I tried explaining this to a couple guys at work. Then I got confused. They laughed at me.

This explanation is a load of rubbish, and it is ridiculous for someone to submit it under the guise of *science* or *statistics*. Statistics is based upon what is objectively known, and the odds for this problem always become 50-50 as soon as one choice is ruled out.

I'm hoping that this video was really intended to be a joke.

enfathom, let your fingers do the Googling.

This is legit. Let me help you:

http://en.wikipedia.org/wiki/Monty_Hall_problem
http://www.cut-the-knot.org/hall.shtml
http://www.letsmakeadeal.com/problem.htm
http://www.usna.edu/MathDept/courses/pre97/sm230/MONTYHAL.HTM

...etc, etc, etc

Awww... your FIRST comment, too! Welcome to VideoSift! Way to make an impression.

it's not 50-50.

there are 2 ways you can win the car by switching, and only one way you can lose the car by switching. that's 2:1 in favor of switching...

enfathom: Really, a quick search would have saved you the embarrassment... An interesting problem and yeah, quite counterintuitive.

wow cool! goats!!

and yea, enfathom, you really need to do some research. here, try these...
http://en.wikipedia.org/wiki/Goats
http://en.wikipedia.org/wiki/Cars
http://en.wikipedia.org/wiki/Math

its all there

Aiyiyi. Enfathom, please listen to these very reasonable sifters. Or, if you're still bent on the statistics end of things, try writing and running a simulation on your computer, or heck, just tearing up some pieces of paper, making a chart to tally your results, and going at it.

What is this, math and science answers by democracy? Now I understand why the country has ended up the way it is. I'm just using the simple principles of statistics as taught and used in the undergrad statistics courses I've been in. I guarantee that anyone of reasonable mathematical credulity would arrive at the 50-50 solution.

The fact is this: when the choice must be made, there are two doors left, and one does not know what is behind either. One is a goat, and one is a car. The chooser cannot say that the door he first chose is any more likely to be a car than the other door. There is literally no other statistical information to deduce any bias towards either one of them. Just for the record, I have to feel sorry for those of you who feel otherwise, and rather than make analogies to people who *think the world is flat* and so forth, direct you to the following URL: http://www.youtube.com/watch?v=pUmVe9qnw7E

We try and make it easy...oh, well.

By the by, just for kicks, Enfathom, would you care to apply one of the simplest ideas of statistics and work the problem out in an experiment in order to verify your claims empirically? Y'know, just to favor us simpletons who couldn't possibly have completed undergraduate and/or graduate courses in statistics?

Oooh! Nice linky rembar. Thanks!

Sure, but just to reassure those of us who may be a wee bit doubtful, can you please go prove to yourself that the world is round? It doesn't matter how many times I prove an answer to myself, it doesn't please anyone else, and either you don't know that or you're trying to do intellectual grandstanding (pretty pathetic).

You could try reasoning it out - it can be mathematically proven too, but only within the limits of knowing how to apply it. Consider two doors, behind one is a goat, behind the other is a car. You can only choose one, but have no way to know which it is. Now, tell me how is that different from the situation presented in the video? Simply, it isn't! You have no way to know what is behind the door chosen primarily, and no way to know what is behind the alternate door.

enfathom, you have not understood the explanation of the problem:

In your comments you state,
"The chooser cannot say that the door he first chose is any more likely to be a car than the other door. There is literally no other statistical information to deduce any bias towards either one of them."

You are correct in this statement if there was only 2 doors when making your initial pick. But the problem is not constrained to the 2 remaining doors. Instead it's all about the initial choice with the 3 doors. The chances for the first pick being a goat is GREATER since the the first choice has a 66% chance of being a goat. This probablity is what makes swapping the better probable choice of getting the car b/c they are most likely to make the bad choice of the goat initially. Since you're most likely to make the wrong choice the remaining door then is the right choice.

What does this have to do with the world being round? Stop trying to build a strawman argument or whatever hand-waving you're trying to do, and deal with the actual issue at hand. The problem does not begin with two doors, it begins with three. And since the greatest chance of choosing a wrong door begins with that initial choice, it therefore becomes to the player's advantage to rechoose, thus exchanging a lesser probability of choosing the right door for a higher probability when one wrong door is eliminated.

Also, since I know you're not going to really read the above, before you go off on your next obnoxiously self-congratulating post, please explain to me how, myself and my fellow sifters aside, mathematics, physics, and computer science professors from (and I'm taking this straight from Google as they come) the University of California San Diego, Hofstra University, the University of Southern Carolina, the University of Chicago, Rice University, the United States Naval Academy, Dartmouth College, the University of Illinois, the University of Alabama, and Stanford University have all managed to slip up and not realize how right you really are? And since of course science and math are not, as you said, democratic, perhaps you should take the time to alert them to their collective silly mistake, as well as correct each of their obviously wrong examples that they have provided on their respective websites. You could change the world of mathematics as we know it!

enfathom,

Just down the hall there, first door on your left. That's right, the VideoSift Troll's Lounge. Go on in and make yourself at home. I'm sure CaptainPlanet420, antimatter, and Quantumushroom are in there already and will be happy to show you around. BillOReilly pops his head in on occasion, but he's just there for the free donuts. He's actually a real teddy bear. And again, welcome! Huh? What was that? Ooooh! You were looking for the YouTube Troll's Lounge! Well, a much larger facility it is! How could you miss it? Oh, never you mind that now. Just go out the way you came and look for the huge neon sign. Trust me, you cannot miss it. Ok... buh-bye! Oh, and if you get lost, just ask someone where you can find lonelygirl15.

The best explanation I've heard of this problem is to imagine that on the game there are 100 doors. One holds a car, the other 99 hold a goat. After you pick your door, the host opens all but two doors and asks if you want to switch. Think about it, you had a 99% chance to pick a goat the first time around, and now the host has narrowed it down to 2 doors. Now enfathom, do you still think there is a 50:50 chance of winning the car?

Welcome, Enfathom!
Way to jump right in.

Nice, jimnms. That's easier to wrap one's brain around.

upvote for the video. Made it a favorite due to the comment section.

I agree with the 50-50's. As the rules are set before you even are asked to choose, the final odds come down to 50/50. Either you win, or you don't. Either you switch, or you don't. The third door is irrelevant to the question because it was always going to be eliminated.

It's like one of those word games where you do a bunch of math concerning $100 for three hotel rooms, and find you've lost $1 into nowhere. Extraneous info thrown in to confuse the question.

Someone should figure out the odds for Deal or No Deal. THAT actually is different from a yes or no choice.

Payback:

The third door is not irrelevant b/c it AFFECTED the initial choice when choosing from the 3 doors. You cannot ignore the 3rd door when it did exist as a valid choice when making the first choice.

Oooh, he hasn't replied after Rembar's slapdown.

All but 1 wrong door will be eliminated whether you chose the right one, or a wrong one. This makes your FINAL choice 50/50. Either right or wrong. Talking about how many choices you COULD have had, before any more than 1 decision disappears, is irrelevant. It was a completely different set of probabilities. Once you alter the terms of the choices, you start anew.

BTW, I'd switch.

Omg, a vid on this same topic was sifted and discussed here many moons ago: http://www.videosift.com/video/Numb3rs-Probability-for-Beginners-Anyone-explain-this

No, Payback, this is a common mistake when performing an analysis of this problem. There can be no independent probabilities, because there is information conveyed that changes one probability set to another, it doesn't just create a new set, through the opening of the non-car door. The fact that the host must intentionally pick a door that does not have a car behind it means that the door elimination is NOT random. Once he alters the terms of choices, the probability set changes to reflect the situation, it doesn't create a new one. If he were to randomly pick a door to open, then a third possibility (the host opens the door with the car behind it before the player even chooses a door) would occur 1/3 of the time. But since we know the host will never do that, we must eliminate that possibility in the second round of choosing, thus leaving us with a 1 out of 2 chance on the new set, and a 1 out of 3 chance on the old set.

Honestly, every single link that's been posted here is great. Check them all out, especially the formal Bayesian analysis. It's really highly informative and insightful.

It's not actually that hard a problem when you think about it. Anyone who says 50/50 seems to have forgotten what happened before the host picks a goat.

When the goat is revealed it doesn't alter the fact that the probability of picking a goat remains at 66%. Picking the car was 33%.

If you are more likely to have picked the goat then switching makes sense. Counter-intuitive certainly but the probability means you should switch.

They'd love you on Deal or No Deal, Payback. ;-) Seriously though, with 100 doors, Monty knows which of the hundred doors has the car behind it. The chance you picked the correct door at the start is 1%. From Monty's perspective, the chance you picked the wrong door is 99%. He knows where the car is. He is not opening doors at random. He is eliminating 98 wrong doors. Now, the door you picked at the start had a 1% chance of being the car. Then, after eliminating 98 wrong doors, there is a 99% chance his remaining door has the car. If, from his perspective, the last door he opens will have the car behind it 99% of the time, why wouldn't you switch doors? With 100 doors, 99% of the time Monty will have a car behind his last door. With 3 doors, 66% of the time Monty will have the car behind his unopened door.

Or, something like that.

Bedtime...

Oh wait! Think of it this way... If you play the 3-door game, Monty has a 66% chance of having the car behind one of his two doors. You would trade your one door for his two doors if given the chance. Him eliminating a known bad one of his two doors doesn't change the odds that the chance of him ultimately having the car is 66%. You are still effectively trading your one door for his two, which is what Deano and rembar and others pointed out above, so if you read this far... sorry for the repetitive redundancy.

Ok... I just know I'm gonna be dreaming of goats. What's new, huh?

plastiquemonkey always causing problems...

Now that you've explained it Fletch I'm confused again

wow...

here's a well-written book explaining the development of statistics and probability thinking:
http://www.amazon.com/Chances-Are-Adventures-Probability/dp/0670034878

for the monty hall problem, just remember there are 2 different ways you can choose a goat door initially (where you'll win if you swap and lose if you don't), and only 1 way you choose a car (where you'll lose if you swap and win if you don't).

1. choose goat A, monty hall opens door to goat B --> SWAP, win car
2. choose goat B, monty hall opens door to goat A --> SWAP, win car
3. choose car, monty hall opens door to goat (A or B) --> SWAP, lose car

that's 2 out of 3...

plastiquemonkey, that's a good explanation.

I wish I could vote twice. Once for the vid, once for enfathom's comments. Welcome to the Sift! (Said in a Sean Connery voice from The Rock).

Fletch: "Oh wait! Think of it this way... If you play the 3-door game, Monty has a 66% chance of having the car behind one of his two doors. You would trade your one door for his two doors if given the chance. Him eliminating a known bad one of his two doors doesn't change the odds that the chance of him ultimately having the car is 66%."

For me, that's a perfect description. Imagine ownership of the doors: you own 1/3, Monty owns 2/3. Wouldn't you want to trade his 2/3 chance for your 1/3? Yep.

I'd rather have a goat. I'm tired of cutting my grass...!

Wow this comment section blew up. Super fantastic post plastiquemonkey, no one can refute that answer. (am I even using the right word?)

Yeah, but quantum physics states that all 3 doors contain both a goat and a car until you open them.

Or something like that.

If your still not seeing how this theory works, go here and run the simulator:

http://delicategeniusblog.com/?page_id=134

The sim has gathered stats from everyone that has played the simulator and 66% of people who swap have won.

It's simple if you just think about it. When picking a door the odds are against you picking the car correct? Your chances of picking a goat is 2 in 3 correct? Just because one of the doors is revealed does not change those odds. So if the odds are that you probably picked a goat and the other goat is revealed the best move is to swap doors.

I just played the simulator and stuck four times straight, winning each time. I then swapped twice and lost both times! Then I realized I forgot to shut off my inverse probability generator. :-/ True story.

enfathom and Payback, thank you for inspiring such a lively comment section. Without you this would have been a boring post.

My husband wrote a program to run this as a simulation; he ran about 5000 iterations, and it showed that the theorum is correct. It basically did it to convince me that he was right about it. My brain still doesn't want to accept it, but I know when to bow in the face of evidence.

It doesn't make sense, but science doesn't always make sense. Look up the Burgess Shale fossils sometime to see more science/nature that doesn't make sense.

likely you already know this, but if you don't, it's worth the click: http://xkcd.com/c246.html

by the way, the KEY to understand this is the following:

monty does NOT open a random door among the two you didn't choose.

monty DOES open a door which does NOT contain a car.

this increases the chances the OTHER door contains it.

i suppose.

problems arises if the goats can shoot fireballs

Actually, everyone is wrong. The correct answer is: not enough information.

If the host prefaces that he will open one of the other doors to reveal a goat once you choose one of the doors, then the monty hall dilemma holds true. However, we are not provided with this underlying assumption and need to call into question the motivation behind the host's actions. For all we know, if you choose a goat door, they reveal it right away and you lose. Meanwhile, if you do pick the right door, they might be giving you a chance to lose it by giving you the option to switch.

Again, you need to be briefed on any anticipated action before the Monty Hall problem holds true. Correct answer: not enough info.

I'm studying for my LSATs, so my brain's on overdrive.

*retract: I just watched the video again. The rules imply that the host always opens a goat door. Carry on.

this is almost a play by play of the huge argument I had at work about this approximately six months ago

I ended up getting some plastic cups and a ping pong ball and played the part of Monty Hall while each coworker that was arguing with me played a contestant.

There was still one person that left unconvinced.
I love it.

Good retract, char. Far more classy than enfathom. /salute

I was taught this and a handful of similar problems as an undergrad.

My professor had previously published and presented similar problems to medical professionals, who have a career where understanding the probability of a correct diagnosis and possible outcomes based on information from imperfect screenings can mean life or death (or just added expense and anxiety). Statistical probability can be very, very counter intuitive at times.

But what if I pick an S,T,R,H,L and E. Do I still have a 66% chance if I switch doors? What if its 2 doors and a curtain. Then what?

eric3579,

Then the only way to win is for your bid to be within $100 of the real price.

awesome video plastique.

The two key facts that keep this from being a 50-50 are:

1. The host is constrained to always open a door revealing a goat

2. The host always makes the offer to switch.

That leaves you with exactly 3 paths to 2 different outcomes. Each path has an equal probability, but since 2 paths lead to the same result, one result happens exactly 2/3 of the time while the other happens 1/3 of the time:

1. choose goat A, monty hall opens door to goat B --> SWAP, win car
2. choose goat B, monty hall opens door to goat A --> SWAP, win car
3. choose car, monty hall opens door to goat (A or B) --> SWAP, lose car

If it helps keep your head from hurting too much, the second choice would be a 50-50 chance of winning if you switch, but the host is CHEATING by knowing where the car is, so you have a 2/3 chance of winning if you switch in the second choice.

One other point: Someone mentioned Deal or No Deal earlier, but that is an entirely different set of probabilities. Keep in mind that in that game, the knowing host doesn't pick which cases to reveal, the clueless player does.

maybe if you are unconvinced you can try to imagine not two goats, but a goat and a llama. With two different animals it could be more easy to understand.

the llama is safer anyway, because it spits, but doesn't throw fireballs like the goat. i don't know what kind of superpowers does the car have.

The car talks and jumps over things. Didn't you ever watch Knight Rider?

sure but here the car is red. it's like a blonde woman: a dumber version. i doubt a red car can talk. unless it's a ferrari, maybe

I was once a ranch hand at a large central Oregon llama ranch and have experienced "green rain" so may times I think I'd rather have the fire balls.

Is this thread dead now?

This thread will not die until Enfathom comes back on here, man's up, admits he was wrong, and upvotes this sift.

basically you're more likely to have chosen the wrong door in the first place. so switching is more likely to win. took me a little while to comprehend but i now concur.

Great post nothing like a little bit of statistics to make people look stupid. I hate statistics.

"This thread will not die until Enfathom comes back on here, man's up, admits he was wrong, and upvotes this sift."


lol

Enfathom, where are you?

Okay it's time to leave him a personal profile comment to invite him back.

I guess I'm just a pessimist/realist. Either you will win, or you won't. Chances are for gambling addicts.

Also, while quantum theory states all the doors are both goats and cars until you open them, chaos theory states once you open the door with the car, a butterfly will fart in China and wipe out Podunk, Iowa with an ice storm of Mocha Chip Frappacino.

Yeah, I upped it...

I would bid a dollar over just to piss him off!

for when enfathom returns: http://www.alaska.net/~clund/e_djublonskopf/Flatearthsociety.htm

you guys gone done and hurt the poor critter's feelings... for shame!

but when he does come back, he really needs to check out http://en.wikipedia.org/wiki/goatscarsandmath

Really, enfathom might have a really good point. If you check out that Flat Earth Society link, you'll learn that it's the Ether that really makes proof of these concepts undeniable.

I think it's easy to get stuck with a 50/50 point of view becuase you're thinking of the chances that an uninformed /biased contestant would have. Yeah, they have two options, and they're choosing one of them. 1/2... etc.

You need to look at the overall probability that the end result will be a car. No?

Isn't "Ether" the liquid in the tubes through which the internets flow?

Basically.

TO SIFT THE VIDEOS THE ETHER MUST FLOW.

i think i have an aunt named ether... shes probably a bitch

Here is my favorite alternative version of the Monty Hall problem.

You have an friend with two children, and you remember that one of them is a boy, but you can't remember if the other is a boy or a girl. What is the probability that the other child is also a boy?

Even more counterintuitive than the Monty Hall problem when you discover that there is only a 1 in 3 chance that the other child is a boy.

I don't follow that, hueco_tanks.

The probability of anyone having a male versus female child is 1:1, ruling out any other influences such as the health of the mother.

So the odds that your friend has only two male children is 1/2 * 1/2 = 1/4, right?

And the base probability that the unknown child is a boy is 1/2.

Where does 1/3 come in?

Yeah, i could barely follow hueco_tanks' example using my crappy grasp of bayesian stats (its also 5am...)

... girl-boy, boy-girl, and boy-boy would be the choices adding to 100% that include at least one boy, because having two girls is not possible. Because you can't have girl-girl, , the probability that you will have girl-boy is 33%, added to that the probability that you'll have boy-girl being 33% and I get a 66% chance the unknown is a girl, while there remains only a paltry 33% for the boy-boy scenerio (and, no chance for girl-girl action).

Thinking of it as two cards face-down; you're told the pool the two cards were drawn from consisted of two aces and two kings. You're shown that one card is a king, so now you know that the chance that you'll have two aces is 0%, while the chance that you'll have ace-king/king-ace is 66%, and getting king-king is 33%.

Wait, maybe I'm wrong here... I'm going to bed.

that only makes sense if the person showing you the card (or your memory for your friend's children) is somehow all-knowing and selective. in the monty hall problem, monty is all-knowing and selective.

if the person showing you a card (in bamdrew's 2-card example) always shows you the card on the right, then the odds are 50-50 for the remaining card.

but the if person showing you a card always shows you a king (in either position, if one exists), then the odds are 66:33 (2:1) the remaining card will be an ace.

can't think of how you can do this with memory of a friend's children, unless your memory remembers boys and tends to ignore girls (deliberately?). that's not how memory works, obviously.

if all you can remember is that the oldest (note the defined ordinal position) child is a boy, then the odds are 50-50 for the 2nd child. similarly if you remember that the oldest is a girl, or that the youngest is a boy, etc. you're really remembering 2 things there: a boy/girl distinction, plus which specific child it belongs to.

but if for some reason you only remember boys (and even then not perfectly -- or you'd remember two boys, if they exist), then you increase the odds that the 2nd child is a girl. the exact odds depend on the sensitivity of your memory's "test" for boys, and the likelihood that you've "forgotten" or "not noticed" a second boy. if your memory tends to notice, for example, 90% of existing boys but 0% of existing girls, the odds are much higher that the remaining mystery child will be a girl.

you really can't make the problem work with memory, because the problem depends on that all-knowing and selective monty hall. the answer to hueco_tanks's question depends on how monty-hall like your memory is...

looris, have you not seen Knight Rider 2000?

Tsk!

knight rider 2000? no lol please tell me they didn't change the color of the car!

btw, hueco_tanks problem does not make any sense, plain and simple.

head.....hurts....!

There are four possible combinations of children. Labelling boys B and girls G, and using the first letter to represent the older child, the sample space is

{BB, BG, GB, GG}.

A two-child family has at least one boy. What is the probability that it has a girl?
In this question the order or age is not important. Therefore the set is:

{BB, BG, GB}

Therefore the probability is 2/3.

Again, the key to this question is that the "rules" of the game make it seem like a 50-50 coin flip, but really weigh heavily in the outcome.

I hate to argue this fine point, but as a probability major, this is a probability problem, not statistics. Hate to argue language

Zephyre, you're right, although I believe it was Enfathom who brought up "statistics" and then I suggested he attempt to test his claims of the probabilities with statistical methods.

@rembar - I know it was him. He's also the guy that suggested the world was round, wtf is he thinking?!?!

Anyway, the simplest way to break this down is you have a 2/3 chance of choosing a goat on your first pick, a 1/3 chance of picking the car, therefore, after one of the goats is revealed to you, you more than likely chose the other goat, and should therefore switch. These are not independent probabilities.

This problem was my introduction to stochastic processes. Blew my mind.

What the hell is this "round" everyone keeps talking about???

There's only a 2/3 probability, statistically-speaking, that the Earth is round.

Sheesh... you people and your "facts".

What the hell are "people"???

Who's Monty Hall? And why does he have so many goats?

So many arguments, so little sense.
Statistics demonstrate conclusively, beyond a shadow of a doubt,
that only a Fool would play the Lotto,
Because the odds against ever possibly winning are astronomical,
And yet each week some Fool proves Statistics wrong by winning the Lotto.

That's because the odds are pretty good somebody will win it, but pretty bad that a specific person will win it.

Anyway, I found a website that gives scientifically and statistically proven Lotto numbers for a very reasonable fee. I wasn't sure whether to believe such nonsense, but my Magic 8-Ball confirmed it. I'm just gonna wait for it to get really big, then you suckers will never see me again!

Ohh I read about this in the book "The Curious Incident of the Dog in the Night-Time". The book had a very nice flow chart explaining it too. Not only that its over all a very good book :-p

That kid problem makes no sense. The problem with it is you are working with two different sets. Not the same set. It's like when you're gambling, say you flip a coin. The probability that the first toss will be heads is 50%. Say you flip head the first time, then you want to flip the coin again. The probability that you flipped heads the first time is 100%. It already happened. The probability that you flip heads the second time is still 50%, though. The first flip does not affect the second.

You're looking at it as if you knew which child was a boy.

Say you have a perfectly representative group of four families, one with each possible grouping of children. Then you send away the family with two girls and look only at the three families with at least one boy; two out of three of them have a girl.
What you're doing is sending away both families whose first child was a girl; of the two remaining families, whose first child was a boy, only one has a girl.

OK?

I want a goat...

but goats stink, and they make poo in your car. that's why you lock them in a little room.

I stopped believing in Logic since the whole Achilles and tortoise scandal. Logic never recovered from that one. From what I've heard she's touring strip clubs with with Dice Clay.

this is a bit bogus. it says the host opens one of the doors that has a goat, it dosnt say that he wont open the door you picked first and then ask you if you want to swap (which obviously you would if it wasnt the car).

well that's a minor bug but you know it assumes it won't do that

i know it's not been long but...

*promote

Re-promoting this video to the front page as a VideoSift Classic. Originally published on Sunday 22nd April 2007 (promotion called by gold star member deputydog)

Ok, I thought about this long and hard. Originally I thought that the answer was that there is no difference between switching or not. I was wrong, and here is how I now understand it.

First you have to understand the question:
"is it better to switch, or does it not make a difference if you switch"

Then you have to understand the situation:
2 Stages
Stage one: 33% chance of picking correctly
Stage two: 50% chance of picking correctly

Then you have to understand that little tiny bit of often overlooked information, the key to understanding the answer:

The door you picked in stage one is probably wrong. You know this because the probability is 33% during stage one. If you don't understand this key point re-read it until it has sunk in. You probably CHOSE WRONG. This is key.

Since you are in stage two, here is your choice:

- stick with the door you have picked (which is probably wrong)

- switch to the other door

This deserves a *promote.

Promoting this video back to the front page; last published Thursday, May 24th, 2007 8:34am PDT - promote requested by Fjnbk.

Again: From Wikipedia

Perhaps the simplest way to understand this apparent paradox is to consider this slight modification to the problem:

1. You select a door
2. The host says you can either open your selected door, or open both of the non-selected doors. Either way, if you reveal the car, you win it.

Of course you would choose to open both of the non-selected doors, because there is a 2/3 chance that the car is behind one of those two doors. Now imagine this:

1. You select a door
2. The host says you can either open your selected door, or open both of the non-selected doors. Either way, if you reveal the car, you win it.
3. Before you can decide, the host opens one of the non-selected doors, revealing a goat.

Again, you would choose to "open" both of the non-selected doors (one of which has already been opened by the host). There is still a 2/3 chance that the car is behind one of those two doors. The fact that the host was kind enough to open one of the non-selected doors for you does not change this probability. This last scenario described is, in fact, precisely the Monty Hall problem.

Yes, it's better to switch than it is to stay. However, the chance of getting the car is still a 50/50 shot, realistically.

The way it was explained to me is the regardless of which door you choose, one door with a goat behind it will be removed which makes it a choice between a goat and a car. The first pick has no impact on the outcome as you treat the goat doors as a single entity. It's just there to create the stay/switch anomaly.
Even counting the third door, looking at the outcomes of each door you get a 50/50 chance regardless which of the three paths you go down. Goat1 leads to a choice between Goat1 and Car, and Goat2 leads to Goat2 and car. Car can lead to either Car and Goat1, or it can lead to Car and Goat2. Each of the four cases leads to an even chance of getting the car. However, if you word it like they do in the problem with staying or switching, it's better to switch, but in effect that's just manipulating it to create a quirky problem.
That's why nobody tries to seriously disprove it, because it's correct in its premise that if you get to pick a door then either stay or switch you're more likely to get the car if you switch. But even so, the chance of getting a car will still be 1/2.

I was also told there's a reason casinos live off of players like me, standing by roulette tables counting the number of times red comes up with the intent on betting black.

No, think of it this way. I have written down next to my computer a number between one and ten million. I want you to guess it. Go ahead, write it down.




Now you can pick: either the number I wrote down is the one you guessed, or it's actually 2,117,352. The only way I can represent to you a number other than the one I wrote down, is if you chose exactly the same number as me. Pretty fucking unlikely. The way the game works, you WANT to choose the wrong number initially, and switch. Unless you're insane, you'll do just that.

With the three doors, it's just harder to see. The chances of you guessing the wrong number are 2 in 3, as opposed to 9,999,999 in 10,000,000. But when I eliminated all other choices, and you stick with your initial pick, you're betting that you correctly identified the right number on your first guess. What are the chances you correctly identified my number, first guess? 33% in the door game. You had a better chance of choosing the wrong door than the right one, which is the objective of the game. Switching is the smart option.

Hang on a second, correct me if I'm wrong here, but the children problem doesn't make any sense.

The choices if you knew one child of the two was a boy are given by a few people as
[BG, GB, BB] giving a 1/3 chance of the second being a boy.
but if you KNOW one is a boy, then BG = GB, it doesn't matter which came first.

Boy-girl and girl-boy are the same thing since you KNOW one is a boy.

Think of it in terms of names. The boy you already know about is named John, and you can't remember if he has a sister named Jane or a brother named McCheesenstein III.

If you sub the names in, the choices are:
John - Jane
Jane - John
John - McCheesenstein III

Since John is in all possible choices, it becomes 50/50 that the second is a boy or a girl.

I think.

[oh, and if you argue that order DOES matter, than BB should appear twice since it could be B1 - B2, or B2 - B1 thus making the choices BG, GB, B(1)B(2), B(2)B(1)]

AAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!


AAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!

I'd like the goat please.

*promote, for a new generation of sifters!

Promoting this video back to the front page; last published Tuesday, January 15th, 2008 5:40pm PST - promote requested by Deano.

>> ^Sketch:
Yeah, but quantum physics states that all 3 doors contain both a goat and a car until you open them.
Or something like that.


Quantum Mechanics wise it should work something like this:
(without the math sort of, more of a thought experiment)

At the beginning Bob shows you the doors to pick from. To you all three doors are in a state of superposition (it could be a goat or car).

Bob of course knows the answer so nothing neat for him.

You pick the door that you think is the car thus "measuring/observing/changing" the doors state. This coincidently affects Bob who's decision was in superposition (deciding which goat to show).

Bob makes his choice and "measures" the door by opening it, allowing you to eliminate one door.

You also know that the door you picked is no longer in superposition as you have picked it and Bob's outside influence destroys it's superposition.

The last door of course to you is still in superposition as it hasn't been opened or "measured" by you in any way.

At the beginning as they all had superposition each door had a 33% chance of being the car and 66% chance of being a goat. When you make your choice for the door Bob's knowledge destroys it's state of superposition and can now only be one thing, not either. The other door is opened as well destroying it's superposition and confirming it's fate.

The last door however, is still in superposition from your point of view. The door still has it's initial setup of 66% goat and 33% door. But, since Bob has measured the system from his point of view it has changed the states the doors were in (on the second opening).

Now that you know which door has one goat and the other door which IS a goat or car you have the last door which is in superposition. As the system changed (and no you can't start a multiple step problem in a closed system and look at the last step and call out 50/50, this is a procedural problem, breaking it up into a separate system would be akin to saying acceleration and a vector are the same thing) it altered the probabilities initially setup. One door can only be a goat or car, making it 33/33. The door that is in superposition though still has it's sate intact therefore the chance is 66/66 respectively.

But, you know you have one goat already out. So by inderect observation you can now see what the probability for each door is. The door measured by Bob has a (goat/car) 33/33 chance of being either. The door in superposition can now be updated without destroying it's state do to the other door. It's chances are now (goat/car) 33/66. I'd pick the door in superposition.

That was a little long winded hopefully I made it understandable.





Time for some cake!

*promote

Promoting this video back to the front page; last published Thursday, December 18th, 2008 6:01pm PST - promote requested by rasch187.

Randomly stumbled across this *quality gem.

Boosting this quality contribution up in the Hot Listing - declared quality by dystopianfuturetoday.

The most important question: What kind of car is it?

Think people!

Wow I had to read this whole thread to actually get it. Now I feel stupid
edit: wow someone did some gravedigging

fastest explanation: write out the possibilities, switching every time
If car is behind door A
Pick A, host removes B/C, switch B/C: lose (B=C)
Pick B, host removes C, switch A: win
Pick C, Host removes B, switch A: win

They've done studies, you know. 60% of the time, it works every time

If world of warcraft has taught me anything it is that random loot is random, % of drop isn't a guarantee. Secondly, the problem with this explenation is that it seems to assume that you'll have multiple chances, which I don't think you do in this scenario. It really doesn't make any difference whatsoever. You are still gambling and you still don't have control over the outcome, even if you up the %.

So roll need for [Car] and greed for [Goat].

People are dumb.

This made me finally get it! Thanks @jimnms
>> ^jimnms:

The best explanation I've heard of this problem is to imagine that on the game there are 100 doors. One holds a car, the other 99 hold a goat. After you pick your door, the host opens all but two doors and asks if you want to switch. Think about it, you had a 99% chance to pick a goat the first time around, and now the host has narrowed it down to 2 doors. Now enfathom, do you still think there is a 50:50 chance of winning the car?

It had me at booby.

*length=5:48

The duration of this video has been updated from unknown to 5:48 - length declared by radx.

Saying you have a fifty percent chance no matter what is like saying everyday there is a 50-50 chance of rain. You end up ignoring math/science in favor of a simple yes/no, on/off, 0/1 binary response.

It's a fun little problem, complicated by the fact that the gameshow host does not choose a door randomly.

If the host did choose a random door of the two remaining, then it would mean that 1/3 of the time he would pick the car, and you would walk away with nothing. If that was so, then your chances of winning the car would increase from 1/3 to 1/2 if he picked a goat, and it wouldn't matter if you switched or not, and your chances would go to 0 if he happened to open the door with the car.


But - the host does not choose randomly - if you did not choose the car, then he will pick the door with the remaining goat. And thus - your chances increase from 1/3 to 2/3 if you switch.

My favourite explanations of the problem:
- Do you want to pick one door, or do you want to pick both the two others? You get both the others if you swap.

- @plastiquemonkey's explanation - you'll pick goat A, goat B or the car - you only lose if you picked the car first, which is 1/3.


And *promote the good old probability problem.

Promoting this video back to the front page; last published Sunday, April 22nd, 2007 9:55pm PDT - promote requested by Zawash.