[BrickJames]

Good vid IAG

[IAG]

Good vid IAG

TY......I think it helps to understand better. Before I became Monty Hall enlightened...lol . I had the same position as Smoke initially. But once you see it in a different way, you will get it. I think the last post I made helps explain it even better than the video , though I haven't watched the video in years.

[JohnGalt2341]

I think the reason why so many people get confused about the Monty Hall problem is because the show only used 3 doors. If the show used 100 or 1000 doors it would become completely obvious to most people that they should change their mind every time. Imagine if Monty Hall showed you a goat behind 998 doors(he could do this every time regardless of your choice) and then told you that of the 2 doors left 1 is a goat and 1 is a car. If you still can't figure out why you should now change your mind you should probably quit gambling unless you enjoy losing.

I just noticed IAG said the same thing in post #35.

[5918mike]

Always switch, if you think otherwise and are trying to explain why, shut the F up

[Sam Odom]

If you still can't figure out why you should now change your mind you should probably quit gambling unless you enjoy losing.

[Smoke]

TY......I think it helps to understand better. Before I became Monty Hall enlightened...lol . I had the same position as Smoke initially. But once you see it in a different way, you will get it. I think the last post I made helps explain it even better than the video , though I haven't watched the video in years.

Let this stand as a testament to how the popularly and improperly taught truth about the Monty Hall Problem is false, for a variety of reasons.


To set an official designation of the rules:
You are a contestant on a gameshow. The host gives you the option of selecting one door from an available three. Behind one door is a new car, behind another is a goat, and behind the remaining door is also a goat. You select one at random. The host now reveals one of the doors you have not selected to be a Goat Door. What is the probability of selecting the Car Door?

The disturbingly frequent misinformed explanation is that switching gives you the greater advantage.

Here is how exactly that line of reasoning is mortally flawed.

The greatest of logical errors is the perception of switching, when in fact you are not switching anything. The first choice is not a genuine choice because the revelation that one of the three doors you are permitted to pick from will be removed. You are in effect simply playing the game. Once you’ve made your misperceived selection, you are given the choice to choose between two options. This is your final decision. The original selection when there were three unknowns does not influence the final decision, because you would have had to choose between two new options regardless of whether you had picked any of the three.

The original choice was one out of three (choose the first door, choose the second door, choose the third door), but the second choice is one out of two (choose one door, choose the other door). The very question itself posed, “do you want to swap (option 1), or do you want to stay (option 2)” in itself defines the new set — the new probability, of 2 choices. Whether or not there had been 2.38 fillion dillion doors is irrespective of whether or not the new question is “stay or swap” at present.

The option between “switch or not switch” are the new options, and now has nothing to do with doors or original previous options.

The oft-cited, poorly-reasoned proof (which Wikipedia does here, is that:

“1. If you pick the Car Door and the host reveals a Goat Door, switching will always lose.
2. If you pick Goat Door A and the host reveals a Goat Door, switching will always win.
3. If you Pick Goat Door B and the host reveals a Goat Door, switching will always win.
Therefore, if you switch, you have a 2/3 chance of winning.This is very flawed logic, akin to suspecting that, if 90% of car accidents in Louisiana involve people with a surname beginning with M, that people with a surname beginning with M should avoid driving in Louisiana.
The fact is, is that when you are asked to select a door in the beginning, you have made a selection that only applies to the original terms. The flawed reasoning operates under the circumstance of the original question imposed upon the second question — but the new question is what sets the standard for calculation, not the original. The reasoning fails to include the fact that the announcer will not open the Car Door AND the announcer will not open the door you have selected, which creates an actual probability of 50/50, because the 3rd door he reveals was never an actual option. Any version of the Monty Hall Problem that includes BOTH of these stipulations AND declares there to be a 66% chance of win by switching bears goat-like intelligence.

Rephrased, the most glaring problem is that there is actually only one chance, 100%, that the Car door is the Car Door. There is a 0% chance that Goat Door A is a Car Door, and a 0% chance that Goat Door B is a Car Door. Of the available options, three doors, you have a chance of 100 divided by 3 options. When one option is eliminated, there are now only two options. That makes the probability of 100% that the Car door is the Car door, and 0% that the Goat Door is the Car door. Now that you only have two options, the probability is 50%.

Even if there were 100 doors, as if often used to rationalize that switching makes better sense, there would STILL only be a 100% probability that the Car Door is the Car Door, and 0% for each of the 99 other Goat Doors. By the purported proof offered by Wikipedia, switching would therefore offer you a 99% chance of getting the door correct by switching, when in fact you are actually only choosing between one door or one other. Regardless of how many past choices you had before, the current question between “switch or swap” (2 options) is what sets the stakes.

Take again for instance if you were asked to select 2 of the 3 doors:

1. You pick the Car Door and a Goat Door, a Goat Door is revealed to be the third. You have a choice between your original selections, a Car or a Goat — a 50/50 shot.
2. You pick a Goat Door and the Car Door, and a Goat Door is revealed as one of YOUR doors. You have a choice between your remaining unopened door or the unselected door — a 50/50 shot.
3. You pick a Goat Door and the other Goat Door. The announcer will unfailingly reveal one of YOUR doors as a goat door. You still have a choice between your remaining unopened door or the unselected door — a 50/50 shot.

Take again for instance if you were to select one door and the host were to reveal that YOUR door is the Goat Door. Your probability is still a 50/50 chance of your new choice correct. The fact here is that you must make a new choice, which is identically probable to the dilemma of whether the host had NOT opened the door you first selected — because the choices were between STAY or SWAP.

Take yet again for instance if you were to select one door and the host were to reveal that one of the doors you had not selected to be a Goat Door. Now, you are turned around and blindfolded, and the prizes behind the doors are rotated behind the two available doors, so that you’re unsure whether the door you had selected is actually still corresponds to your original Car Door guess. The probability does not change, because there are still only two options. The chances are still 100% that one door will be correct and 0% that the other door will be incorrect, resulting in a 50% chance.

Take yet again a visual illustration of your choices: To pick between three doors is like unto throwing a dart at a rotating (a plane rotation) circle that has three equal sections defined as section 1, 2, and 3. It is revealed that the corresponding door the dart hits is your first selection, and following is revealed that one of the unhit doors is a Goat Door. The new circumstance is like unto throwing the dart at a new circle divided half and half into “stay” and “swap” respectively. You are not throwing a dart at a plane-rotation circle with all three options, because you are clearly not going to select the already-opened goat door. The new selection is the identical odds as whether you had not even selected one of them originally.There is no possible way to discern whether the door you have already selected is the Car Door, and LIKEWISE is there no possible way to discern whether the unselected door is the Car Door. However, it IS possible to know whether the third door is a Car Door, because it has already been revealed as a Goat Door. The options are now only between one unknown and another unknown — 50/50.

The reason the purportedly true answer gains acceptance is because it stands to reason only in one case, and not the rest. This is akin to the assertion that the scientific method is the only way to determine truth. In order to determine that very statement, you would have to establish a fact fact without using the scientific method. It’s the same “How do you know the bible is true? God says so! Where? In the bible!” (even though the bible does not say that) argument.

Wikipedia reasons, This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant’s column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, “If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch” (vos Savant, 2006).

This is also incorrect, because the actual probability of one door being 100% the Car Door and the other door having a 0% chance of being the Car Door remains true, regardless of whether the host knows or not. To Switch (option one) or not to Switch (option two) is what is being asked. You are being asked to choose between two actions, the original probability now being completely irrelevant.

Wikipedia also cites a graph showing three doors, where the prizes are actually revealed. It lists the Car Door under a 33% bracket, and brackets the two remaining doors under a collective 66% deliniation (a Venn diagram). The problem here is twofold, at a minimum. Firstly, when you divide 66% by two, you do not have 66% remaining. The two doors do not represent a 66% chance each that averages ([66+66]/2) to 66%, because each door is purportedly a 1-in-3 chance each, or 33% each. Secondly, there is not a 33% chance that the Car Door is the Car Door — it is 100% the Car Door. The other two options are 0% and 0% respectively. If you were to eliminate one of the 0% probability doors, a Goat Door, you’re left with either the 100% option or the 0% option, making it a 50% chance.

The diagram found here uses inverted teacups concealing a diamond. The diagram makes the assertion that switching your original choice when one is revealed. However, the actual choice being made is between one cup or one other cup, a 50/50 chance. The original formula does not apply since one cup has been eliminated. It’s a false positive.

Playing the game yourself and compiling the results is completely arbirtrary because at what point do you cease the experiment? If you were to select the Car Door in your original 33 selections without switching, could you simply just quit, sufficed that staying with your original guess is the 100% sure strategy? It’s the same question of asking whether a coin flipped 50 times resulting in 50 heads will next flip tails

[IAG]

Smoke who wrote that? I haven't read it yet but need to know before I spend the time.

[BrickJames]

Smoke because you originally had 33% chance of picking the right door once they remove them the doors you have a 66% chance if you switch do you get this?

[Sam Odom]

the earth is the center of the universe and all objects revolve around it... and may be flat also

[Foxx]

[Smoke]

Smoke who wrote that? I haven't read it yet but need to know before I spend the time.

not sure. it didn't say. but he is correct.

[Smoke]

Smoke because you originally had 33% chance of picking the right door once they remove them the doors you have a 66% chance if you switch do you get this?

it's all explained in the article I pasted. read it

[IAG]

It's some blogger Ablestmage? ..not sure of qualifications, but my feeling concur with those of "Bill" the comments following his article. I'm going to go with the MIT peeps and most anybody versed in probability.

• Bill says:
  August 9, 2013 at 10:07 pm
  

  It is barely believable to me that people are still debating this issue when it was conclusively settled so many years ago.
  
  Reply
  
  
  
  • ablestmage says:
    August 11, 2013 at 9:12 am
    

    The matter is conclusively settled by my arguments.
    
    Reply
    
    
    
    • Marley52 says:
      August 11, 2013 at 9:57 am
      

      You don’t have any arguments. The only issue that you have conclusively settled is that you don’t understand maths in general and probability theory in particular.
      
      

[Smoke]

Eliminating one of the doors doesn’t change the probability of the original choice, correct. However, you are given a new choice between “stay” and “switch” therefore the previous choice is a completely different probability altogether, and can’t be a variable influencing the new choice

[IAG]

I think the reason why so many people get confused about the Monty Hall problem is because the show only used 3 doors. If the show used 100 or 1000 doors it would become completely obvious to most people that they should change their mind every time. Imagine if Monty Hall showed you a goat behind 998 doors(he could do this every time regardless of your choice) and then told you that of the 2 doors left 1 is a goat and 1 is a car. If you still can't figure out why you should now change your mind you should probably quit gambling unless you enjoy losing.

I just noticed IAG said the same thing in post #35.

+1 Nothing makes it more clear to me than looking at it that way.

[Smoke]

iag...

Three men require lodging, and at the chosen hotel, management asserts the cost for one room’s rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?

[IAG]

Eliminating one of the doors doesn’t change the probability of the original choice, correct. However, you are given a new choice between “stay” and “switch” therefore the previous choice is a completely different probability altogether, and can’t be a variable influencing the new choice

Smoke. Did you look at the 100 door example?

[unde0087]

exactly Smoke, in reality if one was to chose the right door initially then the model falls apart because the math would tell a person that changing your original answer gives you a higher probability yet in reality the probability is 0

[Smoke]

Smoke. Did you look at the 100 door example?

the previous elimination round bears no effect on the most recent choice, to stay or switch. The math involved for Switchmen, like those fooled by the lodging problem, requires the now-irrelevant prior circumstances to remain within the body of evidence, when in reality both prior matters are newly and entirely irrelevant.

[IAG]

iag...

Three men require lodging, and at the chosen hotel, management asserts the cost for one room’s rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?

Hotel has $25 3x$1 to each of three men and $2 to bellhop =$30

[IAG]

exactly Smoke, in reality if one was to chose the right door initially then the model falls apart because the math would tell a person that changing your original answer gives you a higher probability yet in reality the probability is 0

Well of course if if you picked the right one, then switching would turn out to have been the unwise choice...but the question is is it more advantageous to switch in general. If you were picking out of 100 doors and you happen to pick the 1 car out of the 100 doors, then switching would have cost you. but we're going with probability. Of course it won't work every time.

[Smoke]

Hotel has $25 3x$1 to each of three men and $2 to bellhop =$30

The diversion within the question is that the $2 tip is added, rather than (properly) subtracted from the $27 total, to make $25. The question attempts to suggest that the original $30 total is still relevant, rather than the new total of $25 that is the only relevant total for application

[IAG]

the previous elimination round bears no effect on the most recent choice, to stay or switch. The math involved for Switchmen, like those fooled by the lodging problem, requires the now-irrelevant prior circumstances to remain within the body of evidence, when in reality both prior matters are newly and entirely irrelevant.

WADR Smokers, you just keep cutting bits and pieces and pasting stuff from that same blogger's page.
Again,I have to go with MIT and majority of the Mensa type community over a random blogger, but you are entitled to your misguided opinion. . I used to think the same as you so I understand. I do have a feeling someday down the line you will have your AHA moment." I hope so.

[IAG]

The diversion within the question is that the $2 tip is added, rather than (properly) subtracted from the $27 total, to make $25. The question attempts to suggest that the original $30 total is still relevant, rather than the new total of $25 that is the only relevant total for application

Yeah...so I was right...lol. To be fair, I had seen this riddle before.

[Smoke]

WADR Smokers, you just keep cutting bits and pieces and pasting stuff from that same blogger's page.
Again,I have to go with MIT and majority of the Mensa type community over a random blogger, but you are entitled to your misguided opinion. . I used to think the same as you so I understand. I do have a feeling someday down the line you will have your AHA moment." I hope so.

IAG but you still cantdisprove what he is saying.

Is a half-cup of water in a cup that could hold 1 cup, half empty, or half full? Math doesn’t lie, after all. Statisticians, however

[Foxx]

I like this explanation:

1) Consider this: in what cases will switching doors lead to a goat? If one chooses a goat door originally, then switching doors can only lead to a car, as there is only one other goat door and that has been opened. Therefore, the other door must reveal the car. Conversely, if one selects the car door originally, switching doors must lead a goat.

One can therefore see that the only time when switching doors can be wrong is when one selects the car door originally. If one selects a goat door originally, switching must lead to the car. The probability of selecting the car door originally is 1/3. The probability of selecting a goat door originally is 2/3. Therefore the probability of winning the car by changing doors is 2/3. And the probability of losing the car by changing doors is 1/3. Therefore, the rational decision, as Ms. Savant stated, is to change doors in order to double the probability of winning the car.to a goat? If one chooses a goat door originally, then switching doors can only lead to a car, as there is only one other goat door and that has been opened. Therefore, the other door must reveal the car. Conversely, if one selects the car door originally, switching doors must lead a goat.
One can therefore see that the only time when switching doors can be wrong is when one selects the car door originally. If one selects a goat door originally, switching must lead to the car. The probability of selecting the car door originally is 1/3. The probability of selecting a goat door originally is 2/3. Therefore the probability of winning the car by changing doors is 2/3. And the probability of losing the car by changing doors is 1/3. Therefore, the rational decision, as Ms. Savant stated, is to change doors in order to double the probability of winning the car.

1) Consider this: in what cases will switching doors lead to a goat? If one chooses a goat door originally, then switching doors can only lead to a car, as there is only one other goat door and that has been opened. Therefore, the other door must reveal the car. Conversely, if one selects the car door originally, switching doors must lead a goat.
One can therefore see that the only time when switching doors can be wrong is when one selects the car door originally. If one selects a goat door originally, switching must lead to the car. The probability of selecting the car door originally is 1/3. The probability of selecting a goat door originally is 2/3. Therefore the probability of winning the car by changing doors is 2/3. And the probability of losing the car by changing doors is 1/3. Therefore, the rational decision, as Ms. Savant stated, is to change doors in order to double the probability of winning the car.

[Smoke]

I will leave it at the lager but you know he is right

[IAG]

IAG but you still cantdisprove what he is saying.

Is a half-cup of water in a cup that could hold 1 cup, half empty, or half full? Math doesn’t lie, after all. Statisticians, however

Ok now you are really losing me as to your intended point...Optimism/Pessimism analogy? Of course it has been disproved. There is a 10 page thread at EOG....go read that maybe...? I have no idea how you don't see it. I think you have just dug in your heels....this debate has been over for decades.

[5918mike]

As I understand it, your original choice is 1/3 you picked vs the 2/3 you did not pick. Out of the 2/3 you did not pick there is at least one goat, you already know this without the reveal so the reveal changes nothing, they are only showing you what you already know.

For the naysayers, what if there was no reveal, would you switch and take the 2 doors vs the one door? It's the same thing, choosing blind or revealing one goat it's still a choice between 1/3 and 2/3.

[IAG]

I will leave it at the lager but you know he is right

No i don't. I do not even know who it is...and either do you! I think you must be just trolling now my friend. I'm baffled. But I've been there and u are my pal..so let's just agree to disagree. After the great apostrophe debate. I don't have much left.

[RudyRuetigger]

Sammy are you going to teach us the square root of 4 tomorrow??

fukkin unreal

[IAG]

Ok one more. I like this explanation.

[UncleDoc]

Are you the janitor at Harvard like Good Will Hunting?

Ha that is a personal favorite of mine due to the similarities in my own life. Janitor? No. Harvard? No. Harvard of the south? Yes.

[5918mike]

iag...

Three men require lodging, and at the chosen hotel, management asserts the cost for one room’s rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?

You subtract the $2. They each paid $9 total, for a total of $27. $25 went to the hotel for the room and $2 went to the bellhop. So out of the original $30 = $25 hotel, $2 bellhop, $1 each they still have.

[Sam Odom]

Sammy are you going to teach us the square root of 4 tomorrow??

no

the Prime Numbers