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shroomydan
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Re: Quick poll, please vote! [Re: Seuss]
#5833033 - 07/07/06 07:04 PM (17 years, 6 months ago) |
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I don't think the Monty Hall Paradox is valid. After one door is opened revealing a goat, it is no longer valid to view a choice of doors as a choice between three possibilities.
The choice whether to change doors is a choice between two. There is a fifty-fifty chance of a car being behind each door. While the original pick yields a one in three chance, once one goat inhabited door is removed, holding the original door is the logical and mathematic equivalent of choosing it again, this time with a one in two chance of it containing a car. Monty Hall paradox theory treats holding the original door as a single choice among three possibilities, when in actuality there are two choices involved: the original choice among three, and the second choice (hold) between two.
I read the wikipedia link and noticed there is no reference to any scientific journal. I suspect a controlled scientific experiment ( one group switching every time, one group holding every time, and one group switching or holding randomly) would reveal the Monty Hall Paradox as the Monty Hall Myth.
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zappaisgod
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Re: Quick poll, please vote! [Re: shroomydan]
#5833107 - 07/07/06 07:27 PM (17 years, 6 months ago) |
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I gotta think more
Edited by zappaisgod (07/07/06 07:29 PM)
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Asante
Mage


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Re: Quick poll, please vote! [Re: shroomydan]
#5833200 - 07/07/06 07:57 PM (17 years, 6 months ago) |
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You could take a step back and say your chance is one in three throughout the whole procedure: If you pick a door you either got it or you don't, and it doesnt matter if you switch or if Mr Monty pulls open a door: It still remains to be seen if your single pick (switched or not) was correct. Switching is not a second pick because it has not been determined if your first pick was success or failure so if you switch, it is as if your first pick never happened and that makes the switch your first pick.
So Wikipedia and all the clever statistics guys have got it wrong 
Mr Monty can always pull open a door (two goats and one pick after all) so that is without consequence. If you switch without having seen whats behind the door of your initial pick, the switch becomes your first pick because the former has remained without consequence or predictive value. So it in all these cases is a 1 in 3 chance any way you go about it.
-------------------- Omnicyclion.org higher knowledge starts here
Edited by Asante (07/07/06 08:04 PM)
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MckennaManiac420
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Re: Quick poll, please vote! [Re: Seuss]
#5833653 - 07/07/06 10:49 PM (17 years, 6 months ago) |
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Thats interesting Seuss. I didn't grasp it at first but after thinking of monty hall with 9 boxes it made more sense.
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: Asante]
#5834520 - 07/08/06 09:02 AM (17 years, 6 months ago) |
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Quote:
I read the wikipedia link and noticed there is no reference to any scientific journal. I suspect a controlled scientific experiment ( one group switching every time, one group holding every time, and one group switching or holding randomly) would reveal the Monty Hall Paradox as the Monty Hall Myth.
There have been computerized simulations. Which are good enough for me.
Quote:
So Wikipedia and all the clever statistics guys have got it wrong. . . . So it in all these cases is a 1 in 3 chance any way you go about it.
You're disregarding the fact that the host knows where the winning token is. Read the wikipedia section on the 99 door Monty Hall problem.
I found this paper (from the Journal of Experimental Economics) interesting.
More papers can be found here.
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Tao
Village Genius

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Re: Quick poll, please vote! [Re: Seuss]
#5837321 - 07/08/06 11:09 PM (17 years, 6 months ago) |
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Did you recently read The Curious Incident of the Dog in the Night-time?
-------------------- Magash's Grain Tek + Tub-in-Tub Incubator + Magash's PMP + SBP Tek + Dunking = Practically all a newbie grower needs
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zappaisgod
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Re: Quick poll, please vote! [Re: Tao]
#5838138 - 07/09/06 09:40 AM (17 years, 6 months ago) |
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I just went to the computer thing Chuang Tzu linked and played each ten times. I got 50% switching or sticking both. Which is what I would expect. The first choice made in a 1 out of 3 universe is rendered moot when the real choice, to stick or switch, is in a 1 out of 2 universe.
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: zappaisgod]
#5838434 - 07/09/06 11:18 AM (17 years, 6 months ago) |
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Your sample size was much too small to draw any conclusions. I just played it manually 30 times each way and got 50% and 66.7% for sticking and switching respectively. That's also a small sample size but any more would require too much clicking. So I let the computer pick for me 1000 times each way. The end results were 37.8% when sticking and 66.4% for switching—which is more or less exactly as predicted.
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zappaisgod
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Re: Quick poll, please vote! [Re: ChuangTzu]
#5838512 - 07/09/06 11:38 AM (17 years, 6 months ago) |
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Quote:
ChuangTzu said: Your sample size was much too small to draw any conclusions. I just played it manually 30 times each way and got 50% and 66.7% for sticking and switching respectively. That's also a small sample size but any more would require too much clicking. So I let the computer pick for me 1000 times each way. The end results were 37.8% when sticking and 66.4% for switching—which is more or less exactly as predicted.
I know my sample was small, but I just can't follow the logic that gets the 1/3, 2/3 result and frankly distrust the results from the computer simulation. Programming errors and all. I think my statement about the universe in which the true choice occurs is "stick or switch", 50%. We combine your results with my results and we get 50% and 62.5%. Flip a coin 80 times and see what happens. I can actually see a logic that would net a 50%, 66.7% rate. Given that 33.3% is expected from a no Monty game, 50% correct from a Monty + "stick" and a 66.7% for a Monty + "switch", that adds up. Converesly
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: zappaisgod]
#5838610 - 07/09/06 12:13 PM (17 years, 6 months ago) |
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Quote:
frankly distrust the results from the computer simulation. Programming errors and all.
What are the odds that so many different computer programs, written in different languages by different people, would make a random error which not only agreed with each other but also with theory?
Quote:
I can actually see a logic that would net a 50%, 66.7% rate. Given that 33.3% is expected from a no Monty game, 50% correct from a Monty + "stick" and a 66.7% for a Monty + "switch", that adds up.
Actually, it doesn't add up. The sum of the probabilities of all possible choices must equal 100%. If the probability of winning is 66.7% for switching, the probability for winning when sticking must be 100 - 66.7 = 33.3%.
Again, I encourage you to read the wikipedia section on the 100 door Monty Hall problem. Then you'll see why the host's knowledge of what is behind the doors throws off the naive probabilities.
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zappaisgod
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Re: Quick poll, please vote! [Re: ChuangTzu]
#5838773 - 07/09/06 12:59 PM (17 years, 6 months ago) |
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No I did read it. And the probabilities add up because there are 3 100%s being dealt with. First one: No Monty 33% win 67% lose Second: Monty + "stick 50/50 Third: Monty + "switch" 67/33
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: zappaisgod]
#5838871 - 07/09/06 01:36 PM (17 years, 6 months ago) |
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Those probabilities are related though. The first one is irrelevant since we are talking about a Monty Hall situation. The last two are complementary situations—either you switch, or you stay. Only one choice "wins". They aren't separate scenarios. The sum of the probabilities must be 1 because if you switch and win, you would have lost if you had stuck and vice versa. So for each choice which results in a win, there is a corresponding loss for the other choice. When the chance of winning by switching is 66.7%, there must be a 66.7% chance of losing when sticking.
Can you see how in the 100 door problem, switching is advantageous?
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zappaisgod
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Re: Quick poll, please vote! [Re: ChuangTzu]
#5839078 - 07/09/06 02:54 PM (17 years, 6 months ago) |
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You forget, and if you look at the computer link one where you play they end up with 4 results for 2 separate scenarios.
Switch/win vs switch/lose and Stick/win vs stick/lose A total of 200%
Throw in the no Monty (a real occurence)which is a 33 lose 67 win and the switch has to be 67/33 and to maintain the probability across all three.....
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: zappaisgod]
#5839152 - 07/09/06 03:09 PM (17 years, 6 months ago) |
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What are you trying to say?
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Diploid
Cuban


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Re: Quick poll, please vote! [Re: zappaisgod]
#5839780 - 07/09/06 06:08 PM (17 years, 6 months ago) |
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I know my sample was small, but I just can't follow the logic that gets the 1/3, 2/3 result and frankly distrust the results from the computer simulation.
Forget programs and simulations for a second.
There is one explanation of the problem in the wiki that is easy to understand without math and irrefutable...
No matter what choice the player initially makes, at the point when the player is asked if he wants to switch, one of the following three states will be true:
A. The player originally picked the door hiding goat number 1. The game host has shown the other goat.
B. The player originally picked the door hiding goat number 2. The game host has shown the other goat.
C. The player originally picked the door hiding the car. The game host has shown either of the two goats.
Remember that everything before choice-time is irrelevant because, no matter what, at choice-time the game state will ALWAYS be in one of those three states. It doesn't matter how we got here. The important thing to realize is that the game ALWAYS reaches one of these three states at choice-time.
Now, going forward from this point notice that in both case A and B, switching wins, and only in case C does switching lose. That means that switching gives a 2 in 3 chance of wining and staying put gives only a 1 in 3 chance of winning.
-------------------- Republican Values: 1) You can't get married to your spouse who is the same sex as you. 2) You can't have an abortion no matter how much you don't want a child. 3) You can't have a certain plant in your possession or you'll get locked up with a rapist and a murderer. 4) We need a smaller, less-intrusive government.
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ChuangTzu
starvingphysicist



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Re: Quick poll, please vote! [Re: Diploid]
#5839938 - 07/09/06 06:54 PM (17 years, 6 months ago) |
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Another point made in the wiki is that when it's time to choose between stay or switch, you're actually choosing between one door ("stay", 1/3) and the other two doors (switch, 2/3). By the host's action of revealing a goat behind one of the other two doors, he is making it possible for you to switch to the other doors knowing that if one of them has the car he won't reveal it (and thus if there is a goat behind the set of both other doors, you are guaranteed to choose it). This contrasts with your original choice which was completely random.
Edited by ChuangTzu (07/09/06 07:43 PM)
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zappaisgod
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Re: Quick poll, please vote! [Re: Diploid]
#5839939 - 07/09/06 06:54 PM (17 years, 6 months ago) |
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Stop. Here is my issue. In condition three you actually conflate two conditions into one. The two conditions NEED to be separated. 3. You picked the car and Monty reveals goat 1 4. You picked the car and Monty reveals goat 2
This is where I have the problem and why I don't necessarily trust the simulations
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Diploid
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Re: Quick poll, please vote! [Re: zappaisgod]
#5840343 - 07/09/06 09:04 PM (17 years, 6 months ago) |
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3. You picked the car and Monty reveals goat 1 4. You picked the car and Monty reveals goat 2
From a probability point of view, they are the same because both have the same outcome.
What you're saying is something like that the chances of flipping two heads in a row are different than flipping one head, then flipping another head.
-------------------- Republican Values: 1) You can't get married to your spouse who is the same sex as you. 2) You can't have an abortion no matter how much you don't want a child. 3) You can't have a certain plant in your possession or you'll get locked up with a rapist and a murderer. 4) We need a smaller, less-intrusive government.
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ChuangTzu
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Re: Quick poll, please vote! [Re: zappaisgod]
#5840387 - 07/09/06 09:19 PM (17 years, 6 months ago) |
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You could just as well write it like this:
A. The player originally picked the door hiding goat number 1. The game host shows a goat.
B. The player originally picked the door hiding goat number 2. The game host shows a goat.
C. The player originally picked the door hiding the car. The game host shows a goat.
Which goat the host reveals is irrelevant. The situations are determined by the 3 (equiprobable) possible initial choices of the player (goat 1, goat 2, car). All including the extra goat information does is split C into 2 sub-scenarios with a probability of 1/6 each.
I'm interested in your reply to my previous post (which I believe is the crux of the Monty Hall problem).
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Phred
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Re: Quick poll, please vote! [Re: zappaisgod]
#5840476 - 07/09/06 09:42 PM (17 years, 6 months ago) |
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Try thinking of it not in terms of goats and cars, but in the classic Three Card Monte scenario, where the dealer switches around three cards --
Ace of Diamonds (red) Ace of Clubs (black) Ace of Spades (black)
You win if you pick the red ace and lose if you pick either of the two black aces. You make your choice and the dealer then turns over one of the remaining cards and it is a black ace. He then asks you if you want to stick with your first pick or switch.
Even though the two black cards are aces, they are not the SAME ace. One is a club, the other is a spade. Getting back to the Monty Hall scenario, don't think of it as two goats, but as a goat and a pig.
There is nothing wrong with any of the computer simulations, I assure you. You can do the same thing with a friend playing the dealer in a Three Card Monte scenario if you are willing to take the time. The results will be the same. As a matter of fact, I did just that a few decades ago when this seeming paradox was first presented to me.
As has been pointed out, the correct strategy is counter-intuitive. Looking at the problem superficially, almost anyone will say the odds are the same of winning whether you switch or stick. But they odds AREN'T the same because the CHOICE isn't the same.
However, I do believe this has gone on long enough. We are no longer talking politics, but statistics. It's been quite some time since there has been any post in the thread of a political nature, therefore I will lock it and leave it locked until Seuss requests it to be reopened.
Phred
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