The Monty Hall Paradox--eggheads explain?

[Mem]

It is like one of those topographical maps of the moon or mars, and you can't really tell if its a crater or a mountain, and it gets trippy, but once ou see it, you can always see it. Or like a magic eye 3D book. It doesn't make sense it doesn't make sense it doesn't make sense and then - BAM! You've unlocked the concept.

.

Yeah like the picture of the lady that is a skull (no pun intended) when you look at it differently.

 

[Mem]

Yeah like the picture of the lady that is a skull (no pun intended) when you look at it differently.

.

 

[2322]

Yeah but you're 19 and get it. I'm 41 and don't. The over time thing has long passed if your point is relative.

I apologize for being so thick and thank you for your help

Jason, one more thing and I'll probably shut up. It's much more difficult to understand why mathematical concepts behave they way they do, given equal brain capacity, if you know less in math procedures than the next guy. Certain things dawn on you with time. Something like this really needs churning time. Don't expect to understand it right now. You'll see how it works.

In fact, I thank everyone for their time and effort. Sorry to frustrate you. I'm really enjoying this.

 

[2322]

OK, re-reading this....

Regarding the highlighted section below, as I see it, no matter what door you choose, Monty's choice is always affected. It doesn't matter if you choose the car or not because Monty won't know what his choices are until you make yours. All we know is that Monty will always reveal a goat (in that his choice is unaffected), but Monty won't know which goat until you've made your choice.

The rest of the statement below does make sense. I "get" it. It seems to me, if both choices are arbitrary, there should be no reward for switching because, ultimately, the choices are random to begin with. yet it is so and I don't know why. I understand the mechanics, but that it works in the real world, if I played this game for a few months, the results would be true. That bothers me.

mem0101 says that two people can't play at once but I see no reason they can't because it doesn't matter what set of random choices you make, so long as you switch, you will win more often than lose. Even if by coincidence, the two players always chose the same door first, they both should win. This is what I can't get over and I apologize for harping on it.

That's the fallacy most people ended up buying, including myself at first. It seems like it's 50/50 because it seems like the first choice doesn't matter, but it does.

Because the first choice only gives you a 33% chance of picking the car. If you pick the car the first time, Monty's choice is unaffected, and then your decision to switch will always end up with you losing when you picked the car initially.

So you'll lose 33% of the time.

The first choice also gives you a 66% chance of choosing the booby prize. If you pick the booby prize the first time, Monty always shows you the other booby prize, then you switch and you always end up winning when you picked the booby prize initially.

So you'll win 66% of the time.

If you always switch.

Got it?

 

[Gillette]

I'll try one more analogy - I was never that good a tutor it seems.

When you first make your initial choice, all 3 doors are closed. Your chance of your first guess being correct is 33%. If you had picked either of the other 2 doors, your chance would also be 33% for each one, or 67% for the 2 doors you didn't pick, combined.

Now remember - the car doesn't move. Opening another door doesn't reset the odds behind your first choice.

Now one of the remaining doors is opened. There's the goat. The odds to that door drop to 0%. But you had already picked the first door, at 33% odds. So the remaining doors still have a 67% chance of being correct. If there's only one door left, that's a 67% chance for that door.

33:67 = 1:2

I see it but I don't buy it. When you step up on stage you have a 100% chance of being left with 2 options. That's 50/50.

If each door has a 33.333% of being correct the door that is eliminated shouldn't drop to 0% leaving the other with 67%. As it is eliminated so is it's 33.333%, leaving two doors with 33% each or 50/50.

When a science experiment is performed to see which is the determining variable as each variable is controlled for and subsequently discarded the possibility of one of the remaining variables being the determining variable goes up until you are left with only one. Before you reach that last variable there are two, each with an equal chance of being the determining one.

Read my post at the bottom of page 4, as it may provide more clarity and you may have missed it.

Bottom line is, since the first choice isn't 50/50, the second choice isn't either, as long as you don't treat the second one as random.

If you randomize that second choice, it will be 50%.

If you always choose to switch, it will be 66%.

I don't have much more to offer. That is the correct answer, though, and if you have a simulator do it 10,000 times, I assure you it will tell you it won 66% of the time (or don't believe me, I think there are simulators out there you could run yourself if you wanted to prove it to yourself).

I do see the second choice as random. I thought randomness was the very essence of probability. I'll play a bit and think on it.

............................................________
....................................,.-‘”...................``~.,
.............................,.-”...................................“-.,
.........................,/...............................................”:,
.....................,?......................................................\,
.................../...........................................................,}
................./......................................................,:`^`..}
.............../...................................................,:”........./
..............?.....__.........................................:`.........../
............./__.(.....“~-,_..............................,:`........../
.........../(_....”~,_........“~,_....................,:`........_/
..........{.._$;_......”=,_.......“-,_.......,.-~-,},.~”;/....}
...........((.....*~_.......”=-._......“;,,./`..../”............../
...,,,___.\`~,......“~.,....................`.....}............../
............(....`=-,,.......`........................(......;_,,-”
............/.`~,......`-...............................\....../\
.............\`~.*-,.....................................|,./.....\,__
,,_..........}.>-._\...................................|..............`=~-,
.....`=~-,_\_......`\,.................................\
...................`=~-,,.\,...............................\
................................`:,,...........................`\..............__
.....................................`=-,...................,%`>--==``
........................................_\..........._,-%.......`\
...................................,<`.._|_,-&``................`\


edit: Sowwy, but...

Quite alright, Quite Irate.

In The Navy :wink:

 

[Gillette]

Ever hear about the statistician who drowned in a lake with an average depth of 1.2cms?

 

[Guy-jin]

OK, re-reading this....

Regarding the highlighted section below, as I see it, no matter what door you choose, Monty's choice is always affected. It doesn't matter if you choose the car or not because Monty won't know what his choices are until you make yours. All we know is that Monty will always reveal a goat (in that his choice is unaffected), but Monty won't know which goat until you've made your choice.

That's true, looking at it that way. But, like you said, the key is just that you know Monty will always reveal a goat, regardless of what you picked, and when you switch, you will never purposely switch to a goat.

The rest of the statement below does make sense. I "get" it. It seems to me, if both choices are arbitrary, there should be no reward for switching because, ultimately, the choices are random to begin with. yet it is so and I don't know why. I understand the mechanics, but that it works in the real world, if I played this game for a few months, the results would be true. That bothers me.

mem0101 says that two people can't play at once but I see no reason they can't because it doesn't matter what set of random choices you make, so long as you switch, you will win more often than lose. Even if by coincidence, the two players always chose the same door first, they both should win. This is what I can't get over and I apologize for harping on it.

Well, I'm glad it actually helped! I'm no stats teacher, but I do have to teach some stats in what I do, and it can be really hard to explain sometimes.

It might make you feel better to consider it like this: Basically, you are changing it from a game that has a car behind one of three doors to a game that has a car behind two of three doors.

I realize that physically that isn't happening, but by consciously switching every time at the second stage, that's actually what it's like.

As for your final statement, the point is that if two people are selecting doors randomly, the chances of them picking the same door every time are very, very low. So low, in fact, that we'd call them statistically insignificant. So it's not a realistic scenario.

If you were to say that they do it a million times and end up doing that, a statistician would never say, "That's impossible," he'd just say, "That's very, very unlikely."

Hope that helps.

 

[2322]

That's true, looking at it that way. But, like you said, the key is just that you know Monty will always reveal a goat, regardless of what you picked, and when you switch, you will never purposely switch to a goat.

Well, I'm glad it actually helped! I'm no stats teacher, but I do have to teach some stats in what I do, and it can be really hard to explain sometimes.

It might make you feel better to consider it like this: Basically, you are changing it from a game that has a car behind one of three doors to a game that has a car behind two of three doors.

I realize that physically that isn't happening, but by consciously switching every time at the second stage, that's actually what it's like.

As for your final statement, the point is that if two people are selecting doors randomly, the chances of them picking the same door every time are very, very low. So low, in fact, that we'd call them statistically insignificant. So it's not a realistic scenario.

If you were to say that they do it a million times and end up doing that, a statistician would never say, "That's impossible," he'd just say, "That's very, very unlikely."

Hope that helps.

I does. Very helpful.

However....... the doppleganger would only have to choose the same first door and the opposite second door. Rather like the car, there's a 33% chance he'll choose the same door. Not what I would call, "very, very, unlikely." The doppleganger has a 50% chance of choosing the second opposite door and that's not bad. Still improbable, but not necessarily statistically so.:tongue:

 

[Guy-jin]

I does. Very helpful.

However....... the doppleganger would only have to choose the same first door and the opposite second door. Rather like the car, there's a 33% chance he'll choose the same door. Not what I would call, "very, very, unlikely." The doppleganger has a 50% chance of choosing the second opposite door and that's not bad. Still improbable, but not necessarily statistically so.:tongue:

Well, like I said, statistically, they'd get quite literally the same exact score. Try flipping a coin ten times and writing down what you get, then flip it ten more times and write down what you get. Now do it until you repeat a pattern of flips. It's likely to take you quite a while, and that's only ten flips. Now imagine doing it 10,000 times or a million times and comparing and you start seeing just how unlikely it is.

 

[Principessa]

At work, I have an intelligent, youthful assistant. He maintains that the Monty Hall Paradox is true. That is, the counter-intuitive solution is correct. I don't see it. Can someone explain it/convince me? There must be either a simple explanation, or a refutation.

Maybe my brain is too old, and not flexible enough.

Eeekk! A math problem! It looks like prob and stat, I hated that class. :yuck: njqt466 runs and hides under the nearest big dick. :smile::wink:

 

[ManlyBanisters]

Sorry to repeat my own post but I really don't see the problem, when you look at the entire game AS A WHOLE it is quite clear.

There are ONLY 3 possible scenarios, right? Only 3 ways the game can play out.

#1 The player originally picked the door hiding the car. 33.3&#37; likely
#2 The player originally picked the door hiding Goat A. 33.3% likely
#3 The player originally picked the door hiding Goat B. 33.3% likely

From this the following happens:

#1 The player originally picked the door hiding the car. The game host opens another door to show one of the two goats. The player switches and loses (or sticks and wins) The scenario is STILL 33.3% likely

#2 The player originally picked the door hiding Goat A. The game host opens another door to show the ONE other goat. The player switches and wins (or sticks and loses) The scenario is STILL 33.3% likely

#3 The player originally picked the door hiding Goat B. The game host opens another door to show the ONE other goat. The player switches and wins (or sticks and loses) The scenario is STILL 33.3% likely

In the second TWO possibilities there the player switches doors and wins.

The game cannot be seen as two separate parts - it just can't. There are 3 scenarios and in 2 out of 3 switching wins and sticking loses.

 

[rob_just_rob]

I see it but I don't buy it. When you step up on stage you have a 100% chance of being left with 2 options. That's 50/50.

If each door has a 33.333% of being correct the door that is eliminated shouldn't drop to 0% leaving the other with 67%. As it is eliminated so is it's 33.333%, leaving two doors with 33% each or 50/50.

*shrugs* Your probabilities have to add up to 100, Gillette.

 

[Quite Irate]

Additional food for, well, "thought".

Prior probability - Wikipedia, the free encyclopedia

Posterior probability - Wikipedia, the free encyclopedia

Learn Bayesian probability - Wikipedia, the free encyclopedia and you can prove the Monty Hall Problem.

 

[B_dumbcow]

after much careful consideration i think this is why?!?!

say the car is in door 1 and the goats in 2 and 3

not swapping
First door is a car. 33&#37;
Second door is a goat. 33%
Thirs door is a goat. 33%

so if you do not swap, your chances of winning are 33%
you only win if you pick a car first time (33%)

Swapping
First door is car. monty opens a goat door you swap. you lose. 33%
second door is goat. Monty opens the other goat door, so when you swap you can only get the car. 33%
Third door is goat. Monty opens the other goat door, so when you swap you can only get the car. 33%

So if you do swap, your chances of winning are 33% + 33% = 66%
You only win if you pick a goat first time (66%)



does this make sense? it does to me

 

[ZOS23xy]

All seeming extended from a short story called "The Lady or the Tiger?" by Frank R. Stockton, which also generated much ink and books to it's topic.

One door has the lady, and one door has the tiger. The man chooses one, and if he picks the lady, he marries her. He picks the tiger, he gets devorued.

Of course, for the story, the man is in love with the lady, who is the King's daughter. The King does not like the man.

The ending of the story is the question that is the title of the story.

 

[rob_just_rob]

I would be interested to know if the producers of Let's Make a Deal deliberately created this game to take advantage of the paradox.

 

[B_NineInchCock_160IQ]

how is this discussion going on for 7 pages? It was explained in post one.

 

[dong20]

how is this discussion going on for 7 pages? It was explained in post one.

Because some people don't understand the explanation and others can't seem to explain it to them satisfactorily, in all probability.:smile:

 

[ManlyBanisters]

how is this discussion going on for 7 pages? It was explained in post one.

Because some people are refusing to believe the explanation makes sense.

 

[rob_just_rob]

Because some people are refusing to believe the explanation makes sense.

Bingo.