The scams of Statistics...

[dionisos]

The thing about the Monty Hall paradox is that pretty well no one ever changes doors, so there is no way of ever testing it accurately. You can do all the mathematical equations and computer simulations you want, but it's something we can never know for certain.

No, Monty hall is a simple mathematical problem, really, and every body that studied mathematic solve it without any problem.
The difficulty is to make people understand it, not the problem in itself.

See all these simulations, they give all the same result :
http://rosettacode.org/wiki/Monty_Hall_problem#Python

If i can’t convince by rational reasons, i will just play gambling game with people, maybe they will rethink the problem when they will have no money anymore :p

[vegetariantaxidermy]

The thing about the Monty Hall paradox is that pretty well no one ever changes doors, so there is no way of ever testing it accurately. You can do all the mathematical equations and computer simulations you want, but it's something we can never know for certain.

No, Monty hall is a simple mathematical problem, really, and every body that studied mathematic solve it without any problem.
The difficulty is to make people understand it, not the problem in itself.

See all these simulations, they give all the same result :
http://rosettacode.org/wiki/Monty_Hall_problem#Python

If i can’t convince by rational reasons, i will just play gambling game with people, maybe they will rethink the problem when they will have no money anymore :p

I understand it perfectly. I was just explaining why it can't be tested.

[dionisos]

The thing about the Monty Hall paradox is that pretty well no one ever changes doors, so there is no way of ever testing it accurately.

Ok, i don’t understand your argument, what do you mean by "no one ever changes doors" ?,

[vegetariantaxidermy]

The thing about the Monty Hall paradox is that pretty well no one ever changes doors, so there is no way of ever testing it accurately.

Ok, i don’t understand your argument, what do you mean by "no one ever changes doors" ?,

Just what I said. It's been tested, and no one changed doors. If as many people changed doors as didn't, then you could easily compare the percentage of winners from one group with the other. As no one changes doors then how can it be tested? I've seen it done on Mythbusters, and yes, the winners from the 'changing' side far outnumbered those from the side that kept with the original choice. But that's not the same thing. Both sides were aware of the paradox, whereas the show's contestants were not.

[Scott Mayers]

Monty Hall

Never believed in switching would do any difference. I can't make a 100% increased chance to win just by switching.

Just because the host allows me to switch, what difference woudl that do if I allowed myself to switch, before I made my final desition? ..NOTHING!!!

Imagine there are 99 goats and 1 car.

You choose one door, then i reveal to you 98 goats, do you think that by doing it, i increase your probability to get a car, from 1% to 50% ?

Then when you are in a game like it, just ask the host to first open the doors with a goat when he will show you if you win or lose, he will have to open at least 98 doors with a goat, and by doing it, your probability to get the car will increase to 50%. (i hope you see the nonsense)

You are speaking nonsense. You don't understand how the mechanics of the problem works.

I don't see this response here but am guessing it is from another thread? Anyways, the 100 door example as I heard it before to aid in following the original construction of this Monty Hall problem is as follows.

Originally pick a door. You have 1/100 chance to get the car. Then the host reveals 98 of the doors as duds. He always knows where the car is and so if he knows you picked it, he can arbitrarily pick from any of the 98/99 doors remaining. If you didn't pick the car initially, He must remove ALL doors except for the one with the car for the other option. Since in all cases from the start you only have a 1/100 chance, 99/100 of the time, the car will be in the door the host MUST reveal thus assuring without a doubt that it increases your odds to Switch as 99/100 to Win by switching.

This is perfectly rational for 100 doors. But note that the host gets to pick from 98/100 doors to eliminate. This is

100/100 - 1/100[Guest] - 1/100[Host Leaving door unveiled] for either case = 98/100.

But, given only three doors, the host only removes 1 door which is also equal to the quantity remaining.

3/3 - 1/3[Guest] - 1/3[Host Leaving door unveiled] = 1/3.

This 1/3 is the same chance that the original Guest had at the beginning and why in this special case using only 3 doors balances out the apparent logic using more than 3 doors.

In the 100 door sample, the Guest has 100/100 - 1/100[Host Leaving door unveiled] = 99/100[Chance to Win by Switching]

In the 3 door sample, the Guest has 3/3 - 1/3[Host Leaving door unveiled] = 2/3[Chance to Win by Switching]

But,

In the 100 door sample, the Guest has 100/100 - 99/100[Switch Odds to Win] = 1/100 chance to Lose by Staying

and
In the 3 door sample, the Guest has 3/3 - 2/3[Switch Odds to Win] = 1/3 chance to Lose by Staying.

Now Given you have 99/100[Chance to Win by Switching] and 1/100[chance to Win by Staying (= not Switching)] for the 100 doors, the ratio of Stays/Switches for Winning is:

(1/100)/(99/100) = 1/99 odds to Stay over Switching for a Win [So Switching does improve your odds here].

In contrast, you have 2/3[Chance to Win by Switching] and 1/3[Chance to Win by Staying] given the 3 doors. The ratio of Stays/Switches for Winning is:

(1/3)/(2/3) = 1/2 odds to Stay over Switching for a Win [So Switching does not improve your odds here] .

This definitively proves how and why given only three doors, that staying or switching is always only 1/2. The Stays to Switch odds are equal with three doors only. That is, it is also 1/2 odds to Switch over Staying.

Thus whether you stay or switch given only three doors are equivalent odds and so it doesn't matter what you choose.

[dionisos]

n the 3 door sample, the Guest has 3/3 - 1/3[Host Leaving door unveiled] = 2/3[Chance to Win by Switching]

If you have 2/3 chance to win by switching, and 1/3 chance to win by staying, then you have 2 time more chance to win by switching (2/3)/(1/3)=2.

and you have 2 time less chance to win by saying (1/3)/(2/3)=1/2

I agree with all of what you say, before the things about the "odds to Stay over Switching for a Win.", i am unsure of what you mean by it, but if you mean "how many time you have more change to win by staying that by switching", i agree it is 1/2. ((1/3)/(2/3))
But then the "odds to switch over staying for a win", is 2, not 1/2.((2/3)/(1/3))

We are in the good way, i put part of your answer in bold, because you put the result, i don’t understand why you go more far, you have 2/3 chance to win by switching, and 1/3 chance to win by staying, it is exactly what we want.

[Obvious Leo]

To understand this problem correctly you need to forget about the contestant and think about the host of the game. He knows which door the car is behind and he knows which door the contestant has chosen and THEN he opens a door to reveal a goat. The reason why very few contestants switch doors is that they forget about the prior knowledge of the host, who already knows he's going to reveal a goat.

[vegetariantaxidermy]

To understand this problem correctly you need to forget about the contestant and think about the host of the game. He knows which door the car is behind and he knows which door the contestant has chosen and THEN he opens a door to reveal a goat. The reason why very few contestants switch doors is that they forget about the prior knowledge of the host, who already knows he's going to reveal a goat.

They seem to be in full nerd mode, so they are not going to be able to get away from the maths.

[Scott Mayers]

n the 3 door sample, the Guest has 3/3 - 1/3[Host Leaving door unveiled] = 2/3[Chance to Win by Switching]

If you have 2/3 chance to win by switching, and 1/3 chance to win by staying, then you have 2 time more chance to win by switching (2/3)/(1/3)=2.

and you have 2 time less chance to win by saying (1/3)/(2/3)=1/2

I agree with all of what you say, before the things about the "odds to Stay over Switching for a Win.", i am unsure of what you mean by it, but if you mean "how many time you have more change to win by staying that by switching", i agree it is 1/2. ((1/3)/(2/3))
But then the "odds to switch over staying for a win", is 2, not 1/2.((2/3)/(1/3))

We are in the good way, i put part of your answer in bold, because you put the result, i don’t understand why you go more far, you have 2/3 chance to win by switching, and 1/3 chance to win by staying, it is exactly what we want.

I think it is best to be more specific and introduce this specifically with all contents contained in the fractions. I showed how the logic initially appears this way to get a nod from you which I did. But I have to be more specific to show you exactly what each of the denominator and numerator stand for in these cases. I believe the confusion you interpret is regarding how each of the terms using language can confuse what the numerators or denominators mean specifically. Another problem is when or where we must use and additive complement versus a multiplicative one. I'll post soon to demonstrate this with better clarity.

[Obvious Leo]

They seem to be in full nerd mode, so they are not going to be able to get away from the maths.

The point I was seeking to make is that the maths is childishly simple once you understand the nature of the problem you're trying to solve. The host of the game has already shortened the odds for the contestant by taking one of the goats out of the equation so his original guess maintains its original probability of 1/3 while the other unopened door has had its winning probability increased to 2/3.

Interestingly if birds are taught this puzzle they switch their guess immediately. The bloody obvious is better understood by birds, it seems.

[Obvious Leo]

VT. Maths geeks are famous for outsmarting themselves but this is not a maths puzzle, which is why very few people will switch their guess. This is a logic puzzle and humans are notoriously piss-poor at logic. The odds of being correct cannot possibly change just because a door has been opened so if you pick a door then you have a 1/3 chance of being right. The other two doors combined have a 2/3 chance of being right. This cannot change.

[Scott Mayers]

Before even beginning with the math, let us enumerate for all possibilities graphically here to at least show the error. Below is a diagram I created to show all possibilities absolutely. Each option represents the Host placing the car in the door and is shaded as green. Each row represents a possible event. For each place one picks where the car actually is leaves two possibilities in which the host is allowed to reveal what is empty (or a goat). Each original pick is also represented as the option one picks if they stay. The red printed the switches.

Please look over this and assure me if you either agree or not to these possibilities. If you disagree, state why.

Mont_Hall_1.jpg (135.18 KiB) Viewed 3205 times

If you agree to these as all the possible events, count up the events where you switch AND win. If you get 6/12, this is 1/2. Likewise, if you look at all the switches in which you lose, count them up too. Do you see 6/12 = 1/2 here too? How about the Stays? 1/2 too, correct? This demonstrates graphically that whether you switch OR stay, the odds are 1/2 no matter what you could figure prior to using probability. If the math using probability does not match, then this must assure you that it would have to be the math you use to which is causing the trouble. Is this correct?

[Obvious Leo]

Scott. Forget your silly charts and try using using your head. There are three doors behind one of which is a car. You are invited to guess a door and you do so. There is a one in three chance that you have guessed the correct door. There is a two in three chance that one of the other doors is the correct door. That's it. End of story. QED. Problem solved.

Change your fucking guess.

[Scott Mayers]

Scott. Forget your silly charts and try using using your head. There are three doors behind one of which is a car. You are invited to guess a door and you do so. There is a one in three chance that you have guessed the correct door. There is a two in three chance that one of the other doors is the correct door. That's it. End of story. QED. Problem solved.

Change your fucking guess.

Read the damn thing before insulting me here. I already know what the problem is but require a step by step approach. Look above and you'll see precisely ALL possibilities laid out with clarity. If you don't choose to read, then don't comment.

The error is in not recognizing that when you happen to pick the car, the host has not one, but two events that have to be accounted for that are both null. What tricks people is that they assume that they mean the same identical things which is NOT the case. In one unique event after choosing the one with the car, the knowledge of the host allows him to pick either left over doors uniquely in time. So in option one above where the host places the car in door number one AND the Guest picks that door, the host may in one event pick door number two and leave you with the third to switch to, or he could pick door number three leaving you door number two to opt to switch to. These are distinct and separate events.

In the other two cases, there is no other option for the host to actually select from but the one door left without the car. This is why you cannot find any alternative event versions for those. So in total, as illustrated, for each option above, as is with all of them collectively, the events in which the Guest opts to choose to switch are 1/2 and identical to the odds to stay, as 1/2 too.

If you disagree, what do you think is actually wrong with the enumerated events above?

[Obvious Leo]

If you disagree, what do you think is actually wrong with the enumerated events above?

What is wrong with the enumerated events above is that they have nothing to do with the fucking problem. you are just plain WRONG.