Monty Hall has me a little confused

[DougShrapnel]

Ok lets say there are 4 doors to pick from, after you pick a door. 1 door that doesn't contain the prize is shown to you. How does that effect the remaining 2 doors' odds?

I see three options that are possible

1. All 3 remaining doors have a 1/3 chance. Which I think is incorrect.

2. Orginal door keeps it's 1/4 value. If you switch to a new door it gets a 1/2 value and the remaining door gets a 1/4 value.

3. Original door keeps it's 1/4 value and the two remaining doors get a 3/8 value.

I'm guessing the correct response is number 3, but would not be surprised if it is 1 or 2.

Edit to add a question

Is number 1 a baysenian approach?
Is number 2 a Monty hall grouping approach?
What is number 3 called?

[bunny]

Number 3 is called correct

Numbers 1 and 2 are called incorrect approaches

[DougShrapnel]

Post deleted by DougShrapnel

[TomCollins]

[ QUOTE ]
Well now it looks like I'm switching to number 1
C= Car R = Reveal G = Goat
C R G G
C G R G
C G G R
G C R G
G C G R
G R C G
G G C R
G R G C
G G R C

[/ QUOTE ]

You are way off. The odds of getting it right on the first try is 1/4. According to your diagram, it is 1/3.

[DougShrapnel]

Thanks Tom I tried to delete but you were to quick

[maurile]

[ QUOTE ]
3. Original door keeps it's 1/4 value and the two remaining doors get a 3/8 value.

[/ QUOTE ]
Correct. The host will always show you an empty door, whether you pick correctly at first or not. So the fact that he showed you an empty door tells you nothing about whether you picked correctly. Your initial choice therefore still has the same 1/4 of being correct as it always has.

But the door that was shown empty now has a zero chance of being correct. It's 1/4 has to go somewhere. Since it can't go to the door you picked (for the reasons stated above), it goes to the two remaining doors, which each get an additional 1/8.

[KipBond]

[ QUOTE ]
The host will always show you an empty door, whether you pick correctly at first or not.

[/ QUOTE ]

And that is key. The host knows which door the prize is in, and will always show you an non-prize door after you pick a door. That's why your odds increase by switching -- because you are taking advantage of the host's information.

[KipBond]

Has anyone seen the similar game show "Deal or No Deal" (http://www.nbc.com/Deal_or_No_Deal/) on NBC?

Is this a simple EV problem? I've seen one time where the bank actually offered more than the EV -- and the player didn't take it. Every other time, the offer is always less than the EV. Unless I'm calculating the EV wrong -- which I think I might be.

Has anyone analyzed the game structure to find the best strategy? I figured you'd just average the $$ amounts on the board to find your EV, and take the bank offer if it's greater (or close to) that amount. But, if the bank is always offering less than the EV, then in the end, you'll just end up with the case you chose at the beginning -- which before you ever opened any cases, was about $100K (I believe).

So, unlike the Monty Hall game, I think your EV for staying with your first selected case actually increases (or decreases) as other cases are revealed. Right?

[Borodog]

One of the best "gut-level understanding" explanations of the Monty Hall concept that I've seen was to do just this, expand the number of doors, and keep opening doors until you were down to two, the originally chosen door and one other.

The guy who was explaining it did it with playing cards. He took a whole shuffled deck and fanned it out in front of him so only he could see the faces. He asked a guy to draw a card, with the object of trying to draw the Ace of Spades, and lay it face down on the table, unseen.

Then one by one he exposed and discarded cards, but never the Ace of Spades, until he had only one card left. Then he asked the guy if he wanted to switch.

[tolbiny]

The Ev of the game show is to interview and select people who are more likely to gamble, and thus make the game show more excitting/popular/profitable.
Otherwise it should be very easy to calculate the EV of every situation, by just taking the average of all the prizes left.
On the other hand wouldnt' you have to know what tax bracket each prize would put you into?

[BruceZ]

[ QUOTE ]
Ok lets say there are 4 doors to pick from, after you pick a door. 1 door that doesn't contain the prize is shown to you. How does that effect the remaining 2 doors' odds?

I see three options that are possible

1. All 3 remaining doors have a 1/3 chance. Which I think is incorrect.

2. Orginal door keeps it's 1/4 value. If you switch to a new door it gets a 1/2 value and the remaining door gets a 1/4 value.

3. Original door keeps it's 1/4 value and the two remaining doors get a 3/8 value.

I'm guessing the correct response is number 3, but would not be surprised if it is 1 or 2.

Edit to add a question

Is number 1 a baysenian approach?
Is number 2 a Monty hall grouping approach?
What is number 3 called?

[/ QUOTE ]

I find it easiest to think about these problems this way. Say you picked door 1, and he showed you that door 4 was empty. If you have the prize, then he chose door 4 randomly from 3 doors, but if you don't have the prize, then he chose door 4 randomly from just 2 doors, since he can't show your door or the prize door. It is 3/2 times more likely to choose door 4 from 2 doors than from 3 doors. That means that each of the other 2 doors are 3/2 times more likely to hold the prize as your door. Since these must sum to 1, your door is 1/4 (as it was at the beginning), door 2 is 3/8, and door 3 is 3/8.

This is still an application of Bayes' theorem, but with a focus on odds rather than conditional probabilities, which I believe is more intuitive.

Choice 1 would only be correct if door 4 were chosen randomly from all 4 doors, meaning that it was possible for your own door or the prize door to be revealed. This is different from Monty Hall, where the host knows where the prize is and will not open the prize door or your door. Choice 2 is not correct, but with 3 door Monty Hall, your door is 1/3 while the remaining door is 2/3.

[DougShrapnel]

Boro, Kip, and Bruce thanks for the replies. I don't feel so stupid for getting a litlle confused. It appears that in '91 when this problem was first introduced in a major publication a majority of Ph.D.s didn't think it was correct either.

[ctj]

[ QUOTE ]
Boro, Kip, and Bruce thanks for the replies. I don't feel so stupid for getting a litlle confused. It appears that in '91 when this problem was first introduced in a major publication a majority of Ph.D.s didn't think it was correct either.

[/ QUOTE ]

In the game Bridge, this is known as the "Principle of Restricted Choice", and has been well known for some time (at least since the 1960s). By the way, the problem as posed assumes that Monty will always show you an empty door.

Regards,

C.T. Jackson