OT: The Monty Hall Problem

[dare2be]

I'm definitely not super smart or anything, but even a tart should understand that scenario.

Are you calling a PhD, Professor of Mathematics from Florida State University, less than a tart?

 

[SLY]

No, I mean if you look into the problem and start adding added twists and turns to the problem then it is arguable. But when it is laid out to you, cut and dry, what is there to argue? You make a selection and have a 66% chance of picking the wrong door. He takes one of the incorrect choices away, and it becomes 50/50. Going by odds, you would change your answer because originally you had a greater chance of picking the wrong door.

 

[dare2be]

This is the "best" counter-argument that I have read. It's still fallacy, but it almost makes sense:

I still can't see how the odds for the second choice could be anything but
50/50. Sure, Monty deliberately avoided the winning door. But that door
could well be the one you've already chosen. In fact, as I see it, the odds
were really 50/50 even before the first choice was made. Since we know in
advance that Monty is going to open one of the doors, and we also know that he
will always choose a goat, then there was never really a choice of three doors.
In effect, the first choice doesn't count, because whatever choice is made will
eliminate one of the goats. So effectively, the real choice is always going to
be between the other two doors, one of which hides the winner, and there'll be a
1 in 2 chance of choosing it.

 

[ritari330]

That actually makes sense, still wrong, but makes sense as to what they are thinking

 

[Destroydacre]

That actually makes sense, still wrong, but makes sense as to what they are thinking

Makes sense if you don't understand the problem, but I can see why some people would fall for it. It's not 50/50 because the prize was already assigned when there were three doors, not when it's down to two. Similarly, if there were two doors, you picked one and then the prize was assigned and you switched, you'd be right 100% of the time!

 

[juliansteed]

Darn, I wanted people to disagree over this, now it's just a thread over how and why other explinations are wrong

I agree. That's why I simply answered the question without getting into details. Also I referenced "21" to see who would disagree with me in saying it is wrong. That's how you see who really understands the problem and who doesn't but just goes along with their explanation.

 

[juliansteed]

This is the "best" counter-argument that I have read. It's still fallacy, but it almost makes sense:

I still can't see how the odds for the second choice could be anything but
50/50. Sure, Monty deliberately avoided the winning door. But that door
could well be the one you've already chosen. In fact, as I see it, the odds
were really 50/50 even before the first choice was made. Since we know in
advance that Monty is going to open one of the doors, and we also know that he
will always choose a goat, then there was never really a choice of three doors.
In effect, the first choice doesn't count, because whatever choice is made will
eliminate one of the goats. So effectively, the real choice is always going to
be between the other two doors, one of which hides the winner, and there'll be a
1 in 2 chance of choosing it.

The problem here is that this person fails to realize that there is a 2/3 chance that the door Monty choses to reveal is his only option because the car is behind the other unselected door. There is only a 1/3 chance you've already selected the winning door. This argument really isn't much more than what they expect you to think.

 

[dare2be]

ok, how about this? Anything to generate some kind of controversy...someone PM'ed me saying I was being a MD

-----

We've all pretty much got the basic Monty question down now, so let's move on to a slightly more complicated version (which I suppose we should call the Full Monty).

Three contestants, three doors. Each contestant picks a door. No door can be picked by more than one contestant.

Monty opens one (guaranteed to be non-winning) door, and eliminates the player who chose that door.

Now there are two contestants left who each have an unopened door.

Each appears to be in the same position as the contestant in the original problem. Each has chosen a door and seen one of the remaining doors opened.

Shouldn't they *both* think that they should switch to the *other's* door in order to increase their chances of wining from 1/3 to 2/3?

Should they trade doors or not?

 

[juliansteed]

ok, how about this? Anything to generate some kind of controversy...someone PM'ed me saying I was being a MD

-----

We've all pretty much got the basic Monty question down now, so let's move on to a slightly more complicated version (which I suppose we should call the Full Monty).

Three contestants, three doors. Each contestant picks a door. No door can be picked by more than one contestant.

Monty opens one (guaranteed to be non-winning) door, and eliminates the player who chose that door.

Now there are two contestants left who each have an unopened door.

Each appears to be in the same position as the contestant in the original problem. Each has chosen a door and seen one of the remaining doors opened.

Shouldn't they *both* think that they should switch to the *other's* door in order to increase their chances of wining from 1/3 to 2/3?

Should they trade doors or not?

This is an interesting question and 1 that I can see a lot of people who don't accept/understand the Monty Hall problem (not saying that is you) posing.

The difference here is that no matter who picks what door, it is always going to play out the same. 1 contestant will pick the car and 2 will pick goats. As far as we know Monty doesn't care which one it is. So he's always going to have to randomly select between the 2 goats and eliminate 1 of the contestants. By doing this, he increases each of the remaining players' odds from 1/3 to 1/2.

Where this differs from the Monty Hall problem is that 2/3 of the time the 2 remaining doors are not equal and Monty does not have a choice which one to reveal. 2 out of 3 times the player will select 1 of the goats leaving Monty no choice but to reveal the other goat.

 

[ritari330]

You PM'd was me first beme, and I said you were MDing it up

 

[dare2be]

You PM'd was me first beme, and I said you were MDing it up

hey, if I'm gonna MD it up, why would I let facts get in the way :p

 

[Dacks]

Classic problem. I presented it to the high school stats class I once taught and they bombed horribly.

Here's an intuitive example of why you should always switch, that usually convinces people. Instead of cars and goats behind doors, hide a quarter under one of three cards. Perform the 'gameshow'. Next, do the same example with 10 cards and one quarter. After they pick a card, you remove eight of the 'bad cards'. At this point most people know intuitively to switch, and even the most skeptical won't suddenly think their odds jumped from 1/10 to 1/2.

 

[dare2be]

Here is a proposed "solution" to the modified problem I posed...it seems correct but I'm not 100% sure this reasoning is accurate:

It depends:

IF: the THIRD player was eliminated, player 1 should switch and player 2 should stay

LIKEWISE: if the SECOND player was eliminated, p1 should switch and p3 should stay

HOWEVER: if P1 was eliminated, there is no advantage in switching. [err, I think...]

Remember, the KEY word was that each player APPEARS to be "in the same position", but that's not really true. ONLY the first player to pick is "in the same position" -- remember, each player has to pick a different door
[this is a NEW CONSTRAINT that is ADDED by this variation], so players #2 & #3 don't get the same odds as player 1 [more on this later]

At the outset, it CAN BE SHOWN that each player has a 1-in-3 chance of being correct:

* P1 picks a door -- 1-in-3 he's right; players 2 & 3 now pick from doors that COLLECTIVELY have a 2 in 3 chance of being correct.
* P2 picks a door. Since he cannot choose the door that P1 chose [due to that pesky NEW CONSTRAINT], he has 1-in-2 chance of picking what still might be "the winning door", and that "winning door" will be found 2/3rds of the time; 50% of 2/3rds is [amazingly] 1-in-3.
* P3 has no choice -- he is given the final door and STILL has the "same" 1-in-3 as everyone else. [because players 1 & 2 ALSO COLLECTIVELY have a 2/3rds chance of being correct "so far"]

Now, Mr. Hall INTRODUCES NEW INFORMATION which CHANGES THE STATE OF THE SYSTEM
suppose Mr. H eliminates P3 (who had no choice in the matter), where do we stand now?
-- P1 chose 1 from THREE unknowns [1/3 of the 3-out-of-3-potentially-winning doors]
-- P2 chose 1 from TWO unknowns [1/2 of the REMAINING 2-of-3 doors, or 1/3]
-- P3 chose 1 from ONE unknown [ALL of the final 1-of-3 doors, again 1/3] and was eliminated. This gives us NEW INFORMATION about P2's choice:
P2 chose the "good" half of the 2-in-3, which for all practical purposes is now "all" of the 2-in-3 chance that the second and third doors COLLECTIVELY had of being "the correct door". P1 switches provided he can convince P2 to switch [who, if being the "astute" reader of this thread he is, has read THIS message & refuses to do so... ]

Should Mr. H eliminate P2, then P3 should hold firm for the above reason. [this time p3 was "forced" to take the "good" half of the (collective) 2-in-3 chance]

Finally, we get to the case of P1 being eliminated. this NEW INFORMATION tells us something quite interesting:
-- P2 chose 1 from TWO possibilities, which we NOW KNOW have a 100% chance of COLLECTIVELY containing the winning door, not 2/3'rds as before. With this NEW INFORMATION, we NOW KNOW P2 has a 1-in-2 chance
-- likewise, P3 had the other 1-in-2 of what is NOW 100% guaranteed to be behind either the second or third door chosen.

It is this final case where it makes no difference to switch.

 

[dash]

This thread has confirmed for me that I need to see the movie 'The Men Who Stare at Goats'

 

[ritari330]

I heard that movie sucked

 

[juliansteed]

Here is a proposed "solution" to the modified problem I posed...it seems correct but I'm not 100% sure this reasoning is accurate:

It depends:

IF: the THIRD player was eliminated, player 1 should switch and player 2 should stay

LIKEWISE: if the SECOND player was eliminated, p1 should switch and p3 should stay

HOWEVER: if P1 was eliminated, there is no advantage in switching. [err, I think...]

Remember, the KEY word was that each player APPEARS to be "in the same position", but that's not really true. ONLY the first player to pick is "in the same position" -- remember, each player has to pick a different door
[this is a NEW CONSTRAINT that is ADDED by this variation], so players #2 & #3 don't get the same odds as player 1 [more on this later]

At the outset, it CAN BE SHOWN that each player has a 1-in-3 chance of being correct:

* P1 picks a door -- 1-in-3 he's right; players 2 & 3 now pick from doors that COLLECTIVELY have a 2 in 3 chance of being correct.
* P2 picks a door. Since he cannot choose the door that P1 chose [due to that pesky NEW CONSTRAINT], he has 1-in-2 chance of picking what still might be "the winning door", and that "winning door" will be found 2/3rds of the time; 50% of 2/3rds is [amazingly] 1-in-3.
* P3 has no choice -- he is given the final door and STILL has the "same" 1-in-3 as everyone else. [because players 1 & 2 ALSO COLLECTIVELY have a 2/3rds chance of being correct "so far"]

Now, Mr. Hall INTRODUCES NEW INFORMATION which CHANGES THE STATE OF THE SYSTEM
suppose Mr. H eliminates P3 (who had no choice in the matter), where do we stand now?
-- P1 chose 1 from THREE unknowns [1/3 of the 3-out-of-3-potentially-winning doors]
-- P2 chose 1 from TWO unknowns [1/2 of the REMAINING 2-of-3 doors, or 1/3]
-- P3 chose 1 from ONE unknown [ALL of the final 1-of-3 doors, again 1/3] and was eliminated. This gives us NEW INFORMATION about P2's choice:
P2 chose the "good" half of the 2-in-3, which for all practical purposes is now "all" of the 2-in-3 chance that the second and third doors COLLECTIVELY had of being "the correct door". P1 switches provided he can convince P2 to switch [who, if being the "astute" reader of this thread he is, has read THIS message & refuses to do so... ]

Should Mr. H eliminate P2, then P3 should hold firm for the above reason. [this time p3 was "forced" to take the "good" half of the (collective) 2-in-3 chance]

Finally, we get to the case of P1 being eliminated. this NEW INFORMATION tells us something quite interesting:
-- P2 chose 1 from TWO possibilities, which we NOW KNOW have a 100% chance of COLLECTIVELY containing the winning door, not 2/3'rds as before. With this NEW INFORMATION, we NOW KNOW P2 has a 1-in-2 chance
-- likewise, P3 had the other 1-in-2 of what is NOW 100% guaranteed to be behind either the second or third door chosen.

It is this final case where it makes no difference to switch.

I disagree with this solution. According to this solution, when P3 is eliminated all of its odds of being the correct door are being transferred to P2 which is wrong. 1/6 (half of the 1/3) should be transferred to each of the remaining doors, making each door 1/2.

The only way this works is if we know that it is a given that P1 is NEVER the player to be eliminated. When this is the case, it essentially puts P1 in the same position as the contestant in the Monty Hall Problem. But obviously this is not the case since there is no mention of this in the problem and the solution even makes reference to the fact that P1 might be eliminated.

 

[Jimmy_the_Tongue]

Fuck Monty Hall...Chuck Barris was were it's at....

 

[dare2be]

I disagree with this solution.

The only way this works is if we know that it is a given that P1 is NEVER the player to be eliminated. When this is the case, it essentially puts P1 in the same position as the contestant in the Monty Hall Problem.

Good catch...this solution always bugged me as being wrong but I couldn't put my finger on it to give a definitive rebuttal. You exposed it for what it is with just that one constraint.

 

[Dacks]

This is an interesting question and 1 that I can see a lot of people who don't accept/understand the Monty Hall problem (not saying that is you) posing.

The difference here is that no matter who picks what door, it is always going to play out the same. 1 contestant will pick the car and 2 will pick goats. As far as we know Monty doesn't care which one it is. So he's always going to have to randomly select between the 2 goats and eliminate 1 of the contestants. By doing this, he increases each of the remaining players' odds from 1/3 to 1/2.

Where this differs from the Monty Hall problem is that 2/3 of the time the 2 remaining doors are not equal and Monty does not have a choice which one to reveal. 2 out of 3 times the player will select 1 of the goats leaving Monty no choice but to reveal the other goat.

Ding ding ding!

 

[juliansteed]

Good catch...this solution always bugged me as being wrong but I couldn't put my finger on it to give a definitive rebuttal. You exposed it for what it is with just that one constraint.

Its been a while since I've taken any math/stats classes so I'm sure there is a better way to explain in mathmatical terms but I'm glad my explanation made sense to you.