The Monty Hall Paradox

Thus endeth the lesson.

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Chai wrote:

spendius wrote:
If you end up with two doors and you don't know what's behind each it's 50-50. End of story. Emotions don't come into it. That's voodoo.

Your explanations mean you get the car every time. Not just mostly.

Where has anyone mentioned emotions?

You don't "end up" with 2 doors, you have 3 doors.

And the Lord spake, saying, "First shalt thou take out the Holy Pin. Then shalt thou count to three, no more, no less. Three shall be the number thou shalt count, and the number of the counting shall be three. Four shalt thou not count, neither count thou two, excepting that thou then proceed to three. Five is right out. Once the number three, being the third number, be reached, then lobbest thou thy Holy Hand Grenade of Antioch towards thy foe, who, being naughty in my sight, shall snuff it.

Okay, you have a bean under three cups and ask somebody to choose the cup with the bean under it.

Once the choice is made you remove one of the cups--obviously one that did not have a bean under it--and ask the person to choose again. You're telling me that the person is still stuck with odds existing when the choice is between three cups?

You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

I suggest we all do an experiment at home using the bean and three cups method where no manipulation of the results can be possible. I am guessing that in say 10 or 20 guesses, the results will be something different than those indicated in the poll so far.

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Foxfyre wrote:

You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

That's exactly right. If you have a 1/3 chance of picking the right one that means you have a 2/3 chance of picking the wrong one. That doesn't change when one of the cups is removed because you didn't choose from the two remaining cups, you chose from three.

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Hey Foxy- mind how you quote.

A casual reader could think I had written that tosh Chai wrote.

You are playing Russian Roulette with a three chambered revolver with two blanks and one with lead which all look the same.

You choose a chamber. Then some ads. Then Monte chooses a chamber, more ads, then Monte puts the gun to his head, more ads, a funeral service say or an office cleaning firm, pulls trigger, cloud of smoke, ads. Monte bounces back. Asks contestant his occupation and marital status and whether he prefers burial to cremation, more ads. Then he asks which chamber.

Always switch then like into doing something else a few blocks away.

It's all about car mystique. Raising their status psychologically above that of the other tools you have about your premises. Subliminally if you have no flash car you're a goat.

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The problem with the bean/cups thing is that the bean is shifted by the quickness of the hand deceives the eye trick.

That can only be done with cars and goats when the ads are running or where the tape is spliced in the editorial suite in which a bunch of fiends are ROTFLTAO.

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We should remember that Foxy started this lot going and a suspicion has dawned in my mind that she was deviously seeking to get AIDs-ers to expose how they not only blind everybody else with their science but themselves as well and that IDers are not that dumb.

She thus provides incontovertible evidence that AIDs-er shouldn't be allowed within a mile of a grade science classroom.

It's okay in universities because universities are specifically designed to screw heads up. Preening is a failsafe guide to a screwed up head.

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Foxfyre wrote:

Chai wrote:
spendius wrote:
If you end up with two doors and you don't know what's behind each it's 50-50. End of story. Emotions don't come into it. That's voodoo.

Your explanations mean you get the car every time. Not just mostly.

Where has anyone mentioned emotions?

You don't "end up" with 2 doors, you have 3 doors.

And the Lord spake, saying, "First shalt thou take out the Holy Pin. Then shalt thou count to three, no more, no less. Three shall be the number thou shalt count, and the number of the counting shall be three. Four shalt thou not count, neither count thou two, excepting that thou then proceed to three. Five is right out. Once the number three, being the third number, be reached, then lobbest thou thy Holy Hand Grenade of Antioch towards thy foe, who, being naughty in my sight, shall snuff it.

Okay, you have a bean under three cups and ask somebody to choose the cup with the bean under it.

Once the choice is made you remove one of the cups--obviously one that did not have a bean under it--and ask the person to choose again. You're telling me that the person is still stuck with odds existing when the choice is between three cups?

You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

I suggest we all do an experiment at home using the bean and three cups method where no manipulation of the results can be possible. I am guessing that in say 10 or 20 guesses, the results will be something different than those indicated in the poll so far.

When you "remove" the cup, it is still in existance.

Again, the only way that cup/door/whatever object would not be counted, is if it never existed.

The cup exists....it must be counted.

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spendius wrote:

We should remember that Foxy started this lot going and a suspicion has dawned in my mind that she was deviously seeking to get AIDs-ers to expose how they not only blind everybody else with their science but themselves as well and that IDers are not that dumb.

Well Spendi.. if you aren't that dumb than you must be that drunk because the math works out to 1 in 3 for choosing correctly in the first round.

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FreeDuck wrote:

Foxfyre wrote:
You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

That's exactly right. If you have a 1/3 chance of picking the right one that means you have a 2/3 chance of picking the wrong one. That doesn't change when one of the cups is removed because you didn't choose from the two remaining cups, you chose from three.

The principle I understand. It even makes sense. But rationally, to me a choice between two things is a choice between two things no matter what the original number of things might have been.

In "Deal or No Deal" for instance, there are 26 cases to choose from. The odds that you picked the million dollar case at the beginning of the game are very much against you. But as case after case is eliminated, the odds that you picked the right case initially steadily improve. When you get down to the last two they think you have a 50-50 chance of guessing right - keep your case or take the one remaining case held by a drop dead gorgeous gal.

I don't see how that equation changes when the initial starting number is three. I do see the rationale as most here (and the article) explains it.

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Foxfyre wrote:

And yes, yes, and yes to everybody who suggested that I read the explanations and look at the pictures, etc. etc. etc. I have done all that and I am still left with the original problem.

1. In the beginning I have three doors to choose from with a 1 in 3 chance to choose the right door.

Yes

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2. It is an absolute certainty that at least one door that I didn't choose will be a goat. If I chose a goat initially, he will open the door with the other goat. If I chose the car initially he will choose one of the two goats. The door with the goat that he opens is then removed from the equation.

Yes, it is 100% certain that there will be one goat. It is only 33% that there will be 2 goats. Monte doesn't open the doors at random however. He peeks until he finds a goat. He has a 2 in 3 chance of the first door he peeks at having a goat. But it is a 100% certain that the second door he looks at will have a goat if the first one doesn't. You are leaving his requirement out of your equation and you can't if you want to do the math correctly.

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3. Now I have a choice between two doors. It really doesn't matter whether I chose right or wrong initially. I now am essentially choosing between two doors.

Yes, but you have to multiply the doors by the original probability. If Monte revealed the first door he peeked at then you have a 50/50 chance with what is left. If Monte revealed the second door he looked at then you have 100% chance that you are wrong.

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4. I now have a 1 in 2 chance of choosing the right door.

No, you don't since the odds of Monte peeking at the car on the first door are only 1 in 3. Rap did the simple math earlier which clearly shows it. 2 out of 3 times you have a 50/50 chance but 1 out of 3 times you have ZERO chance since the car was under the FIRST door Monte peeked at.

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I'm sorry folks, but that is a 50-50 chance no matter how you slice it.

Slice away but you are cutting off your fingers with the way you are slicing.

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Foxfyre wrote:

FreeDuck wrote:
Foxfyre wrote:
You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

That's exactly right. If you have a 1/3 chance of picking the right one that means you have a 2/3 chance of picking the wrong one. That doesn't change when one of the cups is removed because you didn't choose from the two remaining cups, you chose from three.

The principle I understand. It even makes sense. But rationally, to me a choice between two things is a choice between two things no matter what the original number of things might have been.

In "Deal or No Deal" for instance, there are 26 cases to choose from. The odds that you picked the million dollar case at the beginning of the game are very much against you. But as case after case is eliminated, the odds that you picked the right case initially steadily improve. When you get down to the last two they think you have a 50-50 chance of guessing right - keep your case or take the one remaining case held by a drop dead gorgeous gal.

I don't see how that equation changes when the initial starting number is three. I do see the rationale as most here (and the article) explains it.

In Deal or No Deal the cases are essentially chosen at random to reveal them. They do not reveal the ones that do NOT have the $1,000,000 until you only have 2 cases left.

It isn't the starting number that changes the odds it is how the non winning door is revealed that keeps the odds from being 1/2 at the end.

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parados wrote:

Foxfyre wrote:
FreeDuck wrote:
Foxfyre wrote:
You are going to tell me that 2 out of 3 times the person will have chosen the wrong cup initially and has better odds of finding the bean by choosing the other remaining cup, the one that he did not choose initially?

That's exactly right. If you have a 1/3 chance of picking the right one that means you have a 2/3 chance of picking the wrong one. That doesn't change when one of the cups is removed because you didn't choose from the two remaining cups, you chose from three.

The principle I understand. It even makes sense. But rationally, to me a choice between two things is a choice between two things no matter what the original number of things might have been.

In "Deal or No Deal" for instance, there are 26 cases to choose from. The odds that you picked the million dollar case at the beginning of the game are very much against you. But as case after case is eliminated, the odds that you picked the right case initially steadily improve. When you get down to the last two they think you have a 50-50 chance of guessing right - keep your case or take the one remaining case held by a drop dead gorgeous gal.

I don't see how that equation changes when the initial starting number is three. I do see the rationale as most here (and the article) explains it.

In Deal or No Deal the cases are essentially chosen at random to reveal them. They do not reveal the ones that do NOT have the $1,000,000 until you only have 2 cases left.

It isn't the starting number that changes the odds it is how the non winning door is revealed that keeps the odds from being 1/2 at the end.

Sorry, a guess is a guess is a guess.

In Deal or No Deal it doesn't matter how the non correct cases are chosen. As each one is eliminated, your odds improve on having guessed the correct case.

In the Monty Hale scenario, you can look at it that you had a 2 in 3 chance of an incorrect choice when there were three and that does not change when one of the choices is eliminated.

Or you can look at it that when it is demonstrated that you did not make one wrong choice, the odds of your having made a correct choice improve.

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parados wrote-

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Well Spendi.. if you aren't that dumb than you must be that drunk because the math works out to 1 in 3 for choosing correctly in the first round.

Obviously. When did I say it wasn't. It's 2 to 1. I back horses and have been a bookie and a bridge player.

There is no choice anyway when the three doors are closed. The choice is only made when there are two doors. That first choice is irrelevant except for spinning the game out.

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Except for when Monte peeks at the doors the odds go like this..

66.66% of the odds being 50/50 if he finds a goat behind the first door
and 33% chance of them being ZERO when he finds the car behind the first door.
.66 x .50 = .33

.33 + 0 = .33 as your final odds if you stay with your first choice.

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Foxfyre wrote:

The principle I understand. It even makes sense. But rationally, to me a choice between two things is a choice between two things no matter what the original number of things might have been.

When you made your choice, it was a choice between 3 things, not two.

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In "Deal or No Deal" for instance, there are 26 cases to choose from. The odds that you picked the million dollar case at the beginning of the game are very much against you. But as case after case is eliminated, the odds that you picked the right case initially steadily improve. When you get down to the last two they think you have a 50-50 chance of guessing right - keep your case or take the one remaining case held by a drop dead gorgeous gal.

I've never seen this show so I can't comment. But facts is facts and odds is odds, so to speak.

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I don't see how that equation changes when the initial starting number is three. I do see the rationale as most here (and the article) explains it.

I'm confused as to how you can understand the explanation and say it makes sense but still don't see that it's not 50/50. Or maybe I'm misunderstanding what the confusion is?

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Let me try this from a different angle.

There are two events happening here: the first choice and the second choice. If these events were completely independent, then the odds of choosing the correct door from two would be 50/50. But these are not independent events. The second choice depends on having made the first because the door which is opened depends on that choice. So you can't start over with new odds.

Does that help or does it make things worse?

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FreeDuck wrote-

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. But these are not independent events

They are independent events if the contestant has the choice of switching.

As I said earlier- at the point where the contestant is asked to stay or switch the first door might as well be on the moon.

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Does that help or does it make things worse?

It helped me to reinforce my policy of not going into any dark woods with Americans who are not called Foxy.

Which is French government policy as well.

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All you have to do to resolve the dilemma in your mind is to check the poll results.

Also, the difference between "Let's Make a Deal" and "Deal or No Deal" is that Monty peeks in "Let's Make a Deal."

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Ok, how about from this angle.

Say you have a non-balanced coin -- one side comes up 1/3 more often than the other. When you flip it, you will get one of two outcomes, but the odds are not 50/50. That's basically what's happening here -- you're flipping between your first choice (1/3) and not your first choice (2/3).

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spendius wrote:

FreeDuck wrote-

Quote:
. But these are not independent events

They are independent events if the contestant has the choice of switching.

The constestant has the choice of switching a case that has a 33% chance of a car for one that has a 66% chance of a car. Choose wisely grasshopper.

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As I said earlier- at the point where the contestant is asked to stay or switch the first door might as well be on the moon.

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Does that help or does it make things worse?

It helped me to reinforce my policy of not going into any dark woods with Americans who are not called Foxy.

Which is French government policy as well.

Hey, Spendi> Let's play a game. I will deal three cards and you try to pick the Ace of Spades. After you choose I will turn over one card that isn't the Ace of spades but then you have to keep your card. I will even pay you 6-5 odds. How much do you think you will win? Since there are only 2 cards left at the ends your odds would be 50/50, wouldn't they? So keeping your card you should rake in the money at 6-5 odds in your favor. C'mon, let's play. How much money do you have?

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The Monty Hall Paradox

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