 15 pages, let me make this very clear: if you write a computer program that automatically switches the choice after the door is revealed, the computer wins 66% of the time. Period. So we know switching is the correct strategy.

So, your only task in this thread is to figure why you are wrong. Squeegee gave the perfect explanation, btw, and rasmas gave an excellent intuitive way to look at it.