Pffft!!! Face it, guy, YOU STILL WILL NOT SCORE.

She's married.

I demand a recount. My people were disenfranchised. The keyboard manufacturers all supported my opponent, and created special codes to throw the election. Early exit polls, while amazingly divergent from all pre-election polls, indicated that I was leading, even in respect to the whole Nivengate incident, which I had not campaigned on. Barbara Boxer is weeping, even as I type this. Also, I have it on good authority that bbeck caused the tsunamis by rolling a gigantic d12 along the San Andreas fault.

Won't someone think of the children!!!!

The text below is from the second link on the door puzzle page. It's a short, intuitively understandable explanation. I thought I'd invented it myself, but apparently not...

THE REDUCTIO AD ABSURDUM ARGUMENT:

Change the problem to one of 100 doors, but otherwise the same. The probability that the correct door will be chosen by the contestant is now only 1%. Now, Monty opens 98 doors, removing them from the game. Hmm, now that remaining door is starting to look awfully attractive...

The Monty Hall thing ...

Start as normal, Monty's going to open a door, but doesn't. He asks instead, "Do you want to keep your door, or trade for both of the other two doors?" Keeping your door, you have a 1:3 chance of the prize. Taking BOTH of the other doors, you have a 2:3 chance. That he shows you the contents behind the door known not to have the prize doesn't change the 2:3 for BOTH.

Don't feel bad about not getting it for a while. It is profoundly anti-intuitive.

Not very counterintuitive if you think about. *shrugs*

Guy T:

That's as good an explanation as any.

htom:

That's a good one too.

Since everyone else appears to agree with the 1/2 - 2/3 arguments, I'll go counterintuitive and call bullshit.

The theory is that your initial choice has a 1 in 3 chance of being right, and when one is removed you're still 1 in 3, but there's only one left, so, set theory saying probabilities must add up to 1 means that the "other" door HAS to be 2/3 probability of being right.

But they're two separate tests, people - keeping that 1 in 3 as your probability of being right is ONLY true when there are three choices. Once there are two choices, you're right back to 50 50.

In other words, it's a trick question.

While I do understand what you're saying about the odds...the choice isn't both doors. Whether or not he opens one in advance, you still, ultimately, will 'win' the prize behind only one in three of those doors. The fact that you first chose one door, and then switched, simply replaces one with the other, it does not allow for the opportunity to 'win' if it is then revealed that you had, originally, picked the first door.

Now, were the question:

What should you do? Does switching doors mathematically improve your chances of having chosen the million dollars...

Then, yes...2/3

The kicker is that they they have used the word 'winning', and by doing so, they have still provided an only 1 in 3 chance of 'winning'.

Now, if you're talking averages, then you are looking at odds of 2 in 5.

First choice was 1 in 3
Second choice was 1 in2

It seems Patton and I are together on this one.

I was beginning to wonder if I would be the only dissenter in the crowd.

Don't believe it? Run the computer program.

I'm sure you could even set something up in Excel.

Alternatively, read the Dean Esmay essay.

And finally, here's a java game.

href="http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html

BayesTheorm.

Links are bunk. Sry.

I, still, contend that the phrasing of this question changes the answer.

The pivotal word being 'winning'.

htom, I like your explanation, it's short and understandable. The one I liked at the site GuyT mentioned was the Empirical Model. It's really just a version of htom's explanation with a little more detail. It goes like this: (And pay attention Patton & jmflynny)

Assume the goodies are behind Door A. Now there are only three ways the game can proceed:

The contestant picks Door A. Monty opens either Door B or Door C (it doesn't matter which). If the contestant switches to the remaining door, he loses. If he doesn't switch, he wins.

The contestant picks Door B. Monty opens Door C. If the contestant switches to the remaining door (Door A), he wins. If he doesn't switch, he loses.

The contestant picks Door C. Monty opens Door B. If the contestant switches to the remaining door (Door A), he wins. If he doesn't switch, he loses.

All other possibilities are permutations of the above, the only difference being that the goodies are behind Door B or Door C instead of Door A.

Now, since the contestant's first choice is random among the three doors, each of the above three scenarios are equally likely. Two of them result
in wins if the contestant pursues a "switch" policy, though. Therefore, the better choice is to switch.

Didn't Esmay also swear up and down that Kerry was going to win?

I'm unconvinced here. I keep seeing "try it empirically" but I haven't seen it tried empirically.

I'm in with Patton and jmflynny. When the choice *always* exists between two doors (because Monty always opens one) and the prize location is truly unknown (random) then the probability is the same as calling heads/tails on a flipped penny; .5

OK, I finally figured it out, and Esmay is right on this one. He was still wrong on Kerry though.

It is still true that you will be correct 50% of the time with a coin flip. I can also guess gender 50% of the time with a coin flip, but if I see a beard on the face and a cock between the legs, the coin flip may not be the best determination method.

The trick of the problem is that Monty will NEVER pick the winning door to reveal. Our nature is to keep thinking that the randomness is truly random, but it's not.

Chances would be reduced to 50/50 if Monty opened our chosen door first to reveal that it was not the winning door. If we were THEN asked to choose from the two remaining doors, or chances would be a solid 50/50.

But, Monty is not truly choosing between three choices to reveal, he's only choosing between two choices to reveal. By reducing our choices based on what he knows, not on randomness, Monty is fooling us into staying put because our minds are furiously seeing "random" selection and not thinking thru the slight conceit of the generous "reveal".

Not sure I'm saying it well.

I think the problem with CraigC's explication of the problem is that he "rigs" it (or the problem itself is rigged) by having Monty always opens a door without the prize. So it isn't really a random situation, and random probabilities don't apply as we would intuitively think of them. Or not.

If the contestant didn't have the prize, switching is 2/3 proposition. If contestant did have the prize, switching is a 1/2 proposition. The game is rigged to favor switching because Monty never opens the door that had the prize.

I think these latest explanations are convincing me.

It is better to switch. The key is MONTY KNOWS WHERE THE MONEY IS. He will never reveal your door first and he will never reveal the door with the money first. I've been sitting here trying to write an explanation for about 15 minutes here and I've got nothing, so I hope that helps someone.

Paul beat me to it. That might be as good an explanation as you're going to get because even though I think I understand it it's hard for me to explain it in any way that gets around the central problem that nothing is changing behind the other door you didn't pick. All I can say is it's all in the predictability that the reveal won't show the money or what's behind your door.

No, no, no. Paul and Chris, there's nothing rigged about it. That he shows you a door without a prize is the problem as it is given. Besides, whether it's in the context of a problem or the context of the show, he's not going to show you the door with the prize, or else you don't have to choose, and there's no show.

Did that make sense?

CraigC,
I think we're in violent agreement. The problem as given is not a true 1 in 3 random choose, as our minds make us think of it intuitively since that is how it starts out.

I think Joan actually explains this better than I did (or at least differently) and beat me to it, though I didn't see hers before I posted my own hamhanded explanation.

On one hand, this bothers me a bit since Mrs. Niven is a friend. (I used to maintain their computers.) OTGH, this is the sort of behavior Larry and Jerry sent their composite autobiographical protagonist in 'Inferno' to Hell for punishment and redemption.

FWIW, I ran through 100 iterations of the Java game that Fat Kid posted. I stayed put with my choice every time, which should have (accordiing to Dean) given me a 33.33333% win ratio.

My actual win ratio was 41%, food for thought?

Violent debates about this in grad school.

The key is understanding that for any situation you have to completely re-evaluate the odds throughout the process as new information becomes available.

A computer science major/child prodigy was insistent that switching doors made a difference and couldn't grasp the fact that the "switch doors" option only makes sense when looking at the problem initially, before Monty opens a door.

Once Monty opens a door, you have to recalculate your odds. The door he opened no longer makes any difference, and you're at 50/50.

Monty's knowledge makes no difference whatsoever for the sole reason that he doesn't have control over which of the two remaining doors has the prize. Yes, he picks which one to open, but that is irrelevant since one of the two doors you don't pick won't have the money (and be safe for him to open).

If you bet on a horse at 2:1 and there're two other horses at 2:1 (never happen, I know, but play along), then just because one horse drops out doesn't mean yours has a lesser chance of winning.

RobS, that was exactly my point as well.

I did not use the computer, but when you average the choices, the percentage comes out to .41.

On a serious note: Eric, sorry if I offended, but the attitude and behavior at SF cons aren't quite the same as normal life. Said incident happened in the con suite during the middle of a crowded party, so I don't think anyone was trying to be underhanded. Afterwards, I and several others were invited to Niven's room for stingers (that's a drink, ya pervs), including the guy I was staying with for the weekend -- who'd been with me the whole time. Needless to say, that's not how I would act just every day, not even when I was single.

And on a not-so-serious note...

I'm touched, somewhere. Thanks much for the picture, Ace and James, but lambskin clogs my printer. And I'd love to hang it on my wall but I'd have to remove either my Matrix poster or my framed print of a dragon in a nightgown, and those don't come down for anything. Respect my authoritay, fellow geeks!

As for the Monty Hall problem: I first heard it back in my Sr. level statistics class (yep, BA in Math, GEEK) and professionals have argued both sides of the equation. I come down on the choice to switch doors after the first is eliminated. What makes the difference and changes the odds is the fact that the first eliminated door is not picked at random.

Now if you'll excuse me, I have to go mutilate a couple of Beanie Babies to make a griffin for my husband's helmet crest because he needs it for a tournament this weekend. And some of you probably think I'm joking.

Later,
bbeck

I will use the next four years to associate with terrorist leaders and attempt to undermine your geek outreach programs, bbeck, and we will meet again in '08!!!!!

I will also have to get my insane ex to get out and campaign for me, as well, in lieu of an insane wife.

I'm still playing with it, but here are some observations:

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.

Look at it this way...
If you did this 21 times you would pick the correct door 7 times (on average).

If you stuck with that door you would win 7 times no question.

If you switch doors you would win 14 times, simple.......

Let me phrase it in a way that gives you exactly the same odds but seems very obvious the other way.

You have the 3 doors and Monty allows you to choose one. At that point Monty gives you the option of keeping your door or taking both of the other ones! Of course you would take the 2 doors it gives you 2/3 chance to win. This is the exact same odds as it is described in the original problem, you know that when you pick your original door it has odds of 1/3 of being the correct door leaving the othe r two doors a 2/3 chance of being correct.