May 2008

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Comments

ChrisB

I think almost nobody on here has appreciated what Scott is getting at on this one. Yes, if you know your math, it is no mystery that if the game show host effectively says to you, “you can stick with your choice (2/3 prob of being wrong) or you can swap to these two other doors and if the car is there (2/3 prob of being the case) I'll guarantee you a win by opening a goat door”, then swapping would be in your interest.

It is also understandable that if the game show host tells you honestly that he also has no knowledge of where the car is and he says to you, “you can stick with your choice (2/3 prob of being wrong) or you can swap to these other two doors and I’ll randomly open one and what’s behind the remaining door is yours”. If you were wrong on your first choice (2/3 prob) there’s a 1/2 chance that he’ll open a goat door and you win. Thus there is a 1/3 chance of winning if you swap and a 1/3 chance if you stick so deciding to swap before he opens his door gives no advantage.

This last scenario however is not the game. You don’t have to decide on swapping until after he has opened his door. This is where my brain just doesn’t know any more. I’m thinking, if he opens his door and there’s a goat, regardless whether he opened the door randomly with no knowledge or intentionally with knowledge my choice is now stick with my original choice (2/3 prob of being wrong) or go for the last door which if my first choice was wrong must now contain the car with certainty. I’m still thinking therefore that if I can make my choice after seeing the goat I would now be wiser to swap. His prior knowledge is irrelevant surely? We both have now gained some knowledge. The math experts however tell me that this is not the case. Although the situation, after the goat is revealed, is physically the same it is different because the game show host made a random choice.

Before you all rush to tell me, I know that before the game started there was a 1/3 chance that he would reveal the car but psychologically I don’t care about this case as then swapping or sticking makes no difference, as I loose every time presumably.

To illustrate the dilemma better what do you do if you don’t know whether he opened the goat door with knowledge or randomly. Or what would you feel in the famous 52 deck of cards scenario if you select a card at random and he turns over 50 cards at random and the target car isn’t there. I know it’s mathematically wrong but I’d be thinking if the target card was still in the pack it would be extremely unlikely that he wouldn’t find it in 50 trials, maybe that means I’m more likely to have it. Or I might be thinking I had a 51/52 chance of getting the wrong card and now there’s only one card left unturned from his pack so If I chose wrong (very likely) the unturned card is the right one with certainty so I should swap. The problem illustrates how our brains cannot assess risk logically but does that mean we are wrong?

Michael

"If reality were objective, statistics wouldn’t be influenced by knowledge."

Statistics AREN'T. What is influence is PROBABILITY. And it is obvious that probability is influence by knowledge. Isn't it?

Robert

Why did you consistently misspell Monty? It's spelt right in your link.

In neither case is it the actual knowledge that matters. The reasons the probabilities change is that when Monty knows which door the prize is behind, he never opens that door. If Monty doesn't know which door the prize is behind but just happened never to open the door with the prize behind it consistently in an experiment based on this, then the empirically measured probability would be the same as in the "Monty knows" case.

In the case of the double slit experiment, actual knowledge doesn't matter either. If you measure which slit the photon goes through, and then throw that information away or conceal it forever, or simply don't look at it, the interference pattern still disappears. It's the _measurement_ that's important, not the _observation_.

William

For some reason my post seems to have been lost, but to put it simply this example doesn't work since you're talking about a single instance, and statistics don't really work like that. You can talk about how there is only a 1 in 10 000 000 chance of getting hit by a falling 10 ton barrel of yak butter, and so you're safe.... until you're the poor git who's tombstone reads something pithy like, "Yak'd to death.".

Anyway, I only get to choose once, then statistics doesn't apply. If I get to choose an infinite number of times then I'll have nearly an infinite number of cars no matter whether I change or not, and leaving me with a serious parking problem.

Tormod

WHAT!? SERIOUSLY!?!

Do you mean to say that if someone with knowledge DIVULGES something to you, then you are in a position of making an informed decision improving your odds??

WOW!? That's crazy talk!!!!

(Sarcasm, yes)

Anna

I like the links posted by someone (sorry, couldn't find your entry again in the list to mention by name) because it helps prove a point.

I just used this simulator:

http://www.versificator.co.uk/misc/monty-hall-problem.html
and the first time I played, I kept my first choice of door and got the car.

Obviously, if I had many opportunities to play the same game, in the long run I would do better switching doors, but since the game is being played only once... I wouldn't use probability (long-run relative frequency) to make a decision in this case, and the simulator showed that there's nothing wrong with my choice:-).

Chris B

Too many Chrisses, I'd better get more specific.

@Other Chris / Dave, Monty random vs. Monty knowledge *would* affect the probability IF, and only if, it were possible for him to pick the car.

Scott already said Monty didn't get the car. Presumably, if you ran the experiment over and over, 1/3 of the time Monty would open the door and get the car, and you'd be right.

The times you run the experiment and Monty gets a goat, EVERY TIME that happens, the "switch" door still has 2/3 chance of being correct.

And that will ALWAYS happen, as long as Scott says "Monty opened a door and got a goat."

Which, by the way, is *EXACTLY* the same thing as Monty knowing beforehand that he would get a goat--Scott saying "he opened it and got one" is precisely the same thing, math-wise. There was 0% probability of getting the car, because Scott defined the problem by saying "he didn't get the car." How is that different from knowing where the car is?

The whole trick is to not make it more complicated than it is. There are three doors. You get one, with a 1/3 chance of being right. Monty gets 2, with a 2/3 chance of being right. Monty is offering you his 2, instead of your 1. That's the WHOLE math of it. All the rest is distraction to confuse you. As long as Scott never has Monty pick the car, the problem doesn't change.

In other words, it doesn't matter if Monty knows... as long as SCOTT knows. :)

TallDave

"There are 2 doors, so the odds are 1 in 2 that they already have the right door, and 1 in 2 that switching will get them the car."

For a person who walks in when there are only two doors, this is true. However, he is unaware of the prior knowledge of the person who chose a door; he has been indirectly informed by Monte that that door is less likely to be the correct one. An operation has been performed that indicated the other door is more likely to have the prize.

In the coin analogy, this is as though someone has weighted the coin to make it land heads more often.

Another way to consider this puzzle is to pick one card rnadomly from a deck. The odds of you picking the Ace of Spades is one in 52. Your friend helpfully removes 50 of the non-AOS cards leaving only your card and another. There's a 51/52 chance you did not pick the AOS and that the other card is the AOS.

Now, to another friend who walks in, there's a 50/50 chance either card is the AOS. But you know better.

It's a limited information problem. That may seem "subjective," but only the calculation of the odds is subjective. The cards will not come up AOS 50/50 for your friend; your card will be the wrong one 51/52 times.

KD

I've heard of this one before. I always figured the best way to do it is to either do a live experement and see how the odds change or try to do some computer model if you have a good random number generator.

TallDave

"That is interesting enough on its own. (I’ll give a link later that explains the math of it.) But here is the freaky part. You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge.

If reality were objective, statistics wouldn’t be influenced by knowledge. "

No, the freakiness is an illusion. Here's why: if Monte doesn't know which door the prize is behind, he could open the right one instead of a wrong one and oops! there's the prize. That's why knowledge matters here: it's the statistical difference between the likelihood of the scenarios in which Monte opens accidentally opens the right door. There's nothing subjective about it.

Ben

Chris-
You are implicitly assuming that one of Monty's doors contains the car and the other contains one of the two goats. It is also possible that your door has the car, and his two doors both have goats. There is a 1/3 probability that he has two goats, 2/3 that he has one goat and the car. So if he opens a door with a goat (choosing the door randomly), there is a 1/3*0 + 2/3*1 = 2/3 probability that switching will give you the car (where the 1/3*0 is the probability he has two goats * probability you will win if you switch, and 2/3*1 is the probability he has one goat * the probability you will win if you switch). You thus have a 2/3 probability of winning by switching if he shows you a goat, no matter what method he uses to pick the door he opens.

Dave Oblad

Ok, I couldn't resist still yet another comment.

I explained below how the double-slit experiment can be shown to form faster than light
communication via entangle light beams. To explain this I have to suspect that there
exists a common point that connects all other points in the universe. This common
point is the cause of quantum probability. It's pseudo random noise. I will even
suggest that when we get quantum computers working on analysis of this noise we
might find signals outside of Random Probability. Injected signals from ET's.

SETI, while a cool project, reminds me of a bunch of primitive secluded Islanders using
giant man-made ears to listen for remote drum beats. Lack of detecting such signals
forces them to conclude that they are the only humans on the planet earth.

Meanwhile, radio waves imprinted with "I Love Lucy" re-runs are passing through
their bodies. Perhaps, when we re-direct our focus to decoding Quantum Noise, we
might find re-runs of "I Love Fuzy" being broadcast this very moment from a galaxy
far, far away.. just a thought!

What?... I'm not supposed to reveal the Quantum Singularity Continuum yet?.. Ok!
Never mind folks ! (lol)

Best wishes from Dave :^)

Dave

Monty's knowledge DOES affect the probability outcome. When you picked your door originally, you had a 33% chance of being right, while the filed had a 67% chance of being right.

If Monty picked randomly, then he had a 33% chance of picking a car, which he would do 33% of the time. If he got lucky and picked the goat by accident, then the odds of both of the remaining two doors increase to 50-50.

If however Monty KNOWS that he is eliminating a booby door, then the odds on your door remain 33%, but the odds on Monty's singular unopened door increase to 67%, to reflect the original value of the field.

Chris

Scott, I have to echo what some of the others have said... something I missed by reading your post too fast.

Millions of people are wrong in thinking the Monty Hall problem isn't correct, but *you* are wrong in thinking that Monty *needs* to know what's behind the doors. He only has to know what's behind the doors to guarantee he never picks the car. If he tries one of his at random and *doesn't* pick the car, it's just as good as if he'd known.

In other words, when you pick a door, Monty has the other two. So your door has a 1/3 chance, and Monty's two have a 2/3 chance. Full stop.

Knowledge of the contents have nothing to do with it. Monty is telling you you can have your 1 door, or have the best of his 2 doors. All the rest is wording designed to make it counterintuitive, which works VERY well, judging from the constant outcry of the Disbelievers.

If Monty opens one of his and it's a goat, *whether he knew it or not beforehand*, Monty's two doors still have a 2/3 chance. Not knowing about it beforehand means that Monty *could* have opened the car, but you said he rolled the dice and got a goat, which is no different from having *known* he would get a goat. The two doors Monty had gave him a 2/3 chance of having the car and it's still true.

What you say *is* true if sometimes Monty can get the car, because he didn't know what was behind the doors. Makes for a pretty lame TV show and a pretty unhappy contestant. But if you only show the ones where it's a goat after all, then it's just as good as knowing which was which.

Chris

I noticed that a number of posters seem to believe that it's still in the player's interest to switch doors even it Monty picks a random door. I don't think anyone has corrected them yet directly so I'll pipe in.

In that case, the probability of winning has to include the probability that Monty will pick the car. In the original problem, there was no probability in Monty's decision; now that there is it changes the whole thing.

There's a 50% chance that monty will pick a goat and 50% that he'll pick the car. If he picks the goat then you switch and there is a %66 chance that you win the car. But if he picks the car then there is a 0% chance of you winning the car (but you still switch even though it doesn't matter because we've agreed on the 'always switch' rule). So your chance of winning is (50% * 66%) + (50% * 0%) = 33%. Not 66%.

RandyR

I am beginning to think that "JD" is a construct of Scott Adams. Notice it was the very first reply and it was by far the most vehemenantly against the correct Monte Hall analysis than anyone.

Perhaps someone else did write it, but I think Scott harvested from a different place and transcribed it in reponse to his own blog just to yank our chains.

Paul

This is right in your wheelhouse Scott: http://www.physorg.com/news127499715.html

A 13 year old kid correcting NASA's estimates of whether an asteroid would hit Earth.

Mr. Wampus

A more direct correlation to your post about knowledge changing the world is the "Schrodinger's Cat" experiment (read about it here: http://en.wikipedia.org/wiki/Schrodinger's_cat)

Schrodinger devised a thought experiment to illustrate the weirdness of quantum mechanics, one component of which states that a particle simultaneously assumes all of it's potential states and only locks onto a particular state when observed. His experiment envisions having a live cat sealed inside a box with a poison bomb triggered by radiation. A quantum particle is then introduced to the box. The particle of course assumes all of it's potential states including one that radiates. The cat then is both alive and dead inside the box and only assumes a form upon the event of someone opening the box and observing which state the particle ultimately assumed; one that radiates or one that doesn't.

Weird.

jeff

There seems to be a large number of people who are arguing this puzzle is wrong. Is it that you that are too lazy to read or too challenged to comprehend the problem?

The solution/explanation is all over the internet. Just because you don't get it, doesn't mean it ain't true.

Dance, dance, dance.......

steve?

Of course knowledge changes statistics. I might even venture to say the statistics is the mathematics of applied knowledge.

The most basic example I can think of are loaded dice. One assumes that each number on a six sided die has one in six chance of coming up. However, if the die is loaded, one can change one's prediction in light of the new knowledge.

d-rob

Well, you got it wrong that Monte's knowledge is required to improve your odds of winning. All that is needed is for Monte to reveal a goat, it doesn't matter if he did it intentionally or accidentally.

The point is that there's a 2/3 chance that your first guess was wrong. But as long as there's 2 other choices, you have no way of improving your odds. However after Monte shows you one of the wrong choices, there's still a 2/3 chance your initial guess was wrong, and only a 1/3 chance that the other door is wrong. So obviously its better to switch.

There's nothing mysterious about Monte's knowledge. The only reason the players odds wouldn't improve if Monte chose at random is because in that case Monte would have a 1/3 chance of picking the car and ending the game before the player could switch.

See what I mean?

indo

One of the first things u get taught in Statistics is the difference between random and non-random processes, and how the same set of equations (ergo logic) cannot apply to both.

If Monte knows, his selection of a door is non-random; i.e. u've altered the rules of the game: it's not the game that's altered the rules, so to speak.

To put it another way -> if Monte knows, you can also deduce certain things. If you deduce certain things, u've got more knowledge of the process. If you've got more knowledge of the process, u'd be an idiot not to make a better decision than if u didn't have that knowledge in the first place.

No offence ole' lad, but maybe u should treat statistics like u do voting ... it doesn't hurt to say u don't know! ;-)

(PS - Pls don't mind the post, tis not of my will.)

David MacMillan

Everyone has already corrected you in about a zillion different ways - might as well join the fray. Hope that my explanation of the double-slit will be a bit more readable.

There's a 1 in 3 chance you picked the car, and a 2 in 3 chance you picked a goat. Now if you picked a goat, then Monty is telling you where the car is. If you picked the car, he's just confusng you. So there's a 2 in 3 chance that he is telling you where the car is.

Basically you are betting that your original choice was wrong, which is a safe bet to make since there's a 2 in 3 chance that you were, indeed, wrong. But people don't like to think that they might have been wrong.

On to double-slit....

Light is an electromagnetic disturbance, not a particle...think about it like the droplets of water flung off of a wet dog when it shakes itself. The dog has little bits of water flying off of it...a vibrating electron has little bits of energy flying off of it. The flying bit of energy is sustained as an oscillation of electric and magnetic fields (just like the water droplet's individual molecules are bouncing around). However, since magnetic and electric fields have to align in some particular fashion, there are constraints on what path the bit of energy will take.

Left to itself, an individual bit of energy (a photon) will travel based on the overall electromagnetic vector scattering that an infinite number of photons would have. It's not a finite particle. However, if you observe it then you are constraining it to what you see it doing, not what it actually does.

It's really a bit more complicated than that, but it's not evidence that we are a hologram.

Eric

For the skeptics, I wrote a small Java program (a long time ago) that simulates the problem. Scott is right regarding the improvement of odds if you switch. He's incorrect about Monte knowing affecting the odds. The simulator doesn't "know" where the car is. Maybe he knew that was incorrect.

AL

I’ve read quite a few of the comments below and no one seems to agree with Scott about what he said about the Monty Hall problem. Scott is in fact correct about that if you know that the game show host doesn’t know behind which door the prize is and he only gets lucky by revealing the goat, you have a 1/2 chance of winning the prize either way. The reasoning is as follows:

If on your fist choice you happen to pick the door with the prize (probability 1/3) Monty will always reveal a goat. If on the other hand you pick a door with a goat on your first pick (prob. 2/3), Monty will have a prob. of 1/2 of revealing the prize (resulting on your loss). Provided he revealed the goat, you know that either you picked correctly or you picked incorrectly and Monty got lucky and revealed the goat. Both events are equally likely (1/3 prob. each where the remaining 1/3 prob. is that Monty revealed the prize), so provided Monty revealed a goat your chances are 1/2 either way.

Here is the reasoning in terms of conditional probability.

P(prize) = Probability of your first choice being the door with the prize = 1/3
P(goat) = Probability of Monty revealing a goat after your first choice = 1/3 + (2/3 * 1/2) = 2/3
P(prize | goat) = Probability of your first choice being the prize given that Monty has revealed a goat.

P(prize | goat) = P(prize)/P(goat) = (1/3)/(2/3) = 1/2

If Monty knows what is behind each door and deliberately chooses to reveal the goat no matter what, then as has been argued many times below you increase your chances to 2/3 by switching doors.

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