Sheep or a Car

[TheChort]

Originally posted by: Malak

Originally posted by: Schfifty Five
Does that make sense?

I hate to repeat this again and again and again...

If it were that simple, it wouldn't have been so hotly debated. I've said enough on this subject.

Personally, I don't care how you come to your conclusion on this one, or if you ever do for that matter, mostly because I will probably never meet you.

But for that very same reason, my advice to will hopefully seem more sincere:
Stop thinking so highly of yourself and your mental processes. You are not a genius. You are probably above average, but that's about it.Your attempts to seem like some overly intelligent person doesn't fool anybody except yourself.
Look at some of the comments you've made

I am going to ponder on this and get back with you.

you sound like some 60 year old playing a game of chess over snail mail. Take it easy on yourself. There will be plenty of time for boring people by talking like this when you get older.

 

[6000SUX]

The chance is 50/50. I made my own CORRECT simulation, which is of course super-simple, and verified it.

I won't speculate as to how the other one may have been badly programmed, but really the way it is, there is a very simple random routine to place the car behind one of three windows. The turns where it is behind window 3 are not counted (or in my simulation increment a "pass" counter). The ones where it is behind window 1 count as a win, and behind window 2 count as a loss. I've run it several times now with iterations of 100 tests and the win, loss, and pass numbers tend to equal out, as you'd expect.

Marilyn vos Savant, despite having the highest score on a flawed test, isn't necessarily right all the time. Neither is Wikipedia-- it contains numerous errors.

Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

 

[Schfifty Five]

Originally posted by: 6000SUX
The chance is 50/50. I made my own CORRECT simulation, which is of course super-simple, and verified it.

I won't speculate as to how the other one may have been badly programmed, but really the way it is, there is a very simple random routine to place the car behind one of three windows. The turns where it is behind window 3 are not counted (or in my simulation increment a "pass" counter). The ones where it is behind window 1 count as a win, and behind window 2 count as a loss. I've run it several times now with iterations of 100 tests and the win, loss, and pass numbers tend to equal out, as you'd expect.

Marilyn vos Savant, despite having the highest score on a flawed test, isn't necessarily right all the time. Neither is Wikipedia-- it contains numerous errors.

Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

It doesn't matter which door the car is behind since the host knows what is exactly behind each door.

He let's the contestant pick any door, then he will open up one of the remaining two doors that does not contain the car. So it essentially could be any of the 3 doors depending on what the contestant picks first.

It's 33/66, no 50/50.

 

[6000SUX]

Originally posted by: Schfifty Five

Originally posted by: 6000SUX
The chance is 50/50. I made my own CORRECT simulation, which is of course super-simple, and verified it.

I won't speculate as to how the other one may have been badly programmed, but really the way it is, there is a very simple random routine to place the car behind one of three windows. The turns where it is behind window 3 are not counted (or in my simulation increment a "pass" counter). The ones where it is behind window 1 count as a win, and behind window 2 count as a loss. I've run it several times now with iterations of 100 tests and the win, loss, and pass numbers tend to equal out, as you'd expect.

Marilyn vos Savant, despite having the highest score on a flawed test, isn't necessarily right all the time. Neither is Wikipedia-- it contains numerous errors.

Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

It doesn't matter which door the car is behind since the host knows what is exactly behind each door.

He let's the contestant pick any door, then he will open up one of the remaining two doors that does not contain the car. So it essentially could be any of the 3 doors depending on what the contestant picks first.

It's 33/66, no 50/50.

That's sort of a confused explanation, but I understand what's going on. What you said is not what the OP said.

A very simple way to explain this is that the car will only be behind the contestant's originally picked door 1/3 of the time. When it is not (the other 2/3 of the time), it will be behind the door not opened by the host.

If it were stated correctly, I'm surprised people have trouble with it.

 

[Born2bwire]

Originally posted by: Malak

Originally posted by: chrisms
Change choice: 1600 times Wins: 1010 cars (63%)
Losses: 590 goats (37%)

Keep choice: 1501 times Wins: 499 cars (33%)
Losses: 1002 goats (67%)

Do you notice the variance? This is part of my issue. It's a little hard to explain, but for now I'm saying if you switch it's approximately a 58% chance of winning, not 66%. If you don't switch, you have a 33% chance of winning, and thus a 66% chance of losing. I'm not good at explaining things so I won't even bother explaining my confusing thought pattern that let me to 58%, but the tests I've seen show a variance between switching and not switching.

Wow. How do you even come up with this stuff?

 

[SonicIce]

I don't understand why it's not a 50/50 chance. You will have two doors to choose from after the third is removed. How does changing help at all?

 

[ViRGE]

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

 

[Schfifty Five]

Originally posted by: SonicIce
I don't understand why it's not a 50/50 chance. You will have two doors to choose from after the third is removed. How does changing help at all?

READ THE THREAD.

 

[iamaelephant]

Originally posted by: Born2bwire

Originally posted by: Malak

Originally posted by: chrisms
Change choice: 1600 times Wins: 1010 cars (63%)
Losses: 590 goats (37%)

Keep choice: 1501 times Wins: 499 cars (33%)
Losses: 1002 goats (67%)

Do you notice the variance? This is part of my issue. It's a little hard to explain, but for now I'm saying if you switch it's approximately a 58% chance of winning, not 66%. If you don't switch, you have a 33% chance of winning, and thus a 66% chance of losing. I'm not good at explaining things so I won't even bother explaining my confusing thought pattern that let me to 58%, but the tests I've seen show a variance between switching and not switching.

Wow. How do you even come up with this stuff?

Lay off him. The world needs janitors.

 

[6000SUX]

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

 

[JujuFish]

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

 

[Schfifty Five]

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

haha, agreed. 6000SUX, the OP's wording works fine.

 

[6000SUX]

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

 

[6000SUX]

Originally posted by: Schfifty Five

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

haha, agreed. 6000SUX, the OP's wording works fine.

No. It doesn't state all the particulars of the problem. Strictly read, it is exactly as I interpreted it. It's only the stupidity of the problem as stated that made me realize (on my own) what the real problem is.

 

[6000SUX]

It may be an interesting follow-up problem for some of you (the more talented ones, please) to figure out why, if the host always randomly picked one of the other two windows to show, the contestant's chances would be 50/50. In this scenario there would be no advantage to switching.

 

[dxkj]

IF you pick a sheep, and switch, you win a car
You have a 66% chance of having picked a sheep
Thus you have a 66% chance of winning a car if you switch

 

[6000SUX]

Originally posted by: dxkj
IF you pick a sheep, and switch, you win a car
You have a 66% chance of having picked a sheep
Thus you have a 66% chance of winning a car if you switch

I assume you're answering the original problem (or at least the vos Savant problem). That's not the answer to my new problem.

 

[Xylitol]

of course you have an advantage
it changes from 33% to 50% because you now have 2 doors instead of 3 to choose the car

 

[moomoo40moo]

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

uhh, why would the host show the car? that would ruin the WHOLE point of the game.

 

[6000SUX]

Originally posted by: moomoo40moo

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

uhh, why would the host show the car? that would ruin the WHOLE point of the game.

Duhhh, the problem didn't say, but it wouldn't ruin the whole point of the game. It's a hypothetical situation. Actually, showing a goat every time has no point-- its only point in the hypothetical situation is to befuddle people trying to find the answer to the question.

 

[Schfifty Five]

Originally posted by: Xylitol
of course you have an advantage
it changes from 33% to 50% because you now have 2 doors instead of 3 to choose the car

wrong it goes from 33->66%

 

[JujuFish]

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

You're beginning to sound a bit like Malak in your arrogance. It's okay to admit when you're wrong. The OP says that the hosts shows a door with a sheep behind it. The OP does not say that the host picks a door which happens to have a sheep behind it. There is a distinction between the two, and you're either too proud to admit that or you just didn't realize it.

Edit: Changed "goat" to "sheep".

 

[6000SUX]

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

You're beginning to sound a bit like Malak in your arrogance. It's okay to admit when you're wrong. The OP says that the hosts shows a door with a sheep behind it. The OP does not say that the host picks a door which happens to have a sheep behind it. There is a distinction between the two, and you're either too proud to admit that or you just didn't realize it.

Edit: Changed "goat" to "sheep".

Umm, nope. Feel free to admit that you're the arrogant one. Nothing in the OP's vague wording indicated that the host picks a door; it merely says, in one hypothetical situation, that the host, "who knows what's behind the doors, opens another door, say #3, which has a sheep". You've missed the point completely and got the "picking" part ass-backwards.

The host has to pick in order for the 66% advantage to work. If he merely opens a door without regard to what's behind it, there is no advantage. The OP's wording doesn't say that the host opens a door because it contains a sheep; he says he opens a door which happens to contain a sheep.

There aren't any words which show that the host gives any weight to what's behind the door, only that he knows. Maybe, in the OP's stupid scenario, the host always opens door #3 if it's not chosen. The wording is useless.

Don't try to school your betters. I can run logic circles around the likes of you all day. According to the OP's wording, there is no advantage. But of course, he did not correctly state the problem.

 

[Born2bwire]

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: JujuFish

Originally posted by: 6000SUX

Originally posted by: ViRGE

Originally posted by: 6000SUX
Edit: Maybe I didn't read far enough, and the OP worded the post badly. Does the host reveal #2 always, if he knows the car is in #3, and vice versa?

Correct. The host will always reveal a goat, so if the car is behind #1 he'll open #2 or #3(it doesn't matter either way), if it's behind #2 he'll open door #3, and if it's behind #3 he'll open #2. Your error would seem to be that you're not counting when it's behind #3.

Let's be clear: I didn't make an error. The OP did, in stating the problem incorrectly.

That is incorrect. The OP's wording is quite valid.

Anope. It's not. It doesn't say that the host will always intentionally show a goat, even if he can show a car. You're incorrect.

You're beginning to sound a bit like Malak in your arrogance. It's okay to admit when you're wrong. The OP says that the hosts shows a door with a sheep behind it. The OP does not say that the host picks a door which happens to have a sheep behind it. There is a distinction between the two, and you're either too proud to admit that or you just didn't realize it.

Edit: Changed "goat" to "sheep".

Umm, nope. Feel free to admit that you're the arrogant one. Nothing in the OP's vague wording indicated that the host picks a door; it merely says, in one hypothetical situation, that the host, "who knows what's behind the doors, opens another door, say #3, which has a sheep". You've missed the point completely and got the "picking" part ass-backwards.

The host has to pick in order for the 66% advantage to work. If he merely opens a door without regard to what's behind it, there is no advantage. The OP's wording doesn't say that the host opens a door because it contains a sheep; he says he opens a door which happens to contain a sheep.

There aren't any words which show that the host gives any weight to what's behind the door, only that he knows. Maybe, in the OP's stupid scenario, the host always opens door #3 if it's not chosen. The wording is useless.

Don't try to school your betters. I can run logic circles around the likes of you all day. According to the OP's wording, there is no advantage. But of course, he did not correctly state the problem.

The OP doesn't say, "he opens a door which happens to contain a sheep", he says, "he opens another door which has a sheep." Big difference between happens to have and has. The only beef that one could say with the statement is that the OP does not specify whether or not the host has to open a door everytime. That does change the outcome and was one of the main points of debate in the original statement of the problem in print.

 

[dxkj]

Originally posted by: 6000SUX
It may be an interesting follow-up problem for some of you (the more talented ones, please) to figure out why, if the host always randomly picked one of the other two windows to show, the contestant's chances would be 50/50. In this scenario there would be no advantage to switching.

IF the dude picked randomly he would open your door 1/3 of the time, do you still get another chance to switch?

If the dude picked randomly he could open the car one 1/3 of the time, do you get to pick the door if he does that?

if he opened a sheep door, youd have a 66% benefit for switching still, that wouldnt change

if he opened a car door, youd have a 100% benefit since you could take that one

if he opened your door and it was a sheep, youd have a 50% chance

if he opened your door and it was a car, youd have 100% chance

those add up to more than 50%



if he can only choose the two doors you arent picking

then there would be a 66% chance that he would open up a sheep door, and you would have a 66% chance by switching

there would be a 33% chance he would open up a car door, and 100% chance you win by switching.


66% * 66% + 33% = 76.5% WHEEEE!