That's all bunk ace. The fact is that after the one door is eliminated you are down to a 50/50 chance. Oh sure, if you caluculate it from the beginning you have a 2/3 chance of winning by changing doors. But once a door has been eliminated it becomes a new problem. It is 1 in 2. I defy any statistician to refute that. It's like flipping a coin. What are the odds of you to hit heads 10 in a row on the first toss? 1 in 2 to the tenth power. But on the last toss your odds of hitting heads is one in 2.

Enough ranting. So I'm a math geek, sue me.

monty hall

I believe midaz is wrong.

If Monte Hall was to RANDOMLY reveal the prize behind one of the doors (not initially picked) it would reveal the grand prize one third of the time. Then switching would be a 50/50 event if the grand prize was not revealed.

It is the lack of randomness in Monty Hall's revelation that makes switching choices better.

Here's another way of saying it:
If Hall's revelation was random it would show the grand prize 1/3 of the time. That prize would be given to the contestant. If the grand prize was not revealed, the contestant would have a 50/50 chance of picking the correct prize. That would mean a 50/50 chance for the remaining 2/3rds of possibilities. That is another 1/3rd. (.50 x 2/3 = 1/3)

So a random revelation would reveal the prize 1/3 of the time plus a 50/50 shot on the remaining 2/3 chances is 2 out of three.

I am certain I am correct about this. If you stipulate that you will be changing doors FROM THE BEGINNING then the odds that you get the right one are 2 in 3. However, that is not the way it is presented. It is presented as a choice after one door is eliminated and that is 1 in 2. And to anyone that says otherwise.....NYAH NYAH I CAN'T HEAR YOU........

Respectfully, Midaz, you're absolutely wrong. You say you defy any statistician to refute your view; in fact, that's already been done several times. It's also been tested via real life and computer modeling simulations. What's irrefutable is that you win roughly 1/3 of the time if you stay with your door, and roughly 2/3 of the time if you switch.

It does NOT become a 50/50 chance, or a "new game" after one door is eliminated. The game does not change and the odds do not change, because nothing material about the game has changed; most importantly, the money can not and has not moved from it's original location. (if there was a chance that the money could be moved after the door was eliminated, then you'd be right: it's a new game, and the odds are 50/50, assuming that the money moves about half the time and stays put the other half)

There was a 2/3 chance that you guessed wrong at the outset, and opening one of the remaining doors does nothing to change that; there's STILL a 2/3 chance that you guessed wrong at the outset, so you should switch doors.

If you're still convinced you're right, Dean Esmay says on his blog that he'll give you $1000 if you can prove it. But I wouldn't waste your time trying.

Gonna re-post it here:

My choice determines Monty's choice, not the other way around.

Assume Door A is the winner, and I pick it first, then Monty has two doors to choose from for a reveal.

But, if I pick door B, then Monty only has one choice for his reveal.

What I can't know is if my choice gave Monty two out of two to reveal, or one out of two.

But I can know that my odds of picking a door that will move Monty's choice are two out of three.

I need to play Monty's odds, not mine. If I make a second guess, based on the better odds that Monty now has only a 50/50 choice of reveals, I'm better off taking that second guess.

Here's a kick ass link to an online computer simulation that actually runs the numbers while you watch, and provides the computer code if you wish to run the simulation yourself:

http://www.bignell.demon.co.uk/montyhall.htm

Here's a list of links to other sites explaining the solution( there are tons, I only listed a few):

http://www.uh.edu/engines/epi1577.htm
http://www.calpoly.edu/~mcarlton/riddles.html
http://www.remote.org/frederik/projects/ziege/
http://www.niehs.nih.gov/kids/rd3.htm

Ok, at the risk of turning into a comment spammer, here's a link to a site, from the math dept at Rice University, with a BUTTLOAD of links and info about the Monty Hall riddle (all supporting the switch theory).

http://math.rice.edu/~pcmi/mathlinks/montyurl.html

Sorry Midaz!

And sorry for hogging your comments, Ace. I'll stop now. :)

Respectfully dave....

If we know at the beginning that Monty is going to eliminate one incorrect choice that leaves only 2 options. Therefore the odds are 1 in 2. Again, all the assumptions are based on what you do BEFORE the first choice is made. Once that first choice is made it becomes a new problem. Two options, one is right. It is 1 in 2.

Don't ya love it?

Just not in a gay way.....

Midaz,

Once again I beg you to indulge the idea that it is the non-random nature of Hall's reveal that moves the odds in your favor if you switch.

If it (Hall's choice) was random, you'd be right. It and you are not.

Midaz, here is exactly how it is presented:
"So here's the brainteaser: is there any compelling reason to switch doors?"

Yes. Because I have a 2/3 chance of choosing a door with nothing behind it, therefore, a 2/3 chance of decreasing Monty's odds on beating me.

Monty's odds of beating me are 2/3 in the first choice. For the sake of the argument, let's say that after the reveal, his odds of beating me decrease to 50/50, as in, will I change my choice or stand pat. But not 50/50 odds in relation to the two remaining doors, only in relation to my two choices.

If I stand pat, his odds remain at 2/3 to beat me. If I switch, he only has a 50% chance of beating me. THAT's a compelling reason, however the numbers geeks wanna 'splain it.

This will be my last comment about this because it is after all silly.

The choice is still 1 in 2 because we know from the begging that Monty is going to eliminate one of the options. SO NO MATTER WHAT there will be one good door, one bad door. That is 1 in 2 even though there were 3 in the beginning. So when you think about it the odds of you hitting the winning door were never 1 in 3. They were ALWAYS 1 in 2.

My apologies to Ace for being a math geek on his site. Love your site by the way. Just not in a gay way.....

No, Joan. No.

Switching is better because it improves the odds to 2 out of 3. Please see any and all of the above, but especially my first post.

Midaz, there are none so blind as those who refuse to see.

I cannot believe you forced me to do this Birkel....

Given the fact that we know beforehand that 1 incorrect choice will be eliminated it is NEVER a question of 1 in 3. It is 1 in 2. There may as well not be 3 doors. Pretending that there are 3 choices is where the odds get all fouled up. There are only 2. If monty were to eliminate a door randomly you would be correct. But we all know that Monty never gets rid of the winning door. So no matter what you do your choice will come down to 1 winning door and 1 losing door.

I am truely done now and this time I mean it. :)

Several people have argued that once one booby door is revealed, the odds change to 50/50 that you have the right door because one door is "eliminated" and there are only two left, so the odds must change to 50-50. That's statistcally incorrect.

Maybe this will help...

Even before you made your pick, you knew that at least one of the other doors you didn't pick would have a booby prize, right? So, when Monty reveals that, indeed, one of the doors you did not pick has a booby prize, then you learned which door that was...

...but just what, exactly, have you learned about YOUR DOOR?

Why, not a single thing.

Basically, the revelation doesn't change your door's statistical probability because it neither mathematically nor intuitively changes the premise upon which you made your initial selection. The odds on your door is still 1:3.

Heck, if your door was going to have a 50-50 chance of being the right door based upon the fact that one of the 2 doors you DIDN'T pick held a booby prize, then you'd have that same 50-50 chance before you made your pick because, according to the premise, one of the unpicked doors HAS to have a booby prize.

No, the only knowledge you get from the reveal is that you now know the odds that the revealed door has a 0:3 chance, which would mathematically change the odds of the other unpicked door to a 2:3 chance.

Not convinced? Look at it THIS way...

You're given a choice: you can...

a) You can try to pick a prize from a choice of TWO doors

b) You can try to pick a prize from a choice of THREE doors

Well, we all know that (a) has a better statistical probability than (b). However, it's being argued that the stats on (b) change to exactly what (a) starts out with...based upon the revelation of a fact you already know, that one of the unselected doors will have a booby prize! Come on, people, you already KNOW that, so why would the probability on your pick suddenly change after the reveal?

I know it's hard for people to see this because you're mixing random selection with nonrandom selection. If Monty's choices were made at random and the first revealed door had a booby prize, then yes, your chances of having the right door WOULD increase (provided he didn't pick YOUR door first -- random IS random) because that would be an actual "elimination;" essentially the slate would be wiped clean each time and the probability on your choice would change. That just doesn't happen in this scenario because Monty's choices aren't random and hence the slate is not wiped clean from the original number of choices you had.

Later,
bbeck

Midaz, honey, let's play the game according to your insight. How do I mentally approach the three doors?

I choose a door. I could be right or wrong.

Monty reveals a door. I choose to think nothing of HIS mental approach,and instead cast all my hopes on the hope that my either my first choice or my second choice is right. In that sense, everything you decide is 50/50. Yes, or no.

But here's Monty, giving you an inside edge, a tip. The risk is still all in your decision, and you can only decide to switch or stay. That's the 50/50.

But one choice affects the other. Just like life. Monty doesn't know which door he will reveal until AFTER MY CHOICE. To ignore that simple fact for the beacon of seeming logic will leave you on the rocks more times than not.

I hope for your sake you don't play poker.

Heh, this one always gets 'em goin'. :-)

Sigh.

Joan,

You know before you make your choice that Monty is going to eliminate a bad door for you. So you do not have 3 doors to choose from, you really only have 2. I KNOW that is counter intuitive but it is reality. Now if you didn't know that Monty was going to give you the option to switch then perhaps you are right. But we DO know this. I defy anyone to tell me how the winning door gets eliminated after the first choice. We all know how the game works. You will be left with one winner and one booby prize even though there were 3 doors initially. That is the very definition of 1 in 2.

I love you peeps btw. Good chatter all.

Midaz,

I await your refutation of my initial post.

Until then, we can agree that you are wrong but refuse to see it.

Here's how I've explained this to doubters: Instead of three doors, suppose there are a thousand doors. After you've picked one, which has a .001 chance of containing the prize, Monty says, "OK, of the remaining 999 doors, I'm going to eliminate 998 that I know DO NOT contain the prize. That means that the prize is either behind the door you picked originally, or behind the remaining one that I didn't eliminate. Do you want to switch, or stick with your original choice?"

It's exactly the same scenario, except for the exaggerated numbers. In both cases, Monty is elminating all but one of the remaining doors, and the one(s) he eliminated do not contain the prize. It's just easier to see the advantage when the probability of success goes from .001 to .999 than from 1/3 to 2/3.

Yes, Birkel, but I'm not trying to convince YOU.

I'm trying to word it for insight, not proof.

Con-men have always relied on people's intuitive nature to lead them to exactly the wrong conclusion. So look at it like a con man.

The best setup is an honest proposition that leads a person to make a bad choice. Monty truly is hampering his odds of beating you by making a reveal...but because it's our nature to suspect any "freebies" we feel that our odds are not any better than before. So, how does the con man stay in business, if his odds decrease with the reveal? By knowing more about YOU than you know about MATH.

This version of the shell game has been around since Adam first had neighbors.

Joan,

I understand why you wrote it the way you did.

That's part of the "there are none so blind..." comment above.

My comment wasn't meant for you either, my dear Joan.

Birkel,

WE do NOT agree that I am wrong, lol. You keep ignoring the fact that it is a foregone conclusion that one of the booby prizes will be eliminated. There are never 3 doors to choose from, only 2. I know this is counter intuitive. You say 'but there are 3 doors'. No, there are only 2, no matter what the winner will not be eliminated, only a booby prize.

Wow, what a fun way to use up Ace's bandwidth. I think I will have to go hit the tip jar now. Thanks Ace!

Here's another one that will drive you insane, using the same premise.

There are THREE PANCAKES: one is BROWN on both sides; one is BROWN on one side and GOLD on the other side; and the last is GOLD on both sides... like so...

BROWN/BROWN BROWN/GOLD GOLD/GOLD

Suppose you throw all of the pancakes into a box and blindly and randomly choose one, and you notice that the side that is showing (up) is BROWN. What is the probability that the other side of the pancake is also BROWN?

Should I give the answer now?

Laughing at yourself is a good thing. You're still wrong, but you're happier. It's not a win-win but it's close.

OTOH, I'm right AND I'm laughing at you. Mine's a win-win.

Please refute the logic of my first post. Third in the thread. Waiting.

The Monty Haul question is funny in that it made complete asses out of a large number of elitist statisticians who couldn't accept the truth. It is very, very easy to write a program to test it. I did so in C in about 2 minutes... I also wrote one to test the pancake thing I posted.

Birkel,

I already refuted it. You keep looking at this exercise as a choice among 3 doors. There are only 2. Now who is being willfully blind? :)

Dave,

I'll give it a run.

A Brown occurs three times.
Twice where the other side is Brown.
Once with the other side is Gold. (shouldn't it be golden?)

The odds are 2 out of 3.

Dave,

At the risk of being labeled blind again I will posit this. We KNOW it cannot be golden/golden. Therefore it is either the brown/brown or the brown golden. Only two choices there. It is 50/50.

Hey midaz:

Scroll up to my post of 11:08 pm where I explained it with a scenario involving 1000 doors initially, but only two in the end. Would that be a 50-50 choice because you knew in advance that 998 doors would be eliminated?

Would someone please stop arguing about doors and tell Ace he has been instlanched?

I posted this in a comment in the appropriate thread, but he hasn't seen it.

Dave:

Another way to state your question would be, what would the chances be of blindly picking the brown/brown pancake out of a box containing two other (different) ones. The answer is obviously one in three.

Fun stuff, Midaz!

I can know that Monty is going to eliminate a door beforehand. But I don't know WHICH door he will eliminate, AND NEITHER DOES HE. ( capitalized to show that this is a KEY point. heh. ;) )

Neither of us knows until I choose first. That IS the random odds at the beginning. However, if I can know that my choice will affect Monty's choice, what does that tell me about all three doors?

To look at it as a one-step static logic problem is to lose at cards.

Yes, the answer to the pancake one is also 2/3. You have to think of it as drawing 2 sides at once. The second side can be one of three possible sides: BROWN, BROWN, or GOLD(EN).

Hey, midaz... don't worry... I was initially a staunch 1/2er in this debate... http://www.gamedev.net/community/forums/topic.asp?topic_id=205622&PageSize=25&WhichPage=1

...about it. That thread runs at least 7 or 8 pages. It was probably the longest thread I've ever seen on some incredibly active forums if that tells you anything.

I didn't believe it until I wrote a simulation to test it. Once you accept it, it will become clear.

Another way to state your question would be, what would the chances be of blindly picking the brown/brown pancake out of a box containing two other (different) ones. The answer is obviously one in three.

Not exactly. This is a conditional probability. You are given that you have already drawn a brown side up... in other words, one of the pancakes is excluded for you.

I know it sounds crazy, but the probability of drawing a brown on bottom, given that you have drawn one on top, is 2/3.

My first answer to the pancake question was wrong. Out of a hundred plays where the pancake had one brown side, on average 50 would be the brown/brown and 50 would would be brown/golden. So the chance of the other side being brown would be 50%. And out a hundred plays where the top side was golden, roughly 50 would be the golden/golden and 50 would be the brown/golden.

My answer is that regardless of the top color of the selected pancake, odds are 50% the bottom color will be brown.

Oh my god...

This is the same argument I had years ago, for hours on end...

make it stop...

make it stop...

(J/k. You can hash it out all you like.)

We've been debating it pretty furiously over at Patterico

There's at least two ways to prove that switching is twice as likely to produce a win.

Here's the brute force method:

At the start of the game, there are three places where the prize could be located -- Door 1, 2, or 3. These locations are selected at random. The three scenarios are A, B and C. The probability matrix looks like this:

.....1...2...3
A...$...x...x
B...x...$...x
C...x...x...$

Each of these three scenarios (A, B and C) occurs 1/3 of the time, at random. Look what happens as you play:

Switching strategy (where you pick Door 1):
A. Monty opens 2 (half of these times) or 3 (half of these times), you switch, you lose.
B. Monty opens 3, you switch, you win.
C. Monty opens 2, you switch you win.

Keeping strategy (again, where you pick Door 1):
A. Monty opens 2 or 3 (each 1/2 of this 1/3 scenario), you keep your selection, you win.
B. Monty opens 3, you keep, you lose.
C. Monty opens 2, you keep you lose.

Switching: Two wins, one loss.
Keeping: One wins, two losses.

Scenario A (where you picked the right door the first time), will happen 1/3 of the time. Of those situations, Monty will open Door 2 half of the time, and Door 3 half of the time (i.e., out of those 1/3 occurrences of Scenario A). Scenario A will still only happen 1/3 of the time, because the initial starting conditions are chosen at random.

Multiply this result over the selection of other doors (to account for your selection of Door 1, 2 or 3, which is also at random, at 1/3):

Switching strategy (Door 2):
A. Win
B. Lose
C. Win

Keeping strategy (Door 2):
A. Lose
B. Win
C. Lose

Switching strategy (Door 3):
A. Win
B. Win
C. Lose

Keeping strategy (Door 3):
A. Lose
B. Lose
C. Win

Total: Switching wins 6 of 9. Keeping wins 3 of 9.

Max,

Thanks, I noticed that. But I wouldn't say Instalanched. Just instalinked.

The link is an update to a now-buried post which will barely be seen by anyone, posted at 10 pm at night. Completely buried by tomorrow.

But still-- nice to get it anyway.

I guess.

Here's the logical reasoning method of proving that switching is better:

1. At the beginning of the game, when you select a door, it is more likely that you will guess a wrong door than the right one. (Specifically, you will be wrong 2 out of 3 times, but "more likely to be wrong" is close enough for a logic-based non-numerical explanation.)

2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. Another way of saying this is that if your first guess is wrong, then the door that neither you nor Monty pick MUST be the right door.

3. The only situation in which switching your guess would cause you to lose is if your first guess had been correct.

4. However, as stated in point 1, you were more likely to be wrong in your first choice than correct.

5. Since it is more likely that your original guess was wrong, then changing your guess to the only other available door MUST improve your chances of winning.

QED

Not exactly. This is a conditional probability. You are given that you have already drawn a brown side up... in other words, one of the pancakes is excluded for you.

Yeah, so this is a case where midaz's argument on the doors actually applies, right? Since one of the pancakes is eliminated, it's really a choice between two, which means a 50% probability that a particular one will be selected, right? Or what am I missing?

Ace,

The ice is now broken. (I get the impression that Instapundit doesn't like ollie any more than you do.)

At any rate I think you should mention the link in an update to your post below.

max

F

Regarding the Monty Haul problem... switching essentially reverses the odds to the players favor.

The best example I have ever heard was the one posted above using 1000 doors instead of 3.

If the host opens 998 doors, that leaves the 1 that you chose (1/1000 probability) and the one that remains (999/1000 probability. If you switch doors, you use the host's knowledge of the doors against him. You essentially give him your crappy odds and you take his good ones..

We all still luv ya, Ace.

But then, I come every day with or without the Insta-pimp directing me. So, hmmm, I guess I'm not really your target in all this.

And what I wouldn't give for another Ace-a-lanch sometime. You know... like when I start writing consistent updates on my own blog.

But check out my new collaborative effort . It's a sports blog.

ACCBasketBlog.blogspot.com

It just got started but we've got inside sources and stuff.

Since one of the pancakes is eliminated, it's really a choice between two, which means a 50% probability that a particular one will be selected, right? Or what am I missing?

The key is that you have to think of it as selecting 2 sides as opposed to one pancake. You know intuitively that the GOLD/GOLD pancake is no longer a possibility. What you don't know is which brown side you are looking at.

You could be looking at any of the capitalized sides below:

BROWN/brown brown/BROWN gold/BROWN.

If you look at the sides that are "attached" to the sides that you MAY be looking at, you will see that 2/3 of them are brown.

Did someone say monte-carlo simulations proved this?

That would be proof for me-- evidence trumps thought experiments.

But my problem with the solution has always been similar to Midaz'. The thing is, there's ALWAYS going to be two doors without prizes; so, if Monty Haul shows you one, so what? You knew it was there.

And yeah, I would go down Midaz' thinking that this becomes really a 50-50 choice.

But-- if monte carlo simulations prove it, well then, case closed.

Max,

An update? Just to say I was insta-linked?

I used to do that. But it's pretty self-serving. I did it before because it was new and exciting for me.

You know what they say-- when you get into the endzone, behave as though you've been there before.

Yes, the answer to the pancake one is also 2/3. You have to think of it as drawing 2 sides at once. The second side can be one of three possible sides: BROWN, BROWN, or GOLD(EN).

I think the flaw in your logic is that you're considering "sides" of pancakes indivdually, separate from the pancake they belong to. There are initially 3 brown sides and 3 golden sides, and since you've eliminated two of the golden sides, you're left with two brown sides and one golden side, therefore the odds are 2/3 that the underside is brown. But you're not picking from among "brown, brown or golden".

Restate the problem this way: After selecting the pancake with a blindfold on, you are told by a bystander that the pancake you selected is either the brown/brown or the brown/golden one. The bystander asks you, "what is the probability that you're holding the brown/brown pancake?" The answer seems pretty obviously 50% to me.

Or try another restatement: After selecting the pancake with a blindfold on, you are told by a bystander that the pancake you selected is either the BROWN/BROWN or the GOLDEN/GOLDEN one. The bystander then asks you, "what is the probability that you're holding the BROWN/BROWN pancake?"

What is your reply?

Yes, Ace, it is definitely provable at the above-linked university websites.

10:04 PM post by Dave

And the pancake explanation is fundamentally the same scenario as Monte Hall's. Substitute "prize" and "no prize" for color.

Best regards.

I was leaning towards Midaz until George posted every possible scenario. But there’s still one lingering doubt I can’t get out of my head.

Let’s say I pick Door A. Meanwhile, Alternate Universe Me, playing exactly the same game with the money behind the same door, picks Door B. In both universes, Monty reveals Door C to be a loser. How is it possible for both Me and Alternate Universe Me to have a 2/3 chance of winning if we switch?

Racer X,

No.

A brown side is connected to a brown side in two out of three cases. (Please come to my place to bet on cards or dice or pancakes. I could use the income. /snark)

One Brown side that could be selected is connected to a Brown opposite side.
Another Brown side is attached to a Brown.
Another Brown is connected to a Gold(en).

Certainly the Brown side in question is one of three Brown sides. The odds must therefore be something out of three. You cannot seriously argue that the odds are 0/3, 1/3 or 3/3 so there is only one option left.

Probability:
odds of something occurring/ # of total things

The number of total things in the pancake example: Three possible Brown pancakes.

The odds must be out of three.

chris,

You moved the goalposts.

There was no reason for Hall to reveal Door C in the example where Door A was the winner and you had chosen Door A. Hall could reveal Door B half of the time. Your examples are not symmetrical so the fact the odds don't sum to 1.00 doesn't mean much.

Let’s say I pick Door A. Meanwhile, Alternate Universe Me, playing exactly the same game with the money behind the same door, picks Door B. In both universes, Monty reveals Door C to be a loser. How is it possible for both Me and Alternate Universe Me to have a 2/3 chance of winning if we switch?

You're probably thinking, "out of 99 plays, how could it be that I would win 66 times and Alternate Me would also win 66 times?" The reason it wouldn't happen is that it's not the same game. In 33 cases, both of your initial choices would be wrong and C would be the winner, but Monty wouldn't be able to eliminate it.

Restate the problem this way: After selecting the pancake with a blindfold on, you are told by a bystander that the pancake you selected is either the brown/brown or the brown/golden one. The bystander asks you, "what is the probability that you're holding the brown/brown pancake?" The answer seems pretty obviously 50% to me.

Notwithstanding the fact that the 2/3 probability can easily be verified with a simple program (if you know php or something, give it a shot), you actually touched on my point in your "rewording" of the problem.

The bystander tells you that you are holding one of 2 pancakes, but how do they know? Because the see a brown side up, but they do not know which one.

They could be seeing either of the sides of the brown/brown, or the brown side of the gold/brown. In any of the three cases, there are 3 possible, valid "other sides"... 2 of which are brown.

It's very easy to think of it as drawing a pancake and knowing that it is one of 2. But the fact is that you are drawing 2 sides, the second being dependent on the first.

If you read that very long and extensive thread above (URL I posted) you can see that I even initially believed the 1/2 and argued ferociously (smitty1276 on those forums). It wasn't until I actually wrote about 20 lines of code to test it that I accepted that it was, in fact, 2/3.

As with the Monti Haul problem, once you accept the fact that it is, or might be, 2/3 you start to see how it can be.

Sorry, but all you 1-in-3 or 2-in-3 people are wrong.

It is a fifty-fifty proposition. You and everybody else thinking about 1 in 3 is completely wrong. The odds were never one in three. NEVER. The odds were ALWAYS one in two. You, me, Monty and everybody else knew that he was going to open a door without a prize. That's a given. It doesn't matter where the prize is, Monty is always going to eliminate a choice. That choice was never there to take. It always was and always will be two chances to pick between two doors. PERIOD.

Look at it this way:

There will always be twelve possible outcomes-

1) You pick 1, Monty opens 2, prize is 1.
2) You pick 1, Monty opens 2, prize is 3.
3) You pick 1, Monty opens 3, prize is 1.
4) You pick 1, Monty opens 3, prize is 2.
5) You pick 2, Monty opens 3, prize is 2.
6) You pick 2, Monty opens 3, prize is 1.
7) You pick 2, Monty opens 1, prize is 2.
8) You pick 2, Monty opens 1, prize is 3.
9) You pick 3, Monty opens 1, prize is 3.
10) You pick 3, Monty opens 1, prize is 2.
11) You pick 3, Monty opens 2, prize is 3.
12) You pick 3, Monty opens 2, prize is 1.

Half the time you win, half the time you lose. Always.

See? Monty will NEVER open a door a prize is behind, so that door, whichever one it is, is never a consideration. NEVER. It is never part of the equation.

It is a red herring, designed to mess with your mind. Why? Because the door Monty picks is not a random event. He picks it KNOWING that nothing is behind it.

All this arguing about pancakes, flipping coins, choosing doors, etc. is dealing with RANDOM events. Monty's door opening is NOT a RANDOM event. It is STAGED, because he KNOWS which door to pick.

If Monty didn't know where the prize was, and sometimes he revealed it, then the idea that the prize is randomly located behind three doors is valid.

However, that is not the case. The prize is located behind one of two doors. You just don't know which two doors to choose between until Monty eliminates the third. The third door is ALWAYS going to be a booby prize.

100% of the time.

It cannot be part of the equation.

The answer is 50-50.

The bystander tells you that you are holding one of 2 pancakes, but how do they know? Because the see a brown side up, but they do not know which one.

They could be seeing either of the sides of the brown/brown, or the brown side of the gold/brown. In any of the three cases, there are 3 possible, valid "other sides"... 2 of which are brown.

Now it makes sense. Out of 99 random drawings, the brown/golden one will be selected 33 times, but it's brown side will be facing up only 16.5 of those times. The brown/brown one will also be selected 33 times, but its upside will be brown 33 times. So of the roughly 50 times in which a brown side is facing up, 2/3 (33) of those will be the brown/brown pancake, and 1/3 (16.5~17) will be the brown/golden.

Thanks, I don't need to write code to understand that.

Dogstar, read this previous post of mine:

Here's how I've explained this to doubters: Instead of three doors, suppose there are a thousand doors. After you've picked one, which has a .001 chance of containing the prize, Monty says, "OK, of the remaining 999 doors, I'm going to eliminate 998 that I know DO NOT contain the prize. That means that the prize is either behind the door you picked originally, or behind the remaining one that I didn't eliminate. Do you want to switch, or stick with your original choice?"

It's exactly the same scenario, except for the exaggerated numbers. In both cases, Monty is elminating all but one of the remaining doors, and the one(s) he eliminated do not contain the prize. It's just easier to see the advantage when the probability of success goes from .001 to .999 than from 1/3 to 2/3.

Now answer me this: In the scenario with 1000 doors initially, do you stick or switch?

Birkel, I kind of figured that would be a problem. Now I make no claim to any sort of advanced mathematical knowledge, but accepting the fact that Monty has revealed Door C in this one particular game, how is it that probability is on the side of the switch for two people who select different doors?

Or we can think of it another way. Before the show, Monty says to himself "The money is behind Door A, so I'll reveal Door B unless that's the door he picks, in which case I'll reveal Door C." I haven’t seen a version of the question that excludes this, though I’m admittedly adding onto it which may change things. It seems to me, this would be the closest scenario to reality. So if I pick either Door A or Door C, Monty reveals Door B. Playing the same game only once (as a response to Racer X), how can probability say that both have a better chance of winning by switching?

Stated simply: people keep saying that the reveal tells us nothing new about the door you have chosen so the probability can’t change. But what new information does it give us about the other door we haven’t chosen if Monty has decided ahead of time that unless we pick it, he will reveal Door B first? I really don’t think this is cheating as no version of the problem I’ve seen specifies that Monty MUST have no ideas about which door to reveal beforehand.

Again, if this is mathematically infeasible, I wouldn’t know it, so don’t waste too much energy brushing me off.

That 1,000 door problem is really the killer. The reason Monty hasn't revealed your door in the previous 998 is obvious, but the reason he hasn't revealed door 497 could only because it was randomly chosen out of 1,000 or because it really has the money. As a non-mathematician, I can't tell if this is a real or imagined probability increase though.

Golly DogStar,

What with all the CAPITALIZATION, I'm now convinced. Let's play games of chance together for MONEY. I like my ODDS.

Dogstar, the simple fact is that you will pick the correct door 1 in 3 times (meaning you will pick the wrong door 2/3 times).

If you pick the wrong door (2/3) and switch, you will ALWAYS win. If you pick the right door (1/3) and switch, you will always lose.

Switching doors wins 2/3 of the time.

Also, your breakdown of possible scenarios was flawed. The door is prechosen, not a random event.

Here is how it breaks down. Notice that the door that is opened is irrelevant.

Prize door / Chosen door / Opened Door *=switch wins

1 / 1 / 2 or 3 (single event)
1 / 2 / 3 *
1 / 3 / 2 *
2 / 1/ 3 *
2 / 2 / 1 or 3
2 / 3 / 1 *
3 / 1 / 2 *
3 / 2 / 1 *
3 / 3 / 1 or 2

Notice, of the nine possible scenarios, there is an * next to 6 of them. You win 2/3 of the time by switching doors.

Once again, this is EASILY verifiable. Just write a simple php script or a javascript or a C program or SOMETHING to test it. Or just use cups and a ball or something and try it about 30 times and keep "score".

What difference does it make how many doors Monty opens?

Monty gives you 1000 doors to choose from. One has something good behind it.

You know that, after you make your pick, regardless of which door you pick, he's going to open 998 enpty ones.

So what?

You still have two chances to decide which of two doors to open. All the rest is just show business, and has absolutely nothing to do with math, odds, probability, randomness or anything else.

You computer programmers forget that any program result is a function of the rules written into the program. I've been writing programs since 1977, when I was a freshman in college, and I had to turn in a stack of Fortran cards, and then wait a day to get the result.

The pancake problem I agree with, because it is totally random. Monty didn't pick or discard any of the pancakes.

2/3 for the pancakes, 1/2 for the doors.

I can't tell if this is a real or imagined probability increase though.

It isn't actually a probability increase. In that instance there was ALWAYS a 999/1000 chance that the prize was behind a door that you did NOT pick.

He is going to open 998 doors, regardless of which you pick.

After opening 998 doors, the fact still remains that there is a 999/1000 chance that the prize is behind a door other than the one you chose. Of course there is only one door.... the fact that he opens 998 doors has nothing to do with the odds that you initially made the correct 1/1000 choice.

I t's 1/3-2/3

Monty revealing an empty door tells you nothing.
EVERYONE knew one of the unchosen doors was empty.
This simply squeezes the probability of both unchosen doors into the one left. ( 2/3)

Try this.
Assume you choose door A.
Monty reveals nothing and offers to let you trade for
BOTH B AND C.
This is essentially the same as the original question.
Revealing the empty door won't change it.

Dogstar, no offense intended, but how are you a programmer and don't understand how to code this test?

Really, coding it is even a waste of time. It's this simple,

Here's some pseudocode:

switchWins = 0;
for(i = 0; i {
prizeDoor = random number from 1 to 3
choiceDoor = random number from 1 to 3
if prizeDoor != random door
switchWins++; //switching would choose prize
}

Switching doors wins (switchWins/100000)% of the time.

Dave,

But by the same token, doesn't the 999/1,000 (or 1/2 by what I think is Dogstar's reasoning) probability that you were right by NOT picking Door 497 never change even if it's down to that door and the one you picked?

Oops, that should have been a for-loop for i from 1 to 100000.

You can't count Monty's door. Monty always checks to see where the prize is and which door you picked, and then picks the door nothing is behind. It is not a variable. It is a constant.

There is no randomness associated with Monty's door. It is always empty.

If you write a computer program that assumes Monty's door is always empty, you will get 1/2.

If you write a computer program that assumes Monty's door has A CHANCE of having a prize behind it, which we KNOW IS INCORRECT, you will get your desired switching bonus.

So which program are you going to write?

The door Monty chooses is NOT a constant. It is completely dependent upon YOUR choice. If you choose door 2, he isn't going to open door 2 for example. Furthermore, the door he opens is completely irrelevant.

Look, would you at least agree that you have a 1/3 chance of selecting the correct door and a 2/3 chance of selecting the incorrect door?

Would you also agree that if you choose the wrong door initially, that there is a 100% chance that the prize is behind the remaining door? (This is obvious).

If you managed to answer yes to those 2, you should easily be able to see that if you switch every time, the result is ENTIRELY DEPENDENT UPON YOUR INITIAL CHOICE and NOT AT ALL dependent upon which door is opened for you.

Just switch every time.... if you picked the wrong door initially, which you will 2 out of 3 times, you will win. If you picked the correct door initially, which you will 1 out of 3 times, you will lose.

I can't think of a way to make it any freakin' clearer. This is a well known statistical case, and you should easily be able to find countless formal proofs online if you REALLY need them. At the least, go browse some of the websites dedicated to it.

Here:
http://en.wikipedia.org/wiki/Monty_Hall_problem

Right.

Three doors, 1, 2 and 3. We'll call the door you pick A, and the other two B & C. It doesn't matter whether A is 1, 2 or 3; it works out the same regardless.

One third of the time, the prize is behind door A. Monty can open either B or C (it doesn't matter which), and you can switch if you want. One third of the time you are right to start with, and you will lose if you switch.

One third of the time, the prize is behind door B. Monty must open door C. You are wrong, but you will win if you switch.

One third of the time, the prize is behind door C. Monty must open door B. You are wrong, but you will win if you switch.

Three cases of equal likelihood. In two out of three cases, you will win if you switch; only one time out of three will you win by staying put.

It would be different if Monty picked a door randomly. In that case, two thirds of the time, he would show you an empty door, and you would know that your odds are now 1 in 2. Of course, one third of the time he would show you the prize and you would lose immediately.

The same applies if Monty always picks the door to the left of the one you picked. It's only when he uses his knowledge of where the prize is located that things change.

Dogstar:

Your error is in your list of possible outcomes:

You said, "There will always be twelve possible outcomes-

1) You pick 1, Monty opens 2, prize is 1.
2) You pick 1, Monty opens 2, prize is 3.
3) You pick 1, Monty opens 3, prize is 1.
4) You pick 1, Monty opens 3, prize is 2.
5) You pick 2, Monty opens 3, prize is 2.
6) You pick 2, Monty opens 3, prize is 1.
7) You pick 2, Monty opens 1, prize is 2.
8) You pick 2, Monty opens 1, prize is 3.
9) You pick 3, Monty opens 1, prize is 3.
10) You pick 3, Monty opens 1, prize is 2.
11) You pick 3, Monty opens 2, prize is 3.
12) You pick 3, Monty opens 2, prize is 1."

This is partly true, but look at your last column -- the "prize is" part.

Numbers 1 and 3 both have the prize behind Door 1. However, this will happen 1/3 of the time.

Similarly, numbers 5 and 7 have the prize behind Door 2.

Numbers 9 and 11 have the prize behind Door 3.

If you pick your door at random, then you will pick each door 1/3 of the time. When you pick the RIGHT door (which will happen 1/3 of the time), Monty will pick each of the wrong doors half of THOSE times (i.e., half of the 1/3 of the times you pick the right door, or 1/6 of the total).

Therefore, you can't give equal weight (probability) to all the items in your list. The duplicated items (e.g., 1and 3) each only happen half as often as the others (e.g., 2 and 4).

With the pancakes, it may be easiest to look at it this way...

Label each side of the two pancakes with at least one brown side:

Side A = Top side of brown/brown pancake
Side B = Bottom side of brown/brown pancake
Side C = brown side of brown/gold pancake
Side D = gold side of brown/gold pancake

The pancake you're holding has a brown side up, so you're looking at Side A, B, or C.

If it's Side A, then the other side is brown
If it's Side B, then the other side is brown
If it's Side C, then the other side is gold

(The fact that you picked this from a bag with a gold/gold pancake is a red herring.)

So, the odds are 2:3 that the other side is brown.

Later,
bbeck

Dave:

"The door Monty chooses is NOT a constant. It is completely dependent upon YOUR choice."

No, it's not. Monty ALWAYS chooses an empty door. If Monty chose a random door from the remaining two, he would reveal the prize occasionally.

However, Monty NEVER reveals the prize. Therefore, Monty's choice is not a random event and Monty is fucking with your head.

It's called show biz. It's called a red herring.

No matter what happens, one of the empty doors will never be your choice. It will never be available to you. Monty has fixed the game.

You can't count an event KNOWN to have a constant outcome in a probability equation.

George:

"Numbers 1 and 3 both have the prize behind Door 1. However, this will happen 1/3 of the time. "

It DOES happen 1/3 of the time. It happens in instances 1, 3, 6 and 12. That's four times out of twelve.

I can't believe how hard it is for you two to understand that Monty KNOWS, therefore this is not RANDOM.

What are you talking about? EVERY door is available to you. You can pick ANY OF THE THREE DOORS.

I know... the earth is flat, and the sun orbits the earth, and all of that good stuff, too....

Dogstar,

I know that Monty is not random. In fact, the very fact that Monty knows where the prize is located is the reason that switching will give you 2/3 odds of winning.

When you choose the wrong door at the start of the game, you give Monty ONLY ONE DOOR to open. He cannot open the door you picked. and he cannot open the prize door. So, if you choose a wrong door, there is only one door he can possibly open: the only other wrong door.

As a result, if you picked a wrong door to start with, Monty will, in effect, TELL you which door holds the prize. It can only be the door that you did not initially pick, and it cannot be the door he opens. Therefore, it can only be the remaining door, and therefore swithing to it must win the game.

Now, how often will you have picked the wrong door to start with? Two out of three times. Therefore, picking the wrong door (which happens 2 of 3 times) will FORCE Monty to open the only other wrong door, thereby informing you which door holds the prize.

The only circumstances in which NOT switching will win is if you picked the RIGHT door at the beginning. However, this will only happen 1 out of 3 times. In this situation, Monty can choose either of the other wrong doors. He will open one of these wrong doors half of these times, and the open the other wrong door the other half of these times. Since picking the right door will only happen 1 in 3 times, Monty will open one wrong door 1/6 of the time, and the other wrong door 1/6 of the time.

Therefore, picking a door and switching will win 2 out of 3 times, and not switching will win 1 out of 3 times. The reason this is true is because your picking the right door on your initial guess (even though you do not know it is right yet) will happen 1 in 3 tries. It is better for you to pick the wrong door on your first guess, because that will FORCE Monty to open the other wrong door, thereby telling you which door is the right one. Fortunately for you, you will guess the wrong door 2 out of 3 times.

I think its hopeless, George. This has been proven time and time again and there is a wealth of info about it... but he doesn't seem to be interested in understanding it, only arguing.

Dave:

"What are you talking about? EVERY door is available to you. You can pick ANY OF THE THREE DOORS."

Yup. And when you get done picking, one of the doors you didn't pick is ALWAYS shown to be empty.

No matter which door you picked.

What does that mean, Dave?

It means one of the doors never mattered. It was always going to be empty.

If this is a true random theoretical situation, name me one instance where the prize was actually behind one of the doors Monty opened.

Can't do it? You've just eliminated one of your three possibilities.

The definition of a logical proof is "the showing of a constant, identical result in a totality of circumstances".

Dave, can you prove that Monty's door never had the prize behind it?

Let me put it this way: your initial guess determines the outcome.

If you stick to a correct guess, you win. If you switch from a correct guess, you lose.

Conversely, if you stick to a wrong guess, you lose. And (here's the kicker), if you switch from a wrong guess to the right one, you win.

But, you ask, How do you know which door to switch to if you get your initial guess wrong? Answer: Monty will tell you. It will be the ONLY door you can possibly switch to, because he has opened the only other wrong door for you. Nice Monty.

Guessing right + switching = losing (every time)
Guessing wrong + switching = winning (every time)

We already know that guessing right will happen 1 in 3 times. And we know that guessing wrong will happen 2 in 3 times. This is because there are three places the prize can be, and it is put there at random.

Therefore, switching will lose 1/3 of the time, and win 2/3 of the time.

Let's do it in reverse:

Guessing right + keeping = winning (every time)
Guessing wrong + keeping = losing (every time)

So, keeping your initial guess wins 1/3 of the time and loses 2/3 of the time.

Compare that to switching your initial guess, which wins 2/3 of the time and loses 1/3 of the time.

Thus, switching leads to improved odds of winning (2/3, instead of 1/3).

Try it this way:

There are 100 doors. You choose one. Monty opens 98 doors and asks you if you want to stick or switch.

Do you really think it is a 50-50 that your initial pick is correct?

No way. 99% on the flip; 1% on you being correct first time out.

And when you get done picking, one of the doors you didn't pick is ALWAYS shown to be empty. No matter which door you picked.

This is not exactly true. Not just "one of the doors" will be shown to be empty. One particular door, and no other.

When you get done picking, if you have picked a wrong door (which will happen 2/3 of the time), Monty can ONLY open ONE door. This door is predetermined for Monty. It is determined by (a) the location of the prize (which does not move, and he can't open) and (b) your selection (which he can't open). That leaves only one door.

By choosing the wrong door in the beginning, you have forced Monty's hand. By choosing the wrong door in the beginning, switching to the ONLY OTHER AVAILABLE DOOR will always win. Always (of those 2 out of 3 times, anyway).

Fortunately, this situation will occur 2/3 of the time because you will choose the wrong door 2/3 of the time. Therefore, by switching, you will win 2/3 of the time.

Look, either the problem is a random problem or it isn't.

Does Monty, or does he not, ALWAYS pick an empty door?

If he ALWAYS picks an empty door, the odds MUST be reset after his interference in an otherwise perfectly random scenario.

You are mixing two different probability sets together. That is not logically consistent.

When Monty consciously interferes with an otherwise perfectly random process, the process must restart with the new conditions taken into account.

If Monty occasionally revealed the prize, then your assumption that switching helps is correct.

However, the fact that he NEVER reveals the prize should tell you that the rules of probability have been violated.

There's a word for those who argue the wrong side so passionately

Ig-n-ant

Max-

Again, if Monty merely picks the 98 doors he KNOWS have nothing behind them, you have gained nothing.

If by some miracle he unknowingly opens 98 doors without revealing a prize, then by all means JUMP on his efforts and switch.

After all, he just did all the work for you, didn't he?

But you HAVE TO DIFFERENTIATE between the two suppositions.

Did Monty actually sweat out the trial and error process of randomly chosing 98 doors and successfully eliminating all of them?

Or did he cheat, and just pick the doors he KNEW were empty all along?

Two completely different things.

Birkel-

Watch yourself, boy.

Dogstar,

Let's try that experiment 10 times. You pick the door. I'll be Monty and eliminate 98 doors I KNOW are wrong. You stick with the original door all 10 times. Do you really think you will win five of them?

Maybe, maybe you'll win one. Maybe.

Or try it this way. Go ahead and guess (from Jan 1 to Dec 31) what my birthday is. After you have guessed, I'll eliminate every day except your guess and one day I pick. Do you really think its a 50-50 that your original guess is correct?

The fact that Monty "cheats" by knowing the right door is the whole reason that flipping is the right thing to do. You are putting your completely random guess up against his knowledge of the correct answer.

Just to stir the pot.......

Suppose the initial contestant picks door A. Monty then reveals door B and asks if contestant whichs to switch. At that precise moment the contestant has a massive and fatal heart attack. Not want to end the game Monty goes outside and finds some strange and offers him the choice of door A or door C. How on earth is that NOT a 50/50 proposition. Then new guy only has 2 choices, not 3.

This has been my point all along. The odds change with each choice made. For example the odds of you tossing a coin and getting heads 5 times in a row is 1 in 2 to the 5th power or 1 in 32. But after the inital toss come up heads the odds become 1 in 16. If you hit heads again then the odds become 1 in 8.

And to whomever asked early. Yes, I do play cards and I am quite good at it.

Midaz, Midaz, Midaz, ... Let it GO.

Look, it's been tested over and over and over, and the mathematical and statistical academic communities are now in agreement. Hell, test it yourself if that's what it's going to take to convince you that you're mistaken here. Beating your chest and insisting you're right isn't going to change scientific fact. Let it go.

I BELIVE THIS IS CORRECT- im sorry if someone has already said this.

imagin you have ten boxes to choose from instead of 3. you have a one in 10 chance of choosing the right one correct? now monty removes 8 of the other boxes, leaving the one you chose and one other.
the odds of the box you originally chose still only has a 1 in 10 chance of being correct-that hasnt changed. 1/10
when monty removes all of the other unchosen boxes he is essentially telling you that the one box he doesnt remove is either the box with the prize or it is not 50/50 or 1/2

therefore you should always take the second choice.

if you have 3 boxes:

your first choice will have a 1 in 3 chance of being right. Monty will then remove one of the wrong boxes. there is still 1/3 (33%)chance of your initial choice being correct. however there is a 1/2 (50%)chance that the other box has the prize. so by changing your answer you increase the odds of winning by 17 %

Excellent post, Max! For some reason, that birthday scenario clicked in my head, when the 100 doors thing didn't.

I still think the 2/3 switching probability is wrong.

But like atomic amish, I now think switching increases your odds, from 1 in 3 to 1 in 2.

The key is to start over with the probability calculations after a change is made in the data.

For Max's birthday, my first guess would be right 1/365 of the time, and switching would be right 1/2 the time.

Interesting mental exercise...

Does Monty, or does he not, ALWAYS pick an empty door? If he ALWAYS picks an empty door, the odds MUST be reset after his interference in an otherwise perfectly random scenario. You are mixing two different probability sets together.

The odds would only be "reset" if the prize were moved (or not) at random, from its original starting point, after Monty revealed one wrong door. If the prize location were re-randomized (as between the two remaining doors), then, yes, the odds would be 50-50 that it is behind either of the unopened doors.

But that is not what happens in this game. The prize (according to the definition of the game) does NOT move once the game starts.

The re-randomizing scenario is what you are describing when you say that Monty will always remove one door from the game. He does, but his selection of that door is NOT random. It is influenced by (a) your original selection and (b) his knowledge of the location of the prize.

I think you are confusing two factors: (a) the fact that you always have imperfect knowledge even after Monty reveals one wrong door, and (b) the possibility the prize could be moved after the game starts (which would mean that switching is a 50-50 proposition).

Just because you have imperfect knowledge does not mean that the prize could move. The prize does not move based on what your initial guess was. It is fixed. Your guess influences Monty's actions (completely determining which door he will open if your original guess is wrong), but your guess does not move the prize.

Just because you have imperfect knowledge does not mean that your choice of whether to switch (or not) will win (or not) with the same degree of randomness as if the game started with only two doors.

In short, the only way that it would be 50-50 would be if the location of the prize was re-randomized.

(Incidentally, it would also be 50-50 if Monty's choice of door were also random, but you would have to ignore all the results where he picked the prize door or the same door you picked. In other words, his random selection of a door would only be limited to the situation where you picked the right door to begin with, thus giving him two wrong doors to select from. The game would be random because you have no way of knowing if he opened an empty door out of two possible empty door or if it was the only choice he could make in that game, and any other would have voided the game. But that is a side-issue.)

Not wanting to end the game Monty goes outside and finds some stranger and offers him the choice of door A or door C. How on earth is that NOT a 50/50 proposition?

You have just changed the facts. By taking the man off the street, you have, in effect, re-randomized the location of the prize. You haven't done so by actually re-randomizing its location, but you have deprived the new contestant of the important information that the dead guy had -- namely, what his initial guess was.

If the new contenstant were told what the dead guy had picked, then of course it makes no difference. Switching still gives you an advantage, no matter who does it. But you have to know what the original guess was in order to either switch it or keep it.

But George there is a logical contradiction there. By switching contestants you agree with me that the new guy has a 50/50 chance no matter what provided he isn't given the information that the previous contestant had whittled it down from 3 doors to 2. The thing is they are both presented with the exact same choice. So how can the odds be different, 1 in 2 for the new contestant but 2 in 3 for the old one. Nothing has changed there except for the foreknowledge of the first choice.

aint this fun?

Midaz...

Suppose the initial contestant picks door A. Monty then reveals door B and asks if contestant whichs to switch. At that precise moment the contestant has a massive and fatal heart attack. Not want to end the game Monty goes outside and finds some strange and offers him the choice of door A or door C. How on earth is that NOT a 50/50 proposition. Then new guy only has 2 choices, not 3.

I know it's confusing... conditional probability can get very hairy.

The flaw you have made is that the doors do NOT have an equal probability of holding the prize. One of the doors has a 2/3 probability and the other door (original pick) has a 1/3 probability. These probabilities are meaningless to the new person if he/she has no knowledge of the first pick. Without knowledge of the original pick, it is a 1/2 probability. Knowing that Door A was chosen with a 1/3 probability, however, gives them the knowledge that the other door was one of 2 doors with a combined 2/3 probability.

It ALL COMES DOWN to the fact that if you switch every time, you will win (choose wrong) 2/3 of the time, and lose (choose correctly) 1/3 of the time.

Midaz:

Nothing has changed there except for the foreknowledge of the first choice.

The odds that the prize is behind the door that the dead guy picked are STILL 1 in 3. Just because he died doesn't change that. All that matters is that the prize was placed randomly behind one of the 3 doors, and it did not move. It will always be 1 in 3 that it is behind the door that the dead guy picked. Forever.

For the same reason, there will always be a 2 in 3 chance that the prize is behind one of the other two doors. But one of those two doors has now been revealed as wrong. Therefore, there is a 2 in 3 chance that it is behind the other unopened door. In other words, all of those 2-in-3 odds are now effectively concentrated in that one door (since no one would pick the wrong door that Monty opened).

Keeping the original door that the dead guy picked will have a 1 in 3 chance of winning. Switching to the other door will have a 2 in 3 chance of winning. This is true regardless of who does the choosing.

But the new contestant knows none of this. All he sees are two doors. His ignorance of which door the dead guy picked means that there is a 50% chance that the dead guy picked one of them, and a 50% chance he picked the other.

In other words, the odds of the prize's location haven't changed. But the odds that the new contestant could pick the same door that the dead contestant picked are 50-50.

As a result, half the time the new contestant will pick the same door that the dead guy picked, and half the time he'll pick the other. In other words, half the time he'll "switch," and half the time he won't.

It would be no different than if half the contestants switched and half the contestants did not switch (or if the same contestant switched half the time and kept half the time). Either way, the total wins for the show would be 50-50.

Why? Because if every contestant switched, then the prize would be won 2/3 of the time. If every contestant didn't switch, the prize would be won 1/3 of the time. Exaclty halfway between these two sets of results is: half wins, half loses, or 50-50.

Lots of fun comments, and Dogstar gets my vote for clarity.

Another observation: Knowing something isn't the same as eliminating it. None of the doors are eliminated. I've never seen anyone do it, but I suppose Monty could show me a door I didn't choose, ask me if I wanted to swap, and I could choose the silly booby prize that was just revealed. I am never told that the revealed door is not an option for me.

The way I see it, there's still three doors in the equation, and that's why the odds aren't 50/50. It may be painfully obvious that I would choose among the two remaining, and that's where emotion kicks in over logic.

Monty's giving me knowledge, not eliminating doors.

I'll try one more time ...

Look at the results of a switching strategy:

1. If you initially pick a wrong door, then Monty MUST open the ONLY OTHER wrong door (He can't pick your door, and he can't pick the prize door. In those cases where your pick was wrong, that leaves only one door for Monty.)

2. Therefore, if you pick a wrong door, then whichever other door that Monty doesn't open MUST BE the prize door.

3. Therefore, if you pick a wrong door, and you switch to the door Monty doesn't open, then you MUST win. (Think about this point very carefully.)

4. We know that you will pick a wrong door 2 out of 3 times, and the right door 1 out of 3 times.

5. Therefore, picking a door and switching wins 2 out of 3 times and loses 1 of 3 times.

Look at the results with a "keeping" strategy.

1. If you pick a wrong door, then keeping that door always loses. (This is patently obvious.)

2. We know that you will pick a wrong door 2 out of 3 times, and the right door 1 out of 3 times.

3. Therefore, picking a door and keeping it will win 1 out of 3 times and lose 2 out of 3 times.

You have no control over whether you pick a right or wrong door at the start. You only know that you will be right 1/3 of the time and wrong 2/3 of the time.

But you have control over switching or keeping. Switching will win 2 out of 3 times. Keeping will lose 2 out of 3 times.

That's really all you need to accept to believe that switching is better than keeping.

Please point out any flaw in any one of these steps.

Dogstar gets my vote for clarity

Yeah, for being clearly wrong, maybe.

Maybe this will be simpler:

You have a two-thirds chance of initially selecting one of the two false doors. If you do select a false door, then, if you switch doors after the other false door is revealed, you will necessarily select the true door (b/c you're on one false door already and the other false door has been taken out of play). Thus, if you use the switching strategy, you will win two-thirds of the time because you had a two-thirds chance of initially selecting a false door.

I think the problem for some people (it did for me, initially) is in thinking that the decision whether to switch doors is a discrete event. But it's actually not. The outcome of that decision depends on your prior decision (which of the three doors you initially selected).

Of course, several commenters have posted on the fact that there are a number of computer models of this game to show this is true. If those proposing the odds as 50-50 were correct, you'd think this would play itself out in multiple-iteration simulations, but it doesn't. In fact, the simulations demonstrate that, in fact, the switching strategy wins two-thirds of the time.

I love the fact that every one (me included) is willing to pound sand on what might be the least "important" post of the day.

What? My last 12 brilliant dissertations on the laws of probability didn't convince you? Let me say it again with more fervor.

Must.Not.Give.Ground.On.Critical.Monty.Hall.Issue.

It's fun to see people arguing back on forth on this. I've been asking fellow engineers this question for the last decade and it's amazing to see the arguments that happen. Here's a quick take:

Your chance of being correct with your initial guess is 1/3. Stick your fingers in your ears and don't listen to a darned thing Monty says. Keep your pie hole shut. Just to move the game along while you stand there all catatonic-like, Monty assumes this means you're not going to take the deal. What are the chances that the prize is behind the door? 1/3 as always! How is the fact that Monty was yapping away while your ears were closed and opening some other door over there going to change that? Standing pat is good 1/3 of the time, bad 2/3 of the time. QED.

Suppose there are four doors, and Monty opens only one. Your original choice has a 1/4 probability of winning.

In that case, how many pancakes would the bellboy have to put in the basement?

And don't even THINK about the condoms.

A hotel concierge fills three condoms with pancake batter and trades them to a Monty Hall contestant for 27 dollars ...

three condoms filled with pancake batter?...ahh yes-the infamous 'airtight' aunt jamima special

... I'm not finished with the question.

... after Monty opens the wrong door (revealing a "prize" consisting of the two unaired episodes from the last season of "Who's the Boss?"), the contestant drops dead of a heart attack. He is replaced by another contestant, who refuses to switch his guess when offered the option by Monty, but wins a million dollars anyway.

The dead contestant's heirs file a lawsuit for recovery of the million dollars AND the three batter-filled condoms, which mysteriously ended up in Monty's dressing room.

Now, here's the question: what are the odds that the concierge is gay?

Has anyone ever asked what are the chances of it being either your door or the one that monty opens? its 2/3 right? just as the chance if it not being your door is 2/3. If you picked door a, and then there is a 2/3 chance that its either your door or door c. If door c is revealed to have a goat, then you can say there is a 2/3 chance of it being a million dollars behind your door. Monty Hall's theorem is only correct if you don't take this into account, and the theorem doesn't. It is foolish to suggest that in repeated trials the door you stand in front of originally won't get you the money just as often as the other. Once a door is opened, you are left with a 1 in 2 chance. Simple. Monty is a cretin. End of story.

The easy way to prove this is to just list out all the states this problem can be in.

Your first choice
c - Your choice, an empty box
w - Your choice, a full box.
x - An empty box
o - A box with the money

1 2 3
c x o
x c o
x x w
c o x
x w x
x o c
w x x
o c x
o x c

Right now, 1/3 of the states have you choosing a winner.

Now Monty opens one of the boxes, what does this do to our list? Nothing, there are still the same amount of states to be in, except the X boxes are now open.

So, you *still* have a 1/3 chance of having chosen the winning box, and 2/3 that you havent. Monty opening a box cant change the probability of you choosing the box with the money. Monty would have to randomly choose from all the boxes including yours to reset that.

By Monty eliminating all incorrect doors which the contestant did NOT choose, the odds associated with that door originally (1/3) are distributed EQUALLY over the remaining possibilities.
Thus each possibility takes their original likelihood 1/3 and adds half the eliminated possibility (1/6).
The result is 1/2.

This works equally well with the 998 out of 1000 scenario.
Each of the remaining possibilities takes their original 1/1000 likelihood and adds HALF the eliminated likelihood (499/1000).
The result? 1/2.
And even with 997 out of the 1000. After eliminating the 997 and redistributing the probability, the remaining three possibilities each have an adjusted probability of 1/3.

The argument "once you accept it, it will become clear" is dangerous. That's the cart before the horse. It almost sounds Orwellian.

There's a REASON why academicians who keep producing the 2/3 theorems (drinking their own bathwater along the way and writing supporting computer programs with their faulty assumptions built in) teach in universities instead of holding real jobs and engineers at Boeing and Lockheed successfully send people to the moon.

With the given scenario, if you know how the game is played beforehand, and you know that Monty is going to eliminate an unchosen wrong choice and give you an option to switch, it really makes ZERO difference what your first choice is, and given that 'round two' only leaves two choices (which are 50/50), it really makes no difference whether you switch or not either. Nothing you choose the first time OR the second time makes a hill of beans difference in whether you're going to win or not.