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monty hall goat problem

no way, I am sure that  with goat shown one still has a fifty-fifty. This,  of course, is the ignorant reply but I think that the goat being shown merely proves that you have a fifty fity at the moment rather than your choice still giving a 2/3 chance.

I would love to be proven wrong.

Sam Swanson
Friday, May 16, 2003

Hi Sam

This turns up every month or so on the board, so I'll post the most popular explanations.

http://mathforum.org/dr.math/faq/faq.monty.hall.html
http://www.sjsu.edu/faculty/watkins/mhall.htm
http://www.jimloy.com/puzz/monty.htm

My personal favourite explanation is to imagine that instead of showing you a goat, Monty offers to let you 'keep' your first choice or 'switch'. If you switch, he says, and the prize is behind either of the doors you didn't pick, then you will get it.

It takes a little thought to realise that there is no difference between that and what he actually does. Once you do realise that it becomes clear that switching will always get you the prize when your first guess was wrong (2/3) and never when it was right (1/3).

David Clayworth
Friday, May 16, 2003

Thank you for your reply.

Of course the answer is obvious. 1/3 and 2/3

I (like all probably) love these things for the very way in which the initial instinct is proven wrong by using a little thought.

I must try to apply that to other areas in life...

Sam Swanson
Monday, May 19, 2003

How about this explanation:

There are 4 possible "paths" this game can go - I am referring to the table in the explanation on the techinterview website - let's say it's case 1 ($ behind door 1):

1. you pick door 1, Monty opens door 2
2. you pick door 1, Monty opens door 3
3. you pick door 2, Monty opens door 3
4. you pick door 3, Monty opens door 2

I know paths 1 & 2 seem equivalent, but they are NOT the same path. So all those four paths are just as likely. If you change your mind path 3 & 4 wins. If you don't, path 1 & 2 wins. And it is fifty-fifty in the end.

Clear as mud? It could be explained like this: If paths 1 & 2 were to count as one we would know Monty always chooses door 2 (or door 3) when you first choose door 1. And we don't know that.

Michal Pawlowski
Thursday, May 22, 2003

The paths are not all just as likely. Paths 1 & 2 occur with a probability of 1/6, and the others with a probability of 1/3.

David Clayworth
Friday, May 23, 2003

what happens if it was a more intelligent animal behind the door, that may move or hide as the door was opening? any formulas for that?

richard newhaven
Wednesday, May 28, 2003

The explanation that makes it crystal clear is:

52 cards are face down and your trying to find the ace of hearts. You point to a card. I turn over all the cards except for the card you pointed at and one other (and no sight of the ace of hearts). I offer you the chance to switch your pick.

anon
Friday, June 13, 2003

I love that this comes up a lot. :) I just posted about it on my blog[1] and got a lot of disbelieving people. Even when I provided irrefutable evidence of the statistics: source code which runs 1 million iterations.

Great puzzle. Screwed with my head the first time, too.

[1] http://dotnetguy.techieswithcats.com/archives/003396.shtml

Brad Wilson (dotnetguy.techieswithcats.com)
Tuesday, June 17, 2003

There is no chance of there being an intelligent animal behind the door. This is, after all, a game show.

David Clayworth
Wednesday, June 18, 2003

I disagree with most of what has been said here. The problem, as I see it, depends on when you measure your odds.

If you measure your odds at the very beginning, when you have 3 choices, with the assumption that Monty will not give you a second chance the odds are 1 in 3, or 33%

If you measure your odds at the very beginning with the assumption that Monty will take away a losing door if you're wrong and give you a second chance, the odds are 2 in 3. The fact that Monty takes away a door doesn't change a thing. You either win on the first choice or you get a second chance, thus 2 out of 3 chances to win (or 66%)

If you don't know for sure if Monty is going to take a door away and give you a second choice, then all bets are off. At that point the element of chance has been removed and replaced by Monty's whim.

If you measure your odds AFTER Monty has taken the door away and you are given a second choice, your odds are 1 in 2 at that point (or 50%). This is not rocket science. You have 1 choice and 2 possibilities.

This is one of those problems in which too much information (Monty takes away a door) has no impact on the problem at hand, but actually clouds the solution.

Steve B.
Tuesday, July 15, 2003

Actually, I want to modify my original post. After thinking about this a bit more, I want to provide this insight...

The odds are 50/50. Period!.

I've seen a lot of answers on this problem all over the internet, but they all failed to observe one fact: The actual choice of the doors is only made when there are two doors from which to choose. All of the other stuff (i.e. starting with 3 doors, Monty opens a losing door, etc) is just glitz to make the show fun to watch.

In the end, when the final decision is made the player will always choose one and subsequently open only one of two remaining doors.

It doesn't matter how many doors you start with,
It doesn't matter if Monty opens the doors, closes the doors, or goes to commercial,
It doesn't matter if you got to pick one, two or all of the doors.

The actual selection isn't final until you settle on which door is to be opened. At that time, you have only two doors from which to open. You can select one, then the other, and back to the original one but until you say "open door 'x' ", you haven't made the final decision.

This is somewhat analagous to having a pocket full of coins. You could flip two coins or ten or a hundred, but in the end you decide to flip only one,  thus the odds of landing on heads (or tails) is 50%.

Or another way to look at it, at the toss of a coin, I could call "heads", then "tails", then back to "heads". This does not change my odds.

Or, I could flip two coins knowing that one coin would be removed after the flip (by Monty?), and my odds of landing on my choice remains at 50%.

Steve B.
Wednesday, July 16, 2003

The odds are not 50/50.  You have to take into account prior probabilities.  If Monty were to open a losing door BEFORE you guessed, then you'd be correct.  However, the problem states that you start out with 3 doors, with no other information.

Let's start with what you agree with: you have, initially, a 1/3 chance of picking the correct door.  Assume the doors are labeled A,B,C, and you choose door A.  THIS CREATES A CONDITION.  You now have this situation:

The chance that the prize is behind door A is 1/3.
The chance that the prize is behind door B or C is 2/3.

Monty now eliminates a door in the latter group: door C.
This does not change the fact that your choice of door A has a 1/3 chance of being correct.  It also does not alter the fact that the prize was in either door B or C. 

However, *AFTER* you made your initial choice, door C is removed.  So now you have a set of doors that has a 1/3 chance of containing the winner, and a set of doors with a 2/3 chance of containing the winner.  But wait, each set only has one door in it.  Therefore, door A has a 1/3 chance of being correct, and door B has a 2/3 chance of being correct.

The key to this is that Monty knows exactly where the prize is.  If you know there's a 2/3 chance that B or C has the prize, and then C is eliminated, you now know that there's a 2/3 chance that B has the prize.

Tim H
Thursday, July 17, 2003

Addendum:

the line, "It also does not alter the fact that the prize was in either door B or C.  "

should read,
"It also does not alter the fact that the prize was in either door B or C with 2/3 probability."

Tim H
Thursday, July 17, 2003

This problem keeps coming back, and the correct answer never 'makes sense' at first, if ever.  Some of the explanations above are correct, but even they may not 'make sense' for a while.  Tim H's explanation seems clearest to me.

There are two facts that are easily overlooked:
1) both goats are (presumably) equally unwanted
2) Monty knows more than you do
and most 'common sense' arguments ignore one or both of these facts.

The best way to convince yourself is to get three cards and, keeping the 'prize card' in the same place, try in turn all three possible cards, and the possible outcome(s).

Paul
Thursday, August 07, 2003

Its really an interesting discussion... After reading all the posts i feel that i agree with Steeve.

I quote a post from another link.

There are 4 possible "paths" this game can go - I am referring to the table in the explanation on the techinterview website - let's say it's case 1 ($ behind door 1):

1. you pick door 1, Monty opens door 2
2. you pick door 1, Monty opens door 3
3. you pick door 2, Monty opens door 3
4. you pick door 3, Monty opens door 2

I know paths 1 & 2 seem equivalent, but they are NOT the same path. So all those four paths are just as likely. If you change your mind path 3 & 4 wins. If you don't, path 1 & 2 wins. And it is fifty-fifty in the end.

This clearly prooves that the chances of winning if we switch is just 1/2. I think since this shows all the available paths, all the possibilities, its absolutely clear that the chances are always 1/2...

Doing a random expiriment and getting more results in favour of switiching doesnot show you that its chances are more. No. If you flip a coin 10 times and happened to get all heads, do you say that it has a more chance?? No Never. We cant proove probability by experiment.

Joseph
Monday, September 15, 2003

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