PARADE!
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So I hear there's going to be A Parade in a week in the vicinity of Cairo. So I thought I'd look it up, and found interesting information about this-here marching-band.
http://social.macys.com/parade/?cm_mmc=VanityUrl--parade--n-_-n#/lineup/bands/remove:
anyway the real WTF is the flash animation of the truck shaking (or whatever that's supposed to be) and that stupid taxicab
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Undefined is a great band. I have NaN of their albums.
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It's just shaking because it's as long as 0 bicycles, and the physical problem of being constrained to a 2 dimensional space in a 3 dimensional universe causes instabilities.
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Never before have I seen a flash "website"-style loading screen implemented in HTML. I hope to never see it again.
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When I saw the Subject header, I assumed this was going to be a WTF about the popular Sunday-newspaper supplement, the one where Marilyn vos Savant crashed and burned so badly with the Monty Hall problem years ago.
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@da Doctah said:
the one where Marilyn vos Savant crashed and burned so badly with the Monty Hall problem years ago.
My reading of the Wiki page on the issue indicates that she got the right answer, not the wrong one...??
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@PJH said:
@da Doctah said:
It's a problem that is very sensitive to how you set it up.the one where Marilyn vos Savant crashed and burned so badly with the Monty Hall problem years ago.
My reading of the Wiki page on the issue indicates that she got the right answer, not the wrong one...??
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@Ben L. said:
Never before have I seen a flash "website"-style loading screen implemented in HTML. I hope to never see it again.
Is it the design you object to, or single-page-apps? I ask because SPAs are probably getting more popular. It's because you only need to send front-end components to the web browser once, while the data needs to be updated often.
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@PJH said:
My reading of the Wiki page on the issue indicates that she got the right answer, not the wrong one...??
No, it says she analysed it as an obscure probability problem. That's the wrong answer, whatever 'probability' you come up with. The right answer is to point out that the 'trick' in the Monty Hall problem is a grammatical/linguistic one, not a mathematical one. The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
It's a lot like the 'missing dollar riddle', where the 'trick' is a simple statement of something that's untrue.
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@TDWTF123 said:
The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
@Monty Hall Problem said:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
I guess lying is a linguistic trick now?
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@TDWTF123 said:
No, it says she analysed it as an obscure probability problem. That's the wrong answer, whatever 'probability' you come up with. The right answer is to point out that the 'trick' in the Monty Hall problem is a grammatical/linguistic one, not a mathematical one. The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
It's a conditional probability problem with unequal knowledge (and in fact it is one of the simplest such ones). If you thought people sucked at traditional probability where everyone has all the facts, then conditional partial-knowledge systems are much much worse.You've probably demonstrated that you don't know what you're talking about just by saying it isn't a probability problem.
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@dkf said:
@TDWTF123 said:
No, it says she analysed it as an obscure probability problem. That's the wrong answer, whatever 'probability' you come up with. The right answer is to point out that the 'trick' in the Monty Hall problem is a grammatical/linguistic one, not a mathematical one. The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
It's a conditional probability problem with unequal knowledge (and in fact it is one of the simplest such ones). If you thought people sucked at traditional probability where everyone has all the facts, then conditional partial-knowledge systems are much much worse.Yeah. I like the explanation where the first round has a million doors and the second has two, one of which has the prize. Obviously you'd want to swap then.
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@joe.edwards said:
@TDWTF123 said:
The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
@Monty Hall Problem said:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
I guess lying is a linguistic trick now?I really didn't expect a novel of discussion about Monty Hall. It's very refreshing to learn a new way to get the problem wrong. The most rewarding part of this forum are all of the unexpected side WTFs that are uncovered.
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@boomzilla said:
@joe.edwards said:
@TDWTF123 said:
The trick lies in phrasing the question in such a way that it appears there are three doors involved at one point, even though there aren't really.
@Monty Hall Problem said:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
I guess lying is a linguistic trick now?I really didn't expect a novel of discussion about Monty Hall. It's very refreshing to learn a new way to get the problem wrong. The most rewarding part of this forum are all of the unexpected side WTFs that are uncovered.
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@joe.edwards said:
@Monty Hall Problem said:
Pretty much, yes. The lie's nicely padded out so you don't notice it. That's the root of the 'problem': in fact there are never actually three doors. There's the one you pick, and the one the host leaves you. The one he 'takes away' was never part of the equation to start with.Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
I guess lying is a linguistic trick now?
It's just the same as the bit in the missing-dollar riddle where it says 'so that's $27, plus $2 for the bellhop/waiter/whatever', even though it's not $27 at all, it's $23. The trick is in describing one situation, but then switching to another which sounds the same.
@dkf said:It's a conditional probability problem with unequal knowledge
No, it's set up to look like one, but actually isn't. That's the trick.
@dkf said:You've probably demonstrated that you don't know what you're talking about just by saying it isn't a probability problem.
I didn't say it's not a probability problem, I said it's not an esoteric or difficult one, once you penetrate the ruse: it's a trivially simple probability problem.
Since we're on Wikipedia already, let me quote vos Savant missing the point spectacularly:
@vos Savant said:Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here's a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You'd switch to that door pretty fast, wouldn't you?
Now, allow me to rephrase that in a way that is directly equivalent. There are a million doors, behind which may be prizes. All but two will be opened to demonstrate they contain nothing.
How many doors might the prize be behind?
If you phrase the MHP non-deceptively, it's 'there are three doors, of which one can be ignored because it plays no further part in this problem; a prize lies behind one of the other two; what are your chances of randomly choosing it', and at that point any honest two-thirdser will realise they've fallen for the trap in a schoolyard riddle.
Of course, this trick has been around centuries longer than the Monty Hall show, as something of a mathematicians' joke. I'm not sure many of the prominent two-thirdsers aren't just playing along. The Wikipedia article's hilarious.
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@TDWTF123 said:
Does she really capitalise her name wrong or was that just Wikipedia?
Since she made up the name herself, it's not an option to suggest that she capitalises it wrong; it's capitalised however she says it is. Wikipedia's another matter entirely.If only she were more traditional, she'd simply be Mrs Jarvik.
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@TDWTF123 said:
words
frystare.jpg.docx
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@da Doctah said:
@TDWTF123 said:
According to Wikipedia, it's not a made up name, vos Savant is her mother's maiden name. Since it's Wikipedia, it may or may not be correct.Does she really capitalise her name wrong or was that just Wikipedia?
Since she made up the name herself, it's not an option to suggest that she capitalises it wrong; it's capitalised however she says it is. Wikipedia's another matter entirely.If only she were more traditional, she'd simply be Mrs Jarvik.
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@TDWTF123 said:
There's the one you pick, and the one the host leaves you. The one he 'takes away' was never part of the equation to start with.
Yes it was, because he doesn't get to take it away until after you've made your initial one-in-three choice. So you start with a 1 in 3 chance of picking the car on your first try and the door he takes away eliminates an option known to be wrong, leaving you with a 2 in 3 chance that the as-yet unused door is the one with the car behind it. But of course you knew that.
Perhaps you'd have better luck trying to explain why an aeroplane can't take off from a treadmill.
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@flabdablet said:
@TDWTF123 said:
There's the one you pick, and the one the host leaves you. The one he 'takes away' was never part of the equation to start with.
Yes it was, because he doesn't get to take it away until after you've made your initial one-in-three choice. So you start with a 1 in 3 chance of picking the car on your first try and the door he takes away eliminates an option known to be wrong, leaving you with a 2 in 3 chance that the as-yet unused door is the one with the car behind it. But of course you knew that.
Perhaps you'd have better luck trying to explain why an aeroplane can't take off from a treadmill.
Ok, let's look at this a different way.
We have door 1, door 2, and door 3.
You pick door A, which could be 1, 2, or 3.
Now we remove door B, which is a door other than the one you picked.
You now get to choose between door A and door C (the remaining door).
Did the removal of door B change anything about door A or door C?
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@boomzilla said:
I really didn't expect a novel of discussion about Monty Hall. It's very refreshing to learn a new way to get the problem wrong.
I once failed to explain the problem properly to someone intelligent because I had completely forgotten that a dud door gets opened after you pick one, which is kind of a crucial aspect.
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@flabdablet said:
Perhaps you'd have better luck trying to explain why an aeroplane can't take off from a treadmill.
In order to take off, an the forces that hold an airplane down (gravity) must be at least equalized by the forces that life it up (air lift). These forces must, thus, be at least 1:1, or you can even say "1".
A treadmill, like all physical surfaces, has a friction co-efficient, which reduces the overall effectiveness of the system. Even the best treadmill in the world will introduce some friction drag.
That co-efficient drag multiplies into the "force ratio" described above, bringing it down from 1.
As such, even though a plane SHOULD be able to take off from a treadmill, but because of the friction drag, it can never achieve a "takeoff force" of 1. Instead, the best it can ever hope to achieve is a vanishingly small co-efficient, and end up instead with an insufficient takeoff force of "0.999...", which is less than 1.
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@Lorne Kates said:
As such, even though a plane SHOULD be able to take off from a treadmill, but because of the friction drag, it can never achieve a "takeoff force" of 1.
Thrust of a plane is not generated by the wheels.
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Airplane wheels aren't powered, and the problem assumes the perfect friction-free world only found in physics trivia questions.
Both that question and the Monty Hall question are prone to having all kinds of debates over their answers because it's easy to phrase them wrong when asking to friends (or for people to misread part of the question.) They're also both dumb as hell. If you're going to teach statistics, teach statistics. If you're going to teach physics, teach physics. Don't ask dumb pointless confusing questions.
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@flabdablet said:
Yes it was, because he doesn't get to take it away until after you've made your initial one-in-three choice.
And that makes a difference because...?
It's much easier to understand with the million-door example. Out of the million doors, whichever one you pick doesn't change the opening of 999,998 of the doors. Their opening is not conditional on your answer. It doesn't matter, therefore, whether you pick before or after they're opened. The only two which matter are the one you pick, and the one the host picks.
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@flabdablet said:
why an aeroplane can't take off from a treadmill.
No-one ever takes into account the strong ground-level wind the treadmill would generate :(
But, Randall does of course agree with me about these things being verbal tricks rather than puzzles:
@xkcd guy said:The problem is basically asking “what happens if you take a plane that can’t move and move it?” It might intrigue literary critics, but it’s a poor physics question.
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@TDWTF123 said:
And that makes a difference because...?
Remove door before pick: 50% chance of picking the prize at first, leaving a 50% chance of getting the prize if you swap.
Remove door after pick: 33.3% chance of picking the prize at first, leaving a 66.6% chance of getting the prize if you swap.
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@TDWTF123 said:
But, Randall does of course agree with me about these things being verbal tricks rather than puzzles
In fact, these are very useful problems (well, Monty Hall is, maybe not the treadmill stuff). Real life problems don't present you with a nice equation to solve. It doesn't even pare down the problem to the minimal word problem, like in a math textbook. The first choice actually does affect the choice of the second door (or even the 999,998 doors). It's a good exercise in understanding conditional probabilities (i.e., all probability) and what affects it and what doesn't. I don't want to live in a world where interesting things like that are dismissed as "verbal tricks." It sounds incredibly dull and ignorant.
Maybe if you climbed down from your ivory tower and modeled it in Excel you'd be able to see this.
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@boomzilla said:
@TDWTF123 said:
Unfortunately, the"Monty Hall Problem" *IS* nothing more than a verbal trick. And people who get caught up in the trickery think 2/3 is the correct answer. The Monty Hall Problem is interesting only in that is shows how people can be fooled into allowing irrelevant details to affect their decisions. And it shows that people like to argue even when they have to resort to verbal trickery to support their argument.But, Randall does of course agree with me about these things being verbal tricks rather than puzzles
In fact, these are very useful problems (well, Monty Hall is, maybe not the treadmill stuff). Real life problems don't present you with a nice equation to solve. It doesn't even pare down the problem to the minimal word problem, like in a math textbook. The first choice actually does affect the choice of the second door (or even the 999,998 doors). It's a good exercise in understanding conditional probabilities (i.e., all probability) and what affects it and what doesn't. I don't want to live in a world where interesting things like that are dismissed as "verbal tricks." It sounds incredibly dull and ignorant.
In the beginning there are 3 doors and you have a 1 in 3 chance of picking the door with the prize. After you make your pick Monty Hall opens a door and reveals a goat. This is where people get hung up. Unless you really like goats, you are never going to choose that door, so in reality it has been removed from the equation. Regardless of whether you stay with your original choice or switch to the other door, you still only have 2 doors to choose from. So now you can cut through the verbal trickery and re-state the question:
You have 2 doors. One of them has a new car and one of them has a goat. What are your chances of picking the door with the car.
When worded that way it is impossible to come up with any answer other than you have a 50% chance of picking the new car and it is impossible to get the 2/3 answer without engaging in verbal trickery.
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Given n+2 doors, pick one door at random, and then pick another door at random. The doors may be the same but do not have to be.
For n > 0, there is a larger-than-.5 probability of the two picked doors being different.
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@El_Heffe said:
Regardless of whether you stay with your original choice or switch to the other door, you still only have 2 doors to choose from.
You're not choosing from 2 doors.It's rephrasing time, boys and girls!
There are 500 boxes. One has a diamond, the rest is empty. You select one box. Pick it up, run away with it. It's in your hands. Only 1/500 chance of winning, so your box probably doesn't have the diamond. Best assume it's empty. Suddenly all other boxes are destroyed. Poof. All 499 of them. Ashes.
You are now claiming that there since there's 1 box, there is now a 100% chance it has the diamond. But alas, the probability that it has the diamond hasn't changed regardless of haw many boxes are destroyed, whether I open the 498 other boxes, or 200, or destroy all of them, or some, or one, or cancel the whole event and confiscate the diamond, wherever it was.
In fact, if I destroy ALL boxes, the odds if you getting the diamond are obviously 0, but that odds that you had the diamond was still 1/500.
If you want real results, write a program to do the Mony Hall problem for you. You'll find 2/3 winning chance.
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@dhromed said:
Remove door after pick: 33.3% chance of picking the prize at first
You're begging the question. The chance is either 33 and a third percent, or half, depending on how you've interpreted the scenario. In the latter case, the choices you're offering would be 'fifty percent or half?'Anyway, since we've already demonstrated that it doesn't matter when the door is opened, your 'before pick' and 'after pick' are entirely spurious: how would you tell the difference?
What you have demonstrated, though, very nicely, is that the two situations are not the same and cannot legitimately be directly compared.
QED, it's a trick, not a puzzle.
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@TDWTF123 said:
depending on how you've interpreted the scenario.
It's not open for interpretation.
@TDWTF123 said:
Anyway, since we've already demonstrated that it doesn't matter when the door is opened
You haven't.
@TDWTF123 said:
your 'before pick' and 'after pick' are entirely spurious: how would you tell the difference?
By running the problem a statistically significant amount of times and noticing that it comes out to 2/3.
@TDWTF123 said:
the two situations are not the same and cannot legitimately be directly compared.
By the two situations, you mean "pick before opening" and "pick after opening"? Yes, they're different problems with different probabilities. You keep saying, I think, that both end up at the same 1/2 probability, which is wrong.
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@boomzilla said:
Real life problems don't present you with a nice equation to solve. It doesn't even pare down the problem to the minimal word problem, like in a math textbook. The first choice actually does affect the choice of the second door (or even the 999,998 doors). It's a good exercise in understanding conditional probabilities (i.e., all probability) and what affects it and what doesn't. I don't want to live in a world where interesting things like that are dismissed as "verbal tricks." It sounds incredibly dull and ignorant.
Of course conditional probability is interesting. Hell, tricks and riddles are interesting too. But the MHP is interesting as a trick, not as an exercise in conditional probability because it doesn't involve conditional probability.
The whole point of the MHP is that it is actually every bit as straightforward as it first appears, but it's worded such that there appears to be a 'clever' answer which is non-obvious - only there isn't, because it's a trap.@boomzilla said:
Maybe if you climbed down from your ivory tower and modeled it in Excel you'd be able to see this.
TDEMSYR. But, for what it's worth, it's utterly trivial to model it in Excel or anything else, and any long-enough Monte Carlo run will demonstrate empirically that the answer is not two-thirds.
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@dhromed said:
By the two situations, you mean "pick before opening" and "pick after opening"? Yes, they're different problems with different probabilities. You keep saying, I think, that both end up at the same 1/2 probability, which is wrong.
No, I keep saying that the MHP does not actually involve the scenario that would give two thirds as the answer. We only have the entirely trivial case. The trick is in appearing to make it seem like the conditional probability scenario ever existed.It's like you're arguing about conditional probability in a game of Three Card Monte. Sure, you can calculate odds as if it were random if you like - but to do so is to ignore that it's not a game of chance, but a scam involving sleight of hand.
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@TDWTF123 said:
it doesn't involve conditional probability.
You're saying the odds of picking the prize from three doors is 50%?
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@TDWTF123 said:
I keep saying that the MHP does not actually involve the scenario that would give two thirds as the answer.
Ok, then what's the MHP? Clearly we're working off different descriptions of the problem.
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@dhromed said:
What I've been trying to explain all along is that the point of the MHP is to sound like a problem in conditional probability - or rather, to sound like that if you overthink it - even though it isn't.@TDWTF123 said:
I keep saying that the MHP does not actually involve the scenario that would give two thirds as the answer.
Ok, then what's the MHP? Clearly we're working off different descriptions of the problem.
As I said, it's not that the calculations of conditional probability are wrong given the interpretations people are led towards, but that those interpretations aren't actually what is described.
As I think you said, you've discovered for yourself how sensitive the problem is to the exact wording - yet another sign of a schoolyard trick rather than a genuine puzzle. Consider, for example, the problem of the Seven Bridges of Konigsberg. Is that similarly sensitive to the description of the problem? Are other genuine mathematical puzzles?
The MHP takes a very simple question, and obscures it enough that when you ask anyone intelligent, most people's reaction is to assume it must be some kind of trick question. They then go looking for the 'trick' that isn't there, and the trap is that there's something that looks a lot like one.
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@TDWTF123 said:
As I think you said, you've discovered for yourself how sensitive the problem is to the exact wording - yet another sign of a schoolyard trick rather than a genuine puzzle.
That's normal for probability problems. For any problem, really.
At what point in its description is the MHP ambiguous, according to you?
@TDWTF123 said:
Seven Bridges of Konigsberg. Is that similarly sensitive to the description of the problem?
Of course it is. If you change the condition "cross a bridge only once" to "cross a bridge exactly twice" or "cross a bridge any number of times", the solution changes accordingly.
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@dhromed said:
Of course it is. If you change the condition "cross a bridge only once" to "cross a bridge exactly twice" or "cross a bridge any number of times", the solution changes accordingly.
Of course it's sensitive to large changes, you dolt. I asked whether it's similarly sensitive to slight changes in the wording. It isn't. As an illustration of the difference, the Wikipedia page for the MHP puts the problem as a quotation, with an exact form to follow, while the page for the seven-bridge problem simply says what the problem is in ordinary words.
Let me give you another example. There's a schoolyard joke where someone asks 'If I have six oranges in one hand, and five oranges in the other, what do I have?' and the answer is supposed to be 'big hands'. That's very sensitive to changing the wording. If it was a genuine problem rather than a joke, though, it wouldn't be as long as the essential elements of adding five and six were retained. The hands, the oranges, and the (apparent) use of 'and' for addition are only necessary for the joke.
@dhromed said:At what point in its description is the MHP ambiguous, according to you?
Not ambiguous, misleading. And, obviously, in the part where it's implied that there is a third door relevant to the problem.
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@TDWTF123 said:
@dhromed said:
Remove door after pick: 33.3% chance of picking the prize at first
You're begging the question. The chance is either 33 and a third percent, or half, depending on how you've interpreted the scenario. In the latter case, the choices you're offering would be 'fifty percent or half?'Now you've misused "begging the question" in a novel way.
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@TDWTF123 said:
But the MHP is interesting as a trick, not as an exercise in conditional probability because it doesn't involve conditional probability.
Correct, except that it does. You picked a door. A loser door was removed. When determining the best strategy for switching or staying with your choice, you have to consider the probability of whether you picked the right door already. Going in, you had a 1/3 chance of getting it right, which means that you had a 2/3 chance of getting it wrong. If you picked wrong initially (condition A), then switching wins every time. If you picked right initially (condition B), switching loses every time. But you were more likely to have gotten it wrong (2/3 > 1/3), so you should switch, because 2/3 of the time you win when you switch.
@TDWTF123 said:But, for what it's worth, it's utterly trivial to model it in Excel or anything else, and any long-enough Monte Carlo run will demonstrate empirically that the answer is not two-thirds.
It's always amusing listening to people misunderstand the problem, but if you really want to confuse people, you can go with the Tuesday Birthday problem. That one is hard to keep straight.
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@TDWTF123 said:
@dhromed said:
At what point in its description is the MHP ambiguous, according to you?
Not ambiguous, misleading. And, obviously, in the part where it's implied that there is a third door relevant to the problem.What makes any of the doors irrelevant?
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@TDWTF123 said:
And, obviously, in the part where it's implied that there is a third door relevant to the problem.
The third door is part of the group from which you pick a door. There's no implication, the door is right there in front of you.
@TDWTF123 said:
There's a schoolyard joke where someone asks 'If I have six oranges in one hand, and five oranges in the other, what do I have?' and the answer is supposed to be 'big hands'. That's very sensitive to changing the wording.
That's because "having" has ambiguous meaning. It's not a proper problem statement.
PS
@inb4 TDWTF123 said:
I wrote some hideous code for the problem. The answer, unsurprisingly, is 2/3 win chance when I swap.then obviously your code is wrong
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@boomzilla said:
@TDWTF123 said:
Can I just check that you are in fact a native English speaker before I start taking the piss out of you for that?@dhromed said:
Remove door after pick: 33.3% chance of picking the prize at first
You're begging the question. The chance is either 33 and a third percent, or half, depending on how you've interpreted the scenario. In the latter case, the choices you're offering would be 'fifty percent or half?'Now you've misused "begging the question" in a novel way.
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@boomzilla said:
What makes any of the doors irrelevant?
Why are your questions so much dumber and more boring than Dhromed's?
What makes a door relevant? Hint: it's not that someone says so.
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@TDWTF123 said:
What makes a door relevant? Hint: it's not that someone says so.
You're going to ignore bits of the problem statement without explaining why? The door is right there. You can't miss it.
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@dhromed said:
The third door is part of the group from which you pick a door. There's no implication, the door is right there in front of you.
What about the fire exit? The door to the gents? My front door?The only doors that are relevant are those which, if opened, would affect the outcome of the problem.
@dhromed said:That's because "having" has ambiguous meaning. It's not a proper problem statement.
Yes, you understood the example.
@dhromed said:I wrote some hideous code for the problem. The answer, unsurprisingly, is 2/3 win chance when I swap.
You're still missing the point. You don't need to Monte Carlo it, because the answers to the two problems are both obvious; if there are two doors, a half, and if three, two-thirds. The only question is whether the MHP is the problem you think it is, and that can't be solved using probability.
So, to reiterate, the MHP is not a maths problem. It's a joke/argument-starter/conversational hand-grenade.
QED, I think.
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@dhromed said:
Please refer back to my earlier example of the missing dollar problem. That does indeed involve a direct lie. You cannot trust the interlocutor's summing of the cash in the first, or the number of relevant doors in the second.@TDWTF123 said:
What makes a door relevant? Hint: it's not that someone says so.
You're going to ignore bits of the problem statement without explaining why? The door is right there. You can't miss it.
Begging the question - Wikipedia
