NationStates Jolt Archive

Inspired by a spin off of the .999999r = 1.0 thread I wanted to see how many people get this:

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?

(Edited for a less ambigious version - does not change the puzzle but eliminates speculation of dishonest hosts and certain loopholes:))

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door."
You begin by pointing to door number 1. The host shows you that door number 3 has a goat.


Wiki Link (http://en.wikipedia.org/wiki/Monty_Hall_problem)

Thanks to Dinaverg for the Wiki Link.

So - do you switch or stay?

It doesn't make a difference as far as i can tell.

I've heard this problem before, and it's quite interesting.

The old "Let's Make a Deal" problem. Here's what they want you to think.

You're betting against yourself, really. If you pick it right the first time, they tell you one that's wrong, tempt you to switch, and as long as you didn't pick it right the first time, your chances of success by switching are improved.

Let's say you pick Door Number 1.

If, in fact, the car is behind Door Number 1, they will pick a random door to open and ask if you want to switch.
Stay put: Win the car.
Switch: Zonk.

If it's behind Door Number 2, they open Door Number 3 and ask you if you want to switch.
Stay put: Zonk.
Switch: Win the car.

If it's behind Door Number 3, they open Door Number 2 and ask you if you want to switch.
Stay put: Zonk.
Switch: Win the car.

So really, if you go in with a strategy of "I'm going to switch," you win the car two times out of three. All you have to do is let your initial guess be a wrong one. That's all.

edit: my math was faulty, changed my answer.

edit: I was wrong.

Question, if you immediately choose the door with the car would they reveal that? If they do then the answer is obvious, switch because you have a goat, if not then it's an even split.

33% chance to pick the car immediately, 66% chance of a goat. After one door is eliminated showing a goat you've got a 50/50 split. There's no obvious "correct" choice.

Question, if you immediately choose the door with the car would they reveal that? If they do then the answer is obvious, switch because you have a goat, if not then it's an even split.

33% chance to pick the car immediately, 66% chance of a goat. After one door is eliminated showing a goat you've got a 50/50 split. There's no obvious "correct" choice.

*buzzer*

YOu pick a door. It's a 2/3rds chance the car is not in your door. One of the other doors is opened. It is still a 2/3rds chance the car is not in your door.

*buzzer*

YOu pick a door. It's a 2/3rds chance the car is not in your door. One of the other doors is opened. It is still a 2/3rds chance the car is not in your door.

No, there's a 50/50 chance, unless there's a third unopened door somewhere.

If you go in with a strategy of "Switch always" then you have a 2/3 chance of succeeding. If you have no idea what the fuck you're doing and don't understand that the host gave you a huge advantage, your odds are 50/50.

I have another one for you guys.

A judge agrees to a bargain with a man convicted of murder, since there is questionable evidence in the case. He takes the man back to his cell, gives him two boxes, and says that, in one box, there are 50 blue marbles. In the other, there are 50 red marbles. You may arrange the marbles in the box any way you wish, but there are a few ground rules. First off, neither box may be empty. Second off, all 100 marbles must be in the boxes. If there are any missing, you lose automatically.

The judge will come the next day, open a box at random, and pick out a marble without looking at it. If he draws a red marble, the man is to be executed by beheading. If he draws a blue marble, the man is to go free. The convict has a great idea that will increase his odds greatly. It's no guarantee that he will go free, but it's the best way to improve his chances. What does he do?

My A-Level Further Maths teacher did this problem when we did the Probability module :)

You should always switch, IIRC.

If you go in with a strategy of "Switch always" then you have a 2/3 chance of succeeding. If you have no idea what the fuck you're doing and don't understand that the host gave you a huge advantage, your odds are 50/50.

I have another one for you guys.

A judge agrees to a bargain with a man convicted of murder, since there is questionable evidence in the case. He takes the man back to his cell, gives him two boxes, and says that, in one box, there are 50 blue marbles. In the other, there are 50 red marbles. You may arrange the marbles in the box any way you wish, but there are a few ground rules. First off, neither box may be empty. Second off, all 100 marbles must be in the boxes. If there are any missing, you lose automatically.

The judge will come the next day, open a box at random, and pick out a marble without looking at it. If he draws a red marble, the man is to be executed by beheading. If he draws a blue marble, the man is to go free. The convict has a great idea that will increase his odds greatly. It's no guarantee that he will go free, but it's the best way to improve his chances. What does he do?
Paints the red marbles blue!

Paints the red marbles blue!

OK, no paint. Keep in mind his chance won't be 100 percent. And the marbles can't be altered in any way. If the judge discovers any cheating, the convict is executed on the spot.

Question, if you immediately choose the door with the car would they reveal that? If they do then the answer is obvious, switch because you have a goat, if not then it's an even split.

It does not, otherwise there's no risk

33% chance to pick the car immediately, 66% chance of a goat. After one door is eliminated showing a goat you've got a 50/50 split. There's no obvious "correct" choice.

Actually there is, mathematically....but it only works if, after you chose your door, the host intentionally opens a door he KNOWS would be a goat.

OK, here's how it works, let's say there are doors 1, 2, and 3. You pick door 1. He opens door 3 and shows a goat. you DO NOT KNOW what's behind door 1.

Behind the three doors, in no particular order, are goat 1, goat 2, and car.

When you pick your door, there's an intial 1/3 chance you picked car, 1/3 you picked goat 1, and 1/3 you picked goat 2.

so 1 time out of three you pick goat 1, and host reveals goat 2

1 time out of three you pick goat 2, and host reveals goat 1

1 time out of three you pick car, and host reveals goat 1 OR goat 2.

So there is a:

1/3 chance you picked goat 1 and host revealed goat 2
1/3 chance you picked goat 2 and host revealed goat 1
1/3 chance you picked car and host revealed a goat

Based on this initial calculation, you only had a 1 in 3 chance of getting the car on your first pick, and 2 in 3 chances you picked a goat.

because your odds were only 1/3 of picking the car, it's a 2/3 odds it was in one of the other doors, and you know now which one. Odds are twice as good to pick the other door.

It does not, otherwise there's no risk



Actually there is, mathematically....but it only works if, after you chose your door, the host intentionally opens a door he KNOWS would be a goat.

OK, here's how it works, let's say there are doors 1, 2, and 3. You pick door 1. He opens door 3 and shows a goat. you DO NOT KNOW what's behind door 1.

Behind the three doors, in no particular order, are goat 1, goat 2, and car.

When you pick your door, there's an intial 1/3 chance you picked car, 1/3 you picked goat 1, and 1/3 you picked goat 2.

so 1 time out of three you pick goat 1, and host reveals goat 2

1 time out of three you pick goat 2, and host reveals goat 1

1 time out of three you pick car, and host reveals goat 1 OR goat 2.

So there is a:

1/3 chance you picked goat 1 and host revealed goat 2
1/3 chance you picked goat 2 and host revealed goat 1
1/3 chance you picked car and host revealed a goat

Based on this initial calculation, you only had a 1 in 3 chance of getting the car on your first pick, and 2 in 3 chances you picked a goat.

because your odds were only 1/3 of picking the car, it's a 2/3 odds it was in one of the other doors, and you know now which one. Odds are twice as good to pick the other door.

You're ignoring the fact that he just revealed what was behind one door. Now there's only two possibilities and your initial probability of picking the correct door is no longer relevant. You now have a 50/50 shot and there's no correct answer. Stay or do not, same odds.

No, there's a 50/50 chance, unless there's a third unopened door somewhere.

no, that's the error of the problem.

When you pick you only had a 1/3 chance of being right.

Think of it this way, change the solution a bit.

Let's say you are given an option, three doors, behind two are NOTHING (let's assume goat has NO value), behind one is a car. Doors are numbered 1, 2 and 3.

You pick door 1.

Your host then gives you a choice, stay with door 1, or get the contents of doors 2 AND 3.

Would you still stick with door 1?

The open door is a spectre, it doesn't matter. Basically what this choice does is (assuming there's no value to the goat) gives you the chance either to stay with the door you picked, or choose both doors.

As odd as it sounds, it's twice as likely the car is behind the unpicked door. Basically the host is giving you the option of staying with your 1 door, or chosing both of the two remaining.

If you go in with a strategy of "Switch always" then you have a 2/3 chance of succeeding. If you have no idea what the fuck you're doing and don't understand that the host gave you a huge advantage, your odds are 50/50.

I have another one for you guys.

A judge agrees to a bargain with a man convicted of murder, since there is questionable evidence in the case. He takes the man back to his cell, gives him two boxes, and says that, in one box, there are 50 blue marbles. In the other, there are 50 red marbles. You may arrange the marbles in the box any way you wish, but there are a few ground rules. First off, neither box may be empty. Second off, all 100 marbles must be in the boxes. If there are any missing, you lose automatically.

The judge will come the next day, open a box at random, and pick out a marble without looking at it. If he draws a red marble, the man is to be executed by beheading. If he draws a blue marble, the man is to go free. The convict has a great idea that will increase his odds greatly. It's no guarantee that he will go free, but it's the best way to improve his chances. What does he do?

Hmmm....My first thought is one blue marble in one, and the rest in the other, but I'm not sure...

You're ignoring the fact that he just revealed what was behind one door. Now there's only two possibilities and your initial probability of picking the correct door is no longer relevant. You now have a 50/50 shot and there's no correct answer. Stay or do not, same odds.

No, actually, you're wrong. It's absolutely relevant. You're falling into the trap they designed for you to fall into.

Your intial choice is NOT irrelevant. That's the entire point. You had a 1/3 chance of guessing correctly. You ALWAYS had a 1/3 chance of guessing correctly, the open door never ever changes that.

Monte Carlo method. You switch, I believe.

As odd as it sounds, it's twice as likely the car is behind the unpicked door. Basically the host is giving you the option of staying with your 1 door, or chosing both of the two remaining.

That's not odd, it's impossible, unless there are two cars behind one door it cannot be twice as likely that it is behind one door and not the other.

The host improved your odds from 33% to 50% by eliminating an option.

Hmmm....My first thought is one blue marble in one, and the rest in the other, but I'm not sure...

still wouldn't do anything.

If that's the case, the judge still has a 50/50 chance of picking one box or the other.

If he picks the loaded box, the prisoner has a 98% chance to go free.

If he picks the other box, the prisoner has a 2% chance of going free.

And he has a 50/50 chance of picking both. It's mathematically the same as putting all red in one box and all blue in the other.

still wouldn't do anything.

If that's the case, the judge still has a 50/50 chance of picking one box or the other.

If he picks the loaded box, the prisoner has a 98% chance to go free.

If he picks the other box, the prisoner has a 2% chance of going free.

And he has a 50/50 chance of picking both. It's mathematically the same as putting all red in one box and all blue in the other.

No, one box with only a blue marble. If the judge picks that box the inmate has a 100% chance of going free. If he puts the 49 other blue marbles in with the red he's got a 49% chance of going free if the judge picks that box to take a marble from.

You're ignoring the fact that he just revealed what was behind one door. Now there's only two possibilities and your initial probability of picking the correct door is no longer relevant. You now have a 50/50 shot and there's no correct answer. Stay or do not, same odds.

Oy, read the Wiki, it knows something.

The most common objection to the solution is the idea that, for various reasons, the past can be ignored when assessing the probability. Thus, the first door choice and the host's forced response are ignored. Because there are two doors to choose from, many people jump to the conclusion that there must be a fifty-fifty chance of choosing the right one.

Although ignoring the past works fine for some games, like coin flipping, it doesn't work for all games. In this case what should be ignored is the opening of the door. The player's choice is between the originally picked door and the other two — opening one is simply a distraction. There is only one car. The original choice divides the possible locations of the car between the one door the player picks (1/3 chance) and the other two (2/3 chance). It is already known that at least one of the two doors contains a goat. The revealing of the goat therefore gives the player no additional information about his own door. It doesn't change the 2/3 probability that the car is still in the block of two doors.

That's not odd, it's impossible, unless there are two cars behind one door it cannot be twice as likely that it is behind one door and not the other.

The host improved your odds from 33% to 50% by eliminating an option.

no

he

does

not.

Think of it this way.

What are the odds it's behind door 1? 1/3

Odds behind door 2? 1/3

Odds behind door 3? 1/3

What are the odds it's behind either door 2 OR door 3? 2/3.

When you pick door 1, you only have a 1/3 chance of being right. By voiding door 3, then giving you the option to switch, this functions as giving you the option to pick door 2 AND door 3 at the same time. Statistically it's the same. Instead of simply removing the door, what he does for practical purposes is give you the option to chose what remains behind door 2 and 3 (assuming the goat has no real value...or we replace the goat with empty space).

You choice ONLY had a 1/3 chance of being right. Now if the host did NOTHING more, simply gave you a chance to change your option, didn't reveal anything, then yes, no point in changing, still a 1/3 option of being correct.

However the simple fact remains, you only had a 1/3 chance of being correct, you EVER only had a 1/3 chance of being correct. You're better to switch.

Trust me on this. Or hell, get a friend and try it out yourself.

No, one box with only a blue marble. If the judge picks that box the inmate has a 100% chance of going free. If he puts the 49 other blue marbles in with the red he's got a 49% chance of going free if the judge picks that box to take a marble from.

Isn't it like...100/2 + (49/99)/2?

no

he

does

not.

Think of it this way.

What are the odds it's behind door 1? 1/3

Odds behind door 2? 1/3

Odds behind door 3? 1/3

What are the odds it's behind either door 2 OR door 3? 2/3.

When you pick door 1, you only have a 1/3 chance of being right. By voiding door 3, then giving you the option to switch, this functions as giving you the option to pick door 2 AND door 3 at the same time. Statistically it's the same. Instead of simply removing the door, what he does for practical purposes is give you the option to chose what remains behind door 2 and 3 (assuming the goat has no real value...or we replace the goat with empty space).

You choice ONLY had a 1/3 chance of being right. Now if the host did NOTHING more, simply gave you a chance to change your option, didn't reveal anything, then yes, no point in changing, still a 1/3 option of being correct.

However the simple fact remains, you only had a 1/3 chance of being correct, you EVER only had a 1/3 chance of being correct. You're better to switch.

Trust me on this. Or hell, get a friend and try it out yourself.

By that "logic" and I use the word extremely loosely you have a 2/3 chance of being incorrect by switching or by staying.

By that "logic" and I use the word extremely loosely you have a 2/3 chance of being incorrect by switching or by staying.

You have a two thirds chance of being incorrect, then he opens a door, and you still have a two thirds chance of being incorrect. I reiterate.

In this case what should be ignored is the opening of the door. The player's choice is between the originally picked door and the other two — opening one is simply a distraction.

You have a two thirds chance of being incorrect, then he opens a door, and you still have a two thirds chance of being incorrect. I reiterate.

And you have a 2/3 chance of being incorrect by switching.

No, one box with only a blue marble. If the judge picks that box the inmate has a 100% chance of going free. If he puts the 49 other blue marbles in with the red he's got a 49% chance of going free if the judge picks that box to take a marble from.

oh, I see what you mean.

And I think you're right (although it doesn't give him a 49% chance to go free, it gives him a 49/99 chance (50 read marbels, 49 blue, 99 marbels total in the box), which is actually 49.5%.

so there's a 50% chance he'll pick the box with one marble, which will set him free 100% of the time, and a 50% chance he'll pick the other box, which will set him free 49.5% of the time.

so the math is (1/2)(1/1) + (1/2)(49/99) odds that he will go free

Which gives him a 74.75% chance of being free. I think that this is the correct answer.

And you have a 2/3 chance of being incorrect by switching.

No, because the potential wrong answer has been removed by the host.

By that "logic" and I use the word extremely loosely you have a 2/3 chance of being incorrect by switching or by staying.

only if no door was ever picked.

One door is gone.

You can argue all you want, mathematically I AM correct. Look at the damned link with the post.

Or hell, try it with a friend, use playing cards, try it out 100 times, I promise you, it works, the math is correct. You're falling into the trap that this type of system is designed for. You are wrong. Just because you don't understand it is no reason to get snippy. If you attack my "logic" you're just going to look more stupid.

Do the math yourself. Try it out yourself.

Isn't it like...100/2 + (49/99)/2?

The easiest way to express the math as I did, is a 50/50 chance of him picking the box that will give him freedom 100% of the time, and a 50/50 chance of him picking the box that will give him freedom 49 out of 99 times.

[(1/2)*(1/1)] + [(1/2)*(49/99)]

=

1/2 + 49/198

=

99/198 + 49/198

=

148/198

= approximatly 74.75%

only if no door was ever picked.

One door is gone.

You can argue all you want, mathematically I AM correct. Look at the damned link with the post.

Or hell, try it with a friend, use playing cards, try it out 100 times, I promise you, it works, the math is correct. You're falling into the trap that this type of system is designed for. You are wrong.

Is there an infinite improbability field around this magical door that always has a 2/3 chance of being right? 'Cause it ain't possible. If you ignore the door being revealed as empty and reject the blatantly apparent 50/50 odds and insist upon retaining your 33% chance of being correct from the initial pick, then you have to concede that every door has a 66% chance of being incorrect, and no amount of switching changes that. If you're down to two doors, and they have an equal chance, that's the bloody definition of 50/50.

Is there an infinite improbability field around this magical door that always has a 2/3 chance of being right? 'Cause it ain't possible. If you ignore the door being revealed as empty and reject the blatantly apparent 50/50 odds and insist upon retaining your 33% chance of being correct from the initial pick, then you have to concede that every door has a 66% chance of being incorrect, and no amount of switching changes that. If you're down to two doors, and they have an equal chance, that's the bloody definition of 50/50.

You are not picking ONE door you dolt, you're picking TWO. The door that's unopened and unpicked AND the open, worthless door. You're just not getting that.

You know what, you're right. Everyone else here disagreeing with you is wrong. The link provided by the OP is wrong. The math in that link is wrong.

The entire field of statistics is wrong.

YOU are right because....you said so, apparently.

Here's a thought. Stop arguing, shut up, look at the link in the OP, look at the math, do the math yourself and learn what the hell you're talking about. Or just grab a friend, grab a deck of cards, and run the experiment yourself, read the link:

There is a simple way for people to convince themselves that a switching strategy really does win two out of three times on the average, and that is to simulate the game with playing cards, playing Monty's role. The simulator takes three cards from an ordinary deck to represent the three doors; one 'special' card such as the Ace of Spades should be picked to represent the door with the car, and ordinary cards such as the two red twos picked to represent the goat doors.

The simulator then repeats the following procedure several times, to simulate multiple rounds of the game: One card is dealt at random to the 'player', to represent the door the player picks initially. The simulator, as Monty, then looks at the two cards in his hand and discards a red two (of which he will have at least one.) If the card remaining in the simulator's hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the simulator is holding a red two, the round is recorded as one where staying would have won.

You're wrong. And while you're realizing that, realize also that no matter what the host does, whether he opens a door, jumps up and down, barks like a dog, or take a dump on your head does not change the universal truth that you only had a 1/3 chance of guessing correctly the first time.

NOTHING can change that fact.

they have an equal chance

They do not.

The door that was neither selected nor removed had a chance of being removed, unlike the door you selected. It was not.

There are only two possible reasons its removal would not have occurred:

1. Both doors you did not choose are wrong answers. There is a one-third possibility of this being the case.
2. Only one of the doors you did not choose is the wrong answer. Because a wrong answer was removed, this must have been the door that was removed, and not the door that remains, which therefore must be the right answer. There is a two-thirds possibility of this being the case.

I suppose using X-ray vision is out of the question?

no, that's the error of the problem.

When you pick you only had a 1/3 chance of being right.

Think of it this way, change the solution a bit.

Let's say you are given an option, three doors, behind two are NOTHING (let's assume goat has NO value), behind one is a car. Doors are numbered 1, 2 and 3.

You pick door 1.

Your host then gives you a choice, stay with door 1, or get the contents of doors 2 AND 3.


Not true. You pick door 1, host reveals door 3, and asks if you want to switch to door two.

Either you simply argue that there are two doors, one with a car behind, equalling fifty-fifty chance.

Or you argue that the host is offering you a choice of either everything behind Doors 2 and 3, or everything behind doors 1 and 3. Probability is the same, therefore 50-50 chance.


Switching gets you doors 2 and 3, staying gets you and 1 and 3. No statistical difference between either.

Not true. You pick door 1, host reveals door 3, and asks if you want to switch to door two.

Either you simply argue that there are two doors, one with a car behind, equalling fifty-fifty chance.

Or you argue that the host is offering you a choice of either everything behind Doors 2 and 3, or everything behind doors 1 and 3. Probability is the same, therefore 50-50 chance.


Switching gets you doors 2 and 3, staying gets you and 1 and 3. No statistical difference between either.

and what are the odds that door 1 had nothing in it? 1/3, that's the point.

It

is

not

50/50

read the link. Look at the math. Try it yourself. you're wrong.

You are not picking ONE door you dolt, you're picking TWO. The door that's unopened and unpicked AND the open, worthless door. You're just not getting that.

I'm not picking two, the host is picking one and so am I, the difference is you presume he knows which door is correct. If he does then switching would be prudent, if he does not then there is no obvious correct choice.

If his choice is a goat, then there's two options left, my odds of being correct are now 50%. If his option is a car then I am wrong quite apparently, I had a 66% chance of being incorrect and was, hardly shocking.


1 2 3
| | |
Goat Goat Car <---------Host opens a door I didn't pick.
| |
Open Goat Car <---------Second choice, switch or don't.

There's a goat, and there's a car. There's two doors. I get to pick one.

I'm not picking two, the host is picking one and so am I, the difference is you presume he knows which door is correct.

He does. That is one of the premises of the problem.

He does. That is one of the premises of the problem.

That was not stated in the problem and thus is not considered.

I'm not picking two, the host is picking one and so am I, the difference is you presume he knows which door is correct. If he does then switching would be prudent, if he does not then there is no obvious correct choice.

Great, you admit that if he knows, and choses a door knowing to be a goat, then switching would be prudent.

Now read the OP:

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.

The whole point is he DOES know which is which, and INTENTIONALLY opens the door with the goat. That's the premise of the question.

That was not stated in the problem and thus is not considered.

really?

Read the OP again

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.

And, as a matter of statistics, it's actually irrelevant whether he KNOWS or not, provided that the door he picks is opened, and it reveals a goat. His intent actually doesn't change statistics at all, all that matters is a door that you did not pick is opened, and it's not the car.

Now you WOULD be correct if...say...the door he picked is sealed shut, unopened, giving you no option to open it or know what is behind it.

Yes, if you pick door 1, and he says "ok, lock up door 3, don't open it, and you now can chose 1 or 2" then YES, no matter, stick with either one.

But as long as that door is opened, and as long as it's NOT the car, statistcs don't change. Math doesn't change because we do something intentionally or accidentally. Odds remain the same regardless of what we are thinking.

Nooooooooooooooooooooooooooooooo.

Not this question again. Like none else it has the power to eviscerate any smug feelings of relative intelligence I ever harboured... >.<

really?

Read the OP again

Silly me for reading the supposedly "less ambiguous version" and operating off of it.

Not this question again. Like none else it has the power to eviscerate any smug feelings of relative intelligence I ever harboured... >.<

A younger and more foolish version of myself lost a bet on the correct answer to this question.

It was quite humiliating.

Silly me for reading the supposedly "less ambiguous version" and operating off of it.

oh that one?

the one that says: "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat"

I fail to see how that changes anything. It reads the same way, he's picking a door that you did not pick and it will be a goat.

Read it as many times as you want, both "versions" read exactly the same in this regard. Stop trying to dodge, you're wrong, just admit it and move on, the more you try to fight the more arrogant and less intelligent you seem.

Nooooooooooooooooooooooooooooooo.

Not this question again. Like none else it has the power to eviscerate any smug feelings of relative intelligence I ever harboured... >.<

hehe.

Statistics are wierd, you sorta either understand them or you don't...some just have a head for it as all. I have no problems with those that don't understand them, as long as they don't keep trying to shout to the wind that they do, despite evidence to the contrary.

oh that one?

the one that says:

I fail to see how that changes anything. It reads the same way, he's picking a door that you did not pick and it will be a goat.

Read it as many times as you want, both "versions" read exactly the same in this regard. Stop trying to dodge, you're wrong, just admit it and move on, the more you try to fight the more arrogant and less intelligent you seem.

I'll open another door and reveal a goat, sounds like he knows you just won.

You know, it still doesn't make any sense to me, how a two door problem ends with one being correct 2/3 of the time, infact the exact one you didn't pick initially.

I'll open another door and reveal a goat, sounds like he knows you just won.

only if you're functionally illiterate.

That makes no sense.

You pick a door, you do NOT OPEN IT. The host then picks one of the two you DO NOT PICK, and opens it, he picks a door knowing a goat is behind it.

How the hell do you figure you know for sure what's behind the door you picked?

hehe.

Statistics are wierd, you sorta either understand them or you don't...
No shit, sherlock. :p

some just have a head for it as all. I have no problems with those that don't understand them, as long as they don't keep trying to shout to the wind that they do, despite evidence to the contrary.
Oh, I have more of a head for advanced astrophysics than I have for statistics. :p
But, conveniently enough, I also have the decency to shout that fact to the wind. ;)

You know, it still doesn't make any sense to me, how a two door problem ends with one being correct 2/3 of the time, infact the exact one you didn't pick initially.

because the simple fact is, as you keep being unable to understand, it's NOT a two door problem

only if you're functionally illiterate.

That makes no sense.

You pick a door, you do NOT OPEN IT. The host then picks one of the two you DO NOT PICK, and opens it, he picks a door knowing a goat is behind it.

How the hell do you figure you know for sure what's behind the door you picked?

I don't, you're the one telling me it's a goat all the time.

Oh, I have more of a head for advanced astrophysics than I have for statistics. :p
But, conveniently enough, I also have the decency to shout that fact to the wind. ;)

Then can you explain to me why objects tend to form into eliptical orbits? I would imagine inertia and gravitational forces would pull them into circular orbits over time.

Never did get that.

I don't, you're the one telling me it's a goat all the time.

No, I have no idea what the door you picked it.

All I am saying is that the host picked a different door, and it's a goat. A DIFFERENT DOOR.

Your door may have a car, it may have a goat.

however it'll be a goat 2 times for every 1 it's a car.

I don't, you're the one telling me it's a goat all the time.

Has anyone ever thought about waiting until the goat starts bleating and then picking the door accordingly?

That's what I would do. :)

Should I just shut up and go have fun elsewhere and leave this thread to serious discussion? Yes? Aww. :(

You pick a door
¦
¦ -----------------------¦
Right, 1/3 Wrong 2/3
¦ ¦
¦ Door opened ¦
50% chance of which remaining door prize is behind

------------------ -----------------
Switch Stay Switch Stay
You win You lose You win You lose
Odds=1/6 Odds=1/6 Odds=1/3 Odds=1/3

Equal odds of winning or losing.

Has anyone ever thought about waiting until the goat starts bleating and then picking the door accordingly?

That's what I would do. :)

Should I just shut up and go have fun elsewhere and leave this thread to serious discussion? Yes? Aww. :(

There's someone having a serious discussion here? I think it's fun to get Art's blood pressure up over an absurd mathematical exercise that makes no sense.

2 is actually 3!

Then can you explain to me why objects tend to form into eliptical orbits? I would imagine inertia and gravitational forces would pull them into circular orbits over time.

Never did get that.

I did have the slight suspicion that my sentence didn't really mean what I meant it to mean. *curses English syntax* :p

You pick a door
¦
¦ -----------------------¦
Right, 1/3 Wrong 2/3
¦ ¦
¦ Door opened ¦
50% chance of which remaining door prize is behind

Here's your error.

A wrong door that you did not choose was opened, so now, if in fact you selected the wrong door (2/3 chance), you will win by switching and lose by staying.

Let me explain this in a simple way.

Three doors, behind 2 are a goat, behind 1 is a car. Doors numbered 1 2 and 3.

You pick door 1. Now what are the odds that behind this door is a car? 1 in 3.

You agree with that right? NOTHING else being known, the odds that you picked a car is 1 and 3.

The host says "OK, let's see what's behind door #2!"

Now, after he says that but does NOTHING else, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts walking towards the door and places his hand upon the handle. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts slowly opening the door, but it's dark, and it's not open enough for you to see what's behind it. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the door swings fully open but JUST before you can see what is behind it, someone slips a blindfold over your eyes, so you can't see. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now, the blindfold is slipped off, the other door lies open, and you see before you a young, healthy looking goat.

Your premise is that at every single point along the way, up until you saw the goat, the odds that you picked right was 1 in 3.

And yet...somehow, in some way, when you laid eyes upon that goat, some magical chemistry went on in your eyes, so that seeing the goat somehow changed the laws of statistics in the universe.

No, it's not true at all. Seeing the goat, knowing the goat was there, none of that makes any difference. Your choice had a 1 in 3 chance of getting the car. At no point in that sequence of events did it somehow, mystically, change. That's the point.

Then can you explain to me why objects tend to form into eliptical orbits? I would imagine inertia and gravitational forces would pull them into circular orbits over time.

Never did get that.

There's more than one point of gravity affecting them. If it was an isolated system it'd be a circular orbit (presuming no atmosphere on either object and no non-gravitational interaction. In our system alone we have the sun giving the major force which things orbit, then there's Jupiter further out adding it's gravity to in a small part counter the sun. You've got every planet, asteroid, and bit of space dust adding slight tugs each direction. In the end you get a rather egg shaped path. It's fairly simple really.

Try this:

http://www.addictinggames.com/orbit.html

You can use the map editor to see how adding various objects affects the path of your "comet". Finding a stable orbit in a multi-object system is very hard, which is why space is pretty well clear.

2 is actually 3!

Nope. 3 is always 3, despite your instance that seeing the goat somehow, magically, changes it to 2.

There's more than one point of gravity affecting them. If it was an isolated system it'd be a circular orbit (presuming no atmosphere on either object and no non-gravitational interaction. In our system alone we have the sun giving the major force which things orbit, then there's Jupiter further out adding it's gravity to in a small part counter the sun. You've got every planet, asteroid, and bit of space dust adding slight tugs each direction. In the end you get a rather egg shaped path. It's fairly simple really.

so if we had a solar system with a single planet (assuming that there were no other pieces of dust etc to affect it) then it would fall into a circular orbit? Interesting. I thought objects tended to fall into eliptical orbits due to some nature of movement and gravity.

Let me explain this in a simple way.

Three doors, behind 2 are a goat, behind 1 is a car. Doors numbered 1 2 and 3.

You pick door 1. Now what are the odds that behind this door is a car? 1 in 3.

You agree with that right? NOTHING else being known, the odds that you picked a car is 1 and 3.

The host says "OK, let's see what's behind door #2!"

Now, after he says that but does NOTHING else, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts walking towards the door and places his hand upon the handle. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts slowly opening the door, but it's dark, and it's not open enough for you to see what's behind it. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the door swings fully open but JUST before you can see what is behind it, someone slips a blindfold over your eyes, so you can't see. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now, the blindfold is slipped off, the other door lies open, and you see before you a young, healthy looking goat.

Your premise is that at every single point along the way, up until you saw the goat, the odds that you picked right was 1 in 3.

And yet...somehow, in some way, when you laid eyes upon that goat, some magical chemistry went on in your eyes, so that seeing the goat somehow changed the laws of statistics in the universe.

No, it's not true at all. Seeing the goat, knowing the goat was there, none of that makes any difference. Your choice had a 1 in 3 chance of getting the car. At no point in that sequence of events did it somehow, mystically, change. That's the point.
Hey, I totally understood that!

So now explain to me again the tricky part: why again should I switch my picked door now? Thanks. ;)

There's more than one point of gravity affecting them. If it was an isolated system it'd be a circular orbit (presuming no atmosphere on either object and no non-gravitational interaction. In our system alone we have the sun giving the major force which things orbit, then there's Jupiter further out adding it's gravity to in a small part counter the sun. You've got every planet, asteroid, and bit of space dust adding slight tugs each direction. In the end you get a rather egg shaped path. It's fairly simple really.
Hm, shouldn't it look like the outline of a squiggly, runny fried egg then instead of a tidy ellipse?

You pick a door
¦
¦ -----------------------¦
Right, 1/3 Wrong 2/3
¦ ¦
¦ Door opened ¦
50% chance of which remaining door prize is behind

------------------ -----------------
Switch Stay Switch Stay
You win You lose You win You lose
Odds=1/6 Odds=1/6 Odds=1/3 Odds=1/3

Equal odds of winning or losing.

This is not correct.

Lets say you pick Door 1:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.

Switching will make you win 2/3 times.

Think of it this way:

when you make your first choice there is a 2/3 chance you are wrong. If you are wrong, then the remaining box after the host removes one will definately be the winner. This means you have a 2/3 chance of winning if you switch and a 2/3 chance of losing if you stay.

Basicaly what you are being given an option on is: Choose to open only one box and Choose to open two boxes. Obviously opening two boxes will give a better chance of winning.

Hey, I totally understood that!

So now explain to me again the tricky part: why again should I switch my picked door now? Thanks. ;)

Pretty simple. You have three options, A, B, and C. Two are nuls, worthless, one is the "prize"

You pick A, it has a 1 in 3 chance of being right. It is ALWAYS a 1 in 3 chance of being right.

B is revealed to be a nul, and removed from the option.

So if A has a 1 in 3 chance of being right, B is no longer an option, then C, by mathematic necessity, must have a 2 in 3 chance of being right.

So to increase your odds, you should change your choice to C. You won't ALWAYS win, but you'll win twice as often as you lose.

Basicaly what you are being given an option on is: Choose to open only one box and Choose to open two boxes. Obviously opening two boxes will give a better chance of winning.

THANK YOU, I've been saying that the whole time.

Switching gets you doors 2 and 3, staying gets you and 1 and 3. No statistical difference between either.

Door 1 is 1/3, door 2 is 2/3, door 3 is 0/3.

Let me explain this in a simple way.

Three doors, behind 2 are a goat, behind 1 is a car. Doors numbered 1 2 and 3.

You pick door 1. Now what are the odds that behind this door is a car? 1 in 3.

You agree with that right? NOTHING else being known, the odds that you picked a car is 1 and 3.

The host says "OK, let's see what's behind door #2!"

Now, after he says that but does NOTHING else, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts walking towards the door and places his hand upon the handle. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the host starts slowly opening the door, but it's dark, and it's not open enough for you to see what's behind it. At this point, what are the odds that the car is behind door 1?

1 in 3.

Now the door swings fully open but JUST before you can see what is behind it, someone slips a blindfold over your eyes, so you can't see. At this point, what are the odds that the car is behind door 1?

1 in 3.

Then you shout that you want to change to door 2. What are the odds that the car is behind door 2? 1 in 3.

Then continue.


Now, the blindfold is slipped off, the other door lies open, and you see before you a young, healthy looking goat.

Your premise is that at every single point along the way, up until you saw the goat, the odds that you picked right was 1 in 3.

No, it's not true at all. Seeing the goat, knowing the goat was there, none of that makes any difference. Your choice had a 1 in 3 chance of getting the car. At no point in that sequence of events did it somehow, mystically, change. That's the point.

If the odds of the car being behind Door 1 and Door 2 are the same at the beginning, where does that alter at any stage?

You pick a door
¦
¦ -----------------------¦
Right, 1/3 Wrong 2/3
¦ ¦
¦ Door opened ¦
50% chance of which remaining door prize is behind

------------------ -----------------
Switch Stay Switch Stay
You win You lose You win You lose
Odds=1/6 Odds=1/6 Odds=1/3 Odds=1/3

Equal odds of winning or losing.

*buzzer*

http://upload.wikimedia.org/wikipedia/en/thumb/9/9e/Monty_tree.svg/520px-Monty_tree.svg.png

THANK YOU, I've been saying that the whole time.

You open two either way. You either open 1 and 3, or 2 and 3. No difference.

THANK YOU, I've been saying that the whole time.

I did not realise I was unleashing such a beast with this thread....

Seriously - if you draw it out on paper it is really clear.

Or if you consider the ONLY three options:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.

I thought it would be a simple case of poeple going "Oh! that's a bit counter intuitive".

Guys... just try this out with some cards and a friend.

You open two either way. You either open 1 and 3, or 2 and 3. No difference.

Actually, a difference of about one third.

http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Monty_open_door_chances.svg/248px-Monty_open_door_chances.svg.png

Then you shout that you want to change to door 2. What are the odds that the car is behind door 2? 1 in 3.

If you don't know for certain that the door being opened contains a goat, quite right.


If the odds of the car being behind Door 1 and Door 2 are the same at the beginning, where does that alter at any stage?

At the point door 3 is revealed to not contain the car, and removed as an option.

You open two either way. You either open 1 and 3, or 2 and 3. No difference.

These are the ONLY three possibilities if you switch:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.
Therefore you win 2/3 times if you switch.


These are the ONLY three possibilities if you stay:

Case 1: Prize is behind door one. You stay. You win.
Case 2: Prize is behind door two. You stay. You lose.
Case 3: Prize is behind door three. You stay. You lose.
Therefore you lose 2/3 times if you stay.

Pretty simple. You have three options, A, B, and C. Two are nuls, worthless, one is the "prize"

You pick A, it has a 1 in 3 chance of being right. It is ALWAYS a 1 in 3 chance of being right.

B is revealed to be a nul, and removed from the option.

So if A has a 1 in 3 chance of being right, B is no longer an option, then C, by mathematic necessity, must have a 2 in 3 chance of being right.
That right there? Is where you lost me.
I can see what you're saying (which is a huge step forward in and of itself, believe me) but I just... can't believe it, I guess. It goes against common sense. Against my common sense, at least. :p

Oh well. I make a mean Mousse au Chocolat. That count as anything on the General Skills Scale? :p

let's put this another way:

instead of just one door opening, let's try two:

normally:
you pick door one, the man opens door three.


another visualization:
a man who is wrong most of the time (2/3) says that there's a car behind door 1. A man who is right all the time says there is a goat behind door number 3.

It doesn't take a rocket scientist to ignore the first man and pick door number 2.

let's put this another way:

instead of just one door opening, let's try two:

normally:
you pick door one, the man opens door three.


another visualization:
a man who is wrong most of the time (2/3) says that there's a car behind door 1. A man who is right all the time says there is a goat behind door number 3.

It doesn't take a rocket scientist to ignore the first man and pick door number 2.

Not so good, as it implies luck.

Basically imagine this:

You pick one door.
Your opponent gets the other two.

Who will win 2/3 of the time?

That right there? Is where you lost me.
I can see what you're saying (which is a huge step forward in and of itself, believe me) but I just... can't believe it, I guess. It goes against common sense. Against my common sense, at least.

That's the point of the trick, it does go against common sense. Math doesn't always fit in "common sense", that's why the trick works.

Though you don't have to believe it, let's make a bet you and I the next time you're in the east coast USA.

We'll play this way, I'll take 3 playing cards, two of them clubs, one a diamond, I will look at them. I'll then place all 3 face down.

I will then ask you to point to one. Without flipping it over, I will pick up the remaining two and look at them, without showing you. I will then discard one of the clubs (if there are 2, and I pick up 2 out of three I MUST have picked up one). By logic, we will have two cards left, one in my hand, one face down on the table. One is a diamond, one is a club.

then we will flip over the face down card you picked. If it's a diamond, I'll give you 5 bucks. If it's a club, you give me 5 bucks. We do this 100 times.

By "logic", we will break even. However, I'm pretty sure I'd walk away about 170 bucks richer.

That right there? Is where you lost me.
I can see what you're saying (which is a huge step forward in and of itself, believe me) but I just... can't believe it, I guess. It goes against common sense. Against my common sense, at least. :p

Oh well. I make a mean Mousse au Chocolat. That count as anything on the General Skills Scale? :p

I agree. Makes sense mathematically, but not generally. Effectively like saying 0!=1.

Hey, I totally understood that!

So now explain to me again the tricky part: why again should I switch my picked door now? Thanks. ;)


Hm, shouldn't it look like the outline of a squiggly, runny fried egg then instead of a tidy ellipse?

Inertia, it has a wonderful way of smoothing things out, mostly.

That's the point of the trick, it does go against common sense. Math doesn't always fit in "common sense", that's why the trick works.

Hence the use of 'counterintuitive'.
The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive, although the problem does not yield a logical contradiction.

And boy, is it ever countering some intuits on NS.

I agree. Makes sense mathematically, but not generally. Effectively like saying 0!=1.

Teehee. That was the first thing of this kind I figured out.

n! = (n+1)!/(n+1)

Inertia, it has a wonderful way of smoothing things out, mostly. Ah. Okay.

And boy, is it ever countering some intuits on NS. Hey, hey, easy on the insults. :p

In fact, that's a GREAT way of explaining it easily.

I pick 3 cards from a deck, two are clubs, one is a diamond. I have them in my hand, you can not see them. I allow you to pick one card, we put that card face down.

So I have two cards, you have one. I know what all 3 are, you don't know what any of them are. I tell you that you may either look at the cards in my hand, or the card on the table. If you find a diamond, you win.

If you do not, you lose. You can either keep your one card, or give me that card and take both of mine. You would OBVIOUSLY chose to take the two in my hand, not the one in the table right? Nobody would rationally make a difference choice. Two cards double your chances of finding the diamond

Now, three cards, two are clubs, I have two cards. you KNOW I have club, I MUST have a club. I have to. You know, for a FACT, that in my hand is at least 1 club. You would STILL chose my hand even KNOWING that at least one of those cards is a club.

So instead I show you a club that I have, and throw it away.

Now you have one card, I have one card. You assume now the odds have changed. They didn't, they never did. You KNEW that club was in my hand, you always knew. Me showing it to you and throwing it away changes nothing. All I did is show you the card you KNEW I had. What difference does it make if I show it to you or not? Either way you know it's there.

Why would the situation change if I simply show you the card you knew I had anyway?

That's this situation. You have one door, the game show has two. You know of the two doors the game show has at least one is a goat. What difference does it make that they show it to you. You knew it was there. It doesn't change anything.

You knew that I had at least 1 club, you'd still chose my hand, because the odds were better that way. Why would that ever change once I showed you that club. You KNEW that of the two cards I held, at least one was a club. Why would it matter in ANY way that I showed you a club that you knew I had?

Ah. Okay.

Hey, hey, easy on the insults. :p

Hehehe...It doesn't make any sense and it's somehow an insult...I didn't think I was capable of that.

In fact, that's a GREAT way of explaining it easily.

I pick 3 cards from a deck, two are clubs, one is a diamond. I have them in my hand, you can not see them. I allow you to pick one card, we put that card face down.

So we have two cards in my hand, and one card on the table. I tell you that you may either look at the cards in my hand, or the card on the table. If you find a diamond, you win.

If you do not, you lose. You can either look at the one on the table, or the two in my hand. You would OBVIOUSLY chose to take the two in my hand, not the one in the table right? Nobody would rationally make a difference choice. Two cards double your chances of finding the diamond

Now, three cards, two are clubs, I have two cards. you KNOW I have club, I MUST have a club. I have to. You know, for a FACT, that in my hand is at least 1 club. So I show you a club that I have, and throw it away.

Now I have one card on the table one in my hand. You assume now the odds have changed. They didn't, they never did. You KNEW that club was in my hand, you always knew. Me showing it to you and throwing it away changes nothing. All I did is show you the card you KNEW I had. What difference does it make if I show it to you or not? Either way you know it's there.

Why would the situation change if I simply show you the card you knew I had anyway?

That's this situation. You have one door, the game show has two. You know of the two doors the game show has at least one is a goat. What difference does it make that they show it to you. You knew it was there. It doesn't change anything.

You knew that I had at least 1 club, you'd still chose my hand, why would that ever change once I showed you that club. You KNEW that of the two cards I held, at least one was a club. Why would it matter in ANY way that I showed you a club that you knew I had?

No, no, no, no, no.
You're doing it again. I understand everything up to that point.
It's what comes after that escapes me.
(um, assuming you actually were talking to me >.<)

That's the point of the trick, it does go against common sense. Math doesn't always fit in "common sense", that's why the trick works.

Though you don't have to believe it, let's make a bet you and I the next time you're in the east coast USA.
Honey, just because I say "I can't believe it" doesn't mean I don't believe it. It just means I don't like it. :p

We'll play this way, I'll take 3 playing cards, two of them clubs, one a diamond, I will look at them. I'll then place all 3 face down.

I will then ask you to point to one. Without flipping it over, I will pick up the remaining two and look at them, without showing you. I will then discard one of the clubs (if there are 2, and I pick up 2 out of three I MUST have picked up one). By logic, we will have two cards left, one in my hand, one face down on the table. One is a diamond, one is a club.

then we will flip over the face down card you picked. If it's a diamond, I'll give you 5 bucks. If it's a club, you give me 5 bucks. We do this 100 times.

By "logic", we will break even. However, I'm pretty sure I'd walk away about 170 bucks richer.
Yeah, well, again: I believe you, but I don't understand it.
And you have now also managed to confuse me with the "170 bucks". *doesn't even want to know*

Hehehe...It doesn't make any sense and it's somehow an insult...I didn't think I was capable of that.
I loved it, though. :p

I loved it, though. :p

Yay! :D

No, no, no, no, no.
You're doing it again. I understand everything up to that point.


A repost - but does this make it any clearer?

These are the ONLY three possibilities if you switch:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.
Therefore you win 2/3 times if you switch.


These are the ONLY three possibilities if you stay:

Case 1: Prize is behind door one. You stay. You win.
Case 2: Prize is behind door two. You stay. You lose.
Case 3: Prize is behind door three. You stay. You lose.
Therefore you lose 2/3 times if you stay.

Is this a good explination or does it still confuse? (looking for help in writing a better one)

A repost - but does this make it any clearer?

These are the ONLY three possibilities if you switch:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.
Therefore you win 2/3 times if you switch.


These are the ONLY three possibilities if you stay:

Case 1: Prize is behind door one. You stay. You win.
Case 2: Prize is behind door two. You stay. You lose.
Case 3: Prize is behind door three. You stay. You lose.
Therefore you lose 2/3 times if you stay.

Is this a good explination or does it still confuse? (looking for help in writing a better one)

Okay, I think I'm just really, really thick - but isn't what you describe a different situation? Because am I not already down to two doors by the time I choose if I switch or stay?

Okay, I think I'm just really, really thick - but isn't what you describe a different situation? Because am I not already down to two doors by the time I choose if I switch or stay?

I think pictures are better.

Before:
http://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Monty_closed_doors.svg/160px-Monty_closed_doors.svg.png

After:
http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Monty_open_door_chances.svg/180px-Monty_open_door_chances.svg.png

If that doesn't work out, I guess it's just one of those things you've gotta go with and know it's right, even if you don't quite understand.

I've been trying this since a friend made my brain hurt by describing it (and I am pretty decent with statistics...) and it's more or less panned out. Unfortunately I only have the most tenious grasp and won't try and confuse WYTG by jumping in and explaining it. And I don't want her to get made at me...except for what I'm about to do...


Oh well. I make a mean Mousse au Chocolat. That count as anything on the General Skills Scale? :p

"In one cupboard is a mean Mousse ah Chocolat, in the other two is a swift kick in the nuts..."

I think pictures are better.

Before:
http://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Monty_closed_doors.svg/160px-Monty_closed_doors.svg.png

After:
http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Monty_open_door_chances.svg/180px-Monty_open_door_chances.svg.png

If that doesn't work out, I guess it's just one of those things you've gotta go with and know it's right, even if you don't quite understand.
Are you kidding me? Of course that doesn't work. :rolleyes: I can't even make sense of the numbers under the second set.

But I do like the goat. :p

Seriously, though. I was already content with what I like to call Acceptance Without Understanding back at post #71.
Everybody should save their energy and not try and make me see the statistical light. I appreciate it, but seriously - it's not worth the trouble.

Are you kidding me? Of course that doesn't work. :rolleyes: I can't even make sense of the numbers under the second set.

But I do like the goat. :p

Seriously, though. I was already content with what I like to call Acceptance Without Understanding back at post #71.
Everybody should save their energy to try and make me see the statistical light. I appreciate it, but seriously - it's not worth the trouble. ;)

meh, s'ok, you're cute. You're not expected to think....

*ducks*

Although the chocolate mousse is tempting...

"In one cupboard is a mean Mousse ah Chocolat, in the other two is a swift kick in the nuts..."
Since I don't have any nuts, I'm safe either way. ;p

"In one cupboard is a mean Mousse ah Chocolat, in the other two is a swift kick in the nuts..."

Hmm, personalization of the problem....It could work, no?

meh, s'ok, you're cute. You're not expected to think....

*ducks*

Although the chocolate mousse is tempting...
Ah, I knew those kicks in the nuts would come in handy eventually. :D

Since I don't have any nuts, I'm save either way. ;p

...A swift poke in the ovaries?

Ah, I knew those kicks in the nuts would come in handy eventually. :D

hell, if you wanna fly over the atlantic for that, that's some dedication.

And if I manage to get you THAT angry, I probably deserve it, heh.

But I do like the goat. :p

http://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.pnghttp://i4.photobucket.com/albums/y138/Dragonkirby/Non-Kirby/Goat.png

Since I don't have any nuts, I'm safe either way. ;p
Well, it's your mean Mousse, I assumed you'd be the host. And I'd be set crew and have totally snuck some of the Mousse during rehearsal, so I totally win, statistics be damned! Ha!
Hmm, personalization of the problem....It could work, no?
I thought that, except that no matter what, you get at least one swift kick to the nuts. I think I might be inclined to believe the math people if another one might be coming though...
Ah, I knew those kicks in the nuts would come in handy eventually. :D
See? Host.
...A swift poke in the ovaries?

Poke in the eye?

A repost - but does this make it any clearer?

These are the ONLY three possibilities if you switch:

Case 1: Prize is behind door one. You switch. You lose.
Case 2: Prize is behind door two. You switch. You win.
Case 3: Prize is behind door three. You switch. You win.
Therefore you win 2/3 times if you switch.


These are the ONLY three possibilities if you stay:

Case 1: Prize is behind door one. You stay. You win.
Case 2: Prize is behind door two. You stay. You lose.
Case 3: Prize is behind door three. You stay. You lose.
Therefore you lose 2/3 times if you stay.

Is this a good explination or does it still confuse? (looking for help in writing a better one)Case 1 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you switch, you lose
Case 2 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you stay, you win
Case 3 : Prize is behind door 1, you choose door 2, door 3 is opened, you switch, you win
Case 4 : Prize is behind door 1, you choose door 2, door 3 is opened, you stay, you lose
Case 5 : Prize is behind door 1, you choose door 3, door 2 is opened, you switch, you win
Case 6 : Prize is behind door 1, you choose door 3, door 2 is opened, you stay, you lose
Case 7 : Prize is behind door 2, you choose door 1, door 3 is opened, you switch, you win
Case 8 : Prize is behind door 2, you choose door 1, door 3 is opened, you stay, you lose
Case 9 : Prize is behind door 2, you choose door 2, door 1 or 3 is opened, you switch, you lose
Case 10 : Prize is behind door 2, you choose door 2, door 1 or 3 is opened, you stay, you win
Case 11 : Prize is behind door 2, you choose door 3, door 1 is opened, you switch, you win
Case 12 : Prize is behind door 2, you choose door 3, door 1 is opened, you stay, you lose
Case 13 : Prize is behind door 3, you choose door 1, door 2 is opened, you switch, you win
Case 14 : Prize is behind door 3, you choose door 1, door 2 is opened, you stay, you lose
Case 15 : Prize is behind door 3, you choose door 2, door 1 is opened, you switch, you win
Case 16 : Prize is behind door 3, you choose door 2, door 1 is opened, you stay, you lose
Case 17 : Prize is behind door 3, you choose door 3, door 1 or 2 is opened, you switch, you lose
Case 18 : Prize is behind door 3, you choose door 3, door 1 or 2 is opened, you stay, you win

.

Case 1 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you switch, you lose
Case 2 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you stay, you win
Case 3 : Prize is behind door 1, you choose door 2, door 3 is opened, you switch, you win
Case 4 : Prize is behind door 1, you choose door 2, door 3 is opened, you stay, you lose
Case 5 : Prize is behind door 1, you choose door 3, door 2 is opened, you switch, you win
Case 6 : Prize is behind door 1, you choose door 3, door 2 is opened, you stay, you lose
Case 7 : Prize is behind door 2, you choose door 1, door 3 is opened, you switch, you win
Case 8 : Prize is behind door 2, you choose door 1, door 3 is opened, you stay, you lose
Case 9 : Prize is behind door 2, you choose door 2, door 1 or 3 is opened, you switch, you lose
Case 10 : Prize is behind door 2, you choose door 2, door 1 or 3 is opened, you stay, you win
Case 11 : Prize is behind door 2, you choose door 3, door 1 is opened, you switch, you win
Case 12 : Prize is behind door 2, you choose door 3, door 1 is opened, you stay, you lose
Case 13 : Prize is behind door 3, you choose door 1, door 2 is opened, you switch, you win
Case 14 : Prize is behind door 3, you choose door 1, door 2 is opened, you stay, you lose
Case 15 : Prize is behind door 3, you choose door 2, door 1 is opened, you switch, you win
Case 16 : Prize is behind door 3, you choose door 2, door 1 is opened, you stay, you lose
Case 17 : Prize is behind door 3, you choose door 3, door 1 or 2 is opened, you switch, you lose
Case 18 : Prize is behind door 3, you choose door 3, door 1 or 2 is opened, you stay, you win

.

That's the same thing.

Case 1 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you switch, you lose
Case 3 : Prize is behind door 1, you choose door 2, door 3 is opened, you switch, you win
Case 5 : Prize is behind door 1, you choose door 3, door 2 is opened, you switch, you win

Case 2 : Prize is behind door 1, you choose door 1, door 2 or 3 is opened, you stay, you win
Case 4 : Prize is behind door 1, you choose door 2, door 3 is opened, you stay, you lose
Case 6 : Prize is behind door 1, you choose door 3, door 2 is opened, you stay, you lose


An equal number of "you win" and "you lose" but 2/3rds of the "you win" are paired with "you switch". You have proved yourself wrong.

:eek:

I know how to make everyone else understand it!
Imagine that instead of three doors there are 100 doors.
Imagine that once you picked one of those doors the host would 98 doors w/goats.
Now, which do you pick?
There's a 1/100 chance your door was right if you picked it in the beginning so then the remaining door would have a goat.
If you picked the wrong one in the beginning (99/100 chance) then the door the host didn't open has the car.
So, there is a 99/100 chance that the door the host didn't open has a car.
So you should definitally switch.

So:
There is a 1/3 chance you picked the right door.
There is a 2/3 chance you picked the wrong door.
After the host opens the door, there is still a 2/3 chance you picked the wrong door in the beginning.
If you did pick the wrong door, you should switch.

:eek:

Proving yourself wrong is a Mindfuck, isn't it?

I know how to make everyone else understand it!
Imagine that instead of three doors there are 100 doors.
Imagine that once you picked one of those doors the host would 98 doors w/goats.
Now, which do you pick?
There's a 1/100 chance your door was right if you picked it in the beginning so then the remaining door would have a goat.
If you picked the wrong one in the beginning (99/100 chance) then the door the host didn't open has the car.
So, there is a 99/100 chance that the door the host didn't open has a car.
So you should definitally switch.

That's actually not too bad a way to explain it, methinks. Kinda drives home the point a bit more.

Of course in my heart of hearts I still maintain that when you're down to the two last doors it's a fity-fifty chance of which one has the car. [/stubborn] :p

In the original situation:
If you started with any goat, then switching always gets you the car.
If you started with the car, then switching always gets you a goat.

So we can simplify the process by asking two questions:
"What door do you choose?"
and then:
"Do you want the other prize?"

In this situation:
If you started with any goat, then switching always gets you the car.
If you started with the car, then switching always gets you a goat.

Essentially the same situation, agreed? The results are the same.

So in the second situation, our possibilities are (without switching prize):
Door A - Goat ---> Goat
Door B - Goat ---> Goat
Door C - Car ---> Car

2/3 chance of a goat
1/3 chance of a car

And our possibilities with switching the prize:
Door A - Goat ---> Car
Door B - Goat ---> Car
Door C - Car ---> Goat

1/3 chance of a goat
2/3 chance of a car

Another way of thinking about it: Opening the door doesn't remove it as an option. It's informing you of what's inside the door. Most people immediately eliminate it as a choice because it's a bad idea to choose it and the host doesn't bother asking if you want it.

From Cecil Adams "Worlds Smartest Person" on the subject:

http://www.straightdope.com/classics/a3_189.html

*buzzer*

http://upload.wikimedia.org/wikipedia/en/thumb/9/9e/Monty_tree.svg/520px-Monty_tree.svg.png

Ah, now it makes perfect sense.

:eek:

I am guessing this roughly translates to "Golly! I do believe I have realised switching gives you a 2/3 chance of winning"?

If so is this a reasonable time to expect an apology for the following?

How old are you?

No. And now shut the f up and go back to school.
Or back to your cave.

If you cannot grasp infinitesimal mathematics that is a deficiency of yours that you should remove first before annoying others.

Again, if you cannot grasp that go back to school and finally learn something. And how you see this "Conceptually" is of no relevance to the fact. As I said before, this is no matter of opinion. Also the link to the wikipedia article has been posted numerous times, so you could have looked it up.
(note - quite a few wiki articles have been posted for the Game Show)

This is something so basic, I have to wonder why you need an internet forum to understand this. Did you not have math classes at school? Your insistence in your error instantaneously has me thinking of a cave dweller who just does not know any better.

It was not me who said something stupid and insisted on it. Your claim to be damn good at "this stuff" is compromised somewhat by your little calculation examples.
As to my maths ability - I may not be as good at "this stuff" as I used to be back in university but I think such very basic concepts are still within my understanding.
As for your math ability - "The final 9 in the 10x is 10 times bigger than the final 9 in x."

(my bold)


Particulary given your outburst was in response to two (or three?) posts of mine while you have posted about a dozen times in this and the other thread on this topic (and it has been personally explained to you, and you have linked to (and therefore read i assume) the wiki links), which between the two threads has spanned about as many pages as when you decided to start swearing and insulting my math ability based on one or two posts. And given that this is in all ways a simpler puzzle, which can be easily drawn on diagrams or written down or tried at home with a friend. And also given that you have been treated civily and in a helpful manner by all, with people calmly and politely trying to explain it to you personally in several ways without resorting to abuse or insults?

I get bored of the question and try and solveit using Eny-meany-miny-mo.:D

oh, I see what you mean.

And I think you're right (although it doesn't give him a 49% chance to go free, it gives him a 49/99 chance (50 read marbels, 49 blue, 99 marbels total in the box), which is actually 49.5%.

so there's a 50% chance he'll pick the box with one marble, which will set him free 100% of the time, and a 50% chance he'll pick the other box, which will set him free 49.5% of the time.

so the math is (1/2)(1/1) + (1/2)(49/99) odds that he will go free

Which gives him a 74.75% chance of being free. I think that this is the correct answer.

Bingo. His chances of going free are just shy of 75%. It's the best chance he has.

Another variant of a Car Talk puzzle:

A king has one daughter who's coming of age, and though she wants to get married, he wants to keep her at home because he doesn't trust any man she would marry to succeed him. He subjects every potential suitor to extremely rigorous testing, and in the end, no one has succeeded. Every potential suitor who has failed has been told to pack his things and leave the kingdom forever.

However, from a rural area of the kingdom, a gallant young lad steps up to the challenge. He succeeds in every challenge - slays a dragon, crosses a crocodile-infested moat, and even removes a thorn from a lion's paw. There is one last challenge - a random draw.

Into an opaque bag, the king drops two marbles, a blue one and a red one. If the suitor draws the blue one, he marries the king's daughter; if he draws the red one, he goes into exile (as the king can't risk too much bad PR by just killing the guy.) Nonetheless, after a grand show of dropping both a blue and a red marble into the bag the day before, the king's daughter approaches the lad and tell him the following.

"My father is a cheat. The last two suitors who made it this far lost because he removes the blue marble the night before and replaces it with another red one. No matter what you draw, it will be red." The lad is fully aware that he can't call the king out on his lies, lest he be named a traitor, and he can't tamper with the bag, lest he be named a cheater himself. However, he procceds with the drawing, and the following summer, he marries the king's daughter, and the kingdom was none the wiser that the king was a cheat or that the lad had tricked him. How did he do it?

Bingo. His chances of going free are just shy of 75%. It's the best chance he has.

Another variant of a Car Talk puzzle:

A king has one daughter who's coming of age, and though she wants to get married, he wants to keep her at home because he doesn't trust any man she would marry to succeed him. He subjects every potential suitor to extremely rigorous testing, and in the end, no one has succeeded. Every potential suitor who has failed has been told to pack his things and leave the kingdom forever.

However, from a rural area of the kingdom, a gallant young lad steps up to the challenge. He succeeds in every challenge - slays a dragon, crosses a crocodile-infested moat, and even removes a thorn from a lion's paw. There is one last challenge - a random draw.

Into an opaque bag, the king drops two marbles, a blue one and a red one. If the suitor draws the blue one, he marries the king's daughter; if he draws the red one, he goes into exile (as the king can't risk too much bad PR by just killing the guy.) Nonetheless, after a grand show of dropping both a blue and a red marble into the bag the day before, the king's daughter approaches the lad and tell him the following.

"My father is a cheat. The last two suitors who made it this far lost because he removes the blue marble the night before and replaces it with another red one. No matter what you draw, it will be red." The lad is fully aware that he can't call the king out on his lies, lest he be named a traitor, and he can't tamper with the bag, lest he be named a cheater himself. However, he procceds with the drawing, and the following summer, he marries the king's daughter, and the kingdom was none the wiser that the king was a cheat or that the lad had tricked him. How did he do it?

Well there's more than one way he could have tricked him...

You're ignoring the fact that he just revealed what was behind one door. Now there's only two possibilities and your initial probability of picking the correct door is no longer relevant. You now have a 50/50 shot and there's no correct answer. Stay or do not, same odds.

Actually, your initial probability is relevant. There is a 2/3 chance that you pick a wrong door on your first pick and the show host shows you the other wrong door, giving you a 100% chance to succeed by switching. There is 1/3 chance that you pick the right door and the show host shows you one of the wrong doors, giving you a 0% chance to succeed by switching.

Mathematically:
The probability you pick the wrong one first and pick the right one by switching => 2/3*1
The probability you pick the right one first and pick the right one by switching => 0

Total chance of picking the right one is 2/3

Now let's assume you stay. There is a 2/3 chance that you pick a wrong door on your first pick and the show host shows you the other wrong door, giving you a 0% chance to succeed by staying. There is 1/3 chance that you pick the right door and the show host shows you one of the wrong doors, giving you a 100% chance to succeed by staying.

The probability you pick the right one first and pick the right one by staying => 1/3*1
The probability you pick the wrong one first and pick the right one by staying => 0

Total chance of picking the right one is 1/3


Do you understand now?

Well there's more than one way he could have tricked him...

Pick your favorite. Pick four of your favorites. Go nuts.

Oh, I get it. You pick one door, and the host says, essentially, "Do you want to exchange your one door for my two doors?"
Two doors is better.

I am guessing this roughly translates to "Golly! I do believe I have realised switching gives you a 2/3 chance of winning"?

If so is this a reasonable time to expect an apology for the following?Where did all the posts you quoted after that come from?

A king has one daughter who's coming of age, and though she wants to get married, he wants to keep her at home because he doesn't trust any man she would marry to succeed him. He subjects every potential suitor to extremely rigorous testing, and in the end, no one has succeeded. Every potential suitor who has failed has been told to pack his things and leave the kingdom forever.

However, from a rural area of the kingdom, a gallant young lad steps up to the challenge. He succeeds in every challenge - slays a dragon, crosses a crocodile-infested moat, and even removes a thorn from a lion's paw. There is one last challenge - a random draw.

Into an opaque bag, the king drops two marbles, a blue one and a red one. If the suitor draws the blue one, he marries the king's daughter; if he draws the red one, he goes into exile (as the king can't risk too much bad PR by just killing the guy.) Nonetheless, after a grand show of dropping both a blue and a red marble into the bag the day before, the king's daughter approaches the lad and tell him the following.

"My father is a cheat. The last two suitors who made it this far lost because he removes the blue marble the night before and replaces it with another red one. No matter what you draw, it will be red." The lad is fully aware that he can't call the king out on his lies, lest he be named a traitor, and he can't tamper with the bag, lest he be named a cheater himself. However, he procceds with the drawing, and the following summer, he marries the king's daughter, and the kingdom was none the wiser that the king was a cheat or that the lad had tricked him. How did he do it?He drew a picture of a blue marble. Get it? He drew the blue marble.

If it wasn't a trick question, he pulled one marble out of the bag and held it in a fist so no one could see it. Then he turned the bag over so the other marble fell out. Since one red marble fell out of the bag, the witnesses thought he's holding a blue marble.

Where did all the posts you quoted after that come from?

He drew a picture of a blue marble. Get it? He drew the blue marble.

If it wasn't a trick question, he pulled one marble out of the bag and held it in a fist so no one could see it. Then he turned the bag over so the other marble fell out. Since one red marble fell out of the bag, the witnesses thought he's holding a blue marble.

Good answer. In the puzzler I remember, he actually swallowed the marble and then took the other one out, and in another version, he simply said, "I pick the color opposite of the one that I draw." Any of those works.