NationStates Jolt Archive

The Monty Hall Problem is as follows:

You are a contestant on a game show hosted by Monty Hall.
There are 3 doors before you and you are to pick one.
You know for certain there is a prize behind one door and one door only.
You pick door X.
After you pick door X, door Z is opened to reveal nothing. You are then given an option to stay with door X or switch to door Y. At this point, what do you do?

The following are given for the purpose of this problem:
-You are not psychic.
-You had know way of knowing which door the prize was behind.
-The prize was in place behind one door before you picked door X and it hasn't been moved since.
-Every single time you pick any door at first, you are guaranteed to be shown an empty door and given an option to switch to the remaining closed door.
-You want to win the prize.
-You win the prize if and only if your it lies behind the final door you choose.

Again, the question is what should you do?
A.) Stay with the door you picked (X).
B.) Switch to the other door that's still open (Y). or
C.) It doesn't matter. (50/50)

you have better odds if you switch.

When you first picked a door, you had a one third chance of being right. This means that you had a two thirds chance of being wrong.

As you have been shown that one of the other doors is empty, then there is a two thirds chance (that which there was before of you being wrong) that the prize is behind the third door. (Not the one you originally chose, not the open one) compared to the one third chance of it being behind the door you have already selected.

Therefore switch.

This should explain everything:
http://i5.photobucket.com/albums/y195/travisemergency/monty-hall.gif

Sorry, but that's complete bollocks. For 3, the odds of getting the car are 1 in 3, for 2, the odds of getting the car are 1 in 2. You can't transfer/compound odds in that way.

IF you have 2 windows and are told that the prize is behind one of them, there is a 50% chance that each has the prize. If another window is added, where you are told the prize is definitely not behind it, the odds remain 50/50. Therefore, by removing this window, so that there are again 2 windows, the odds remain 50/50. They do not change to 1/3 and 2/3.

Sorry, but that's complete bollocks. For 3, the odds of getting the car are 1 in 3, for 2, the odds of getting the car are 1 in 2. You can't transfer/compound odds in that way.

IF you have 2 windows and are told that the prize is behind one of them, there is a 50% chance that each has the prize. If another window is added, where you are told the prize is definitely not behind it, the odds remain 50/50. Therefore, by removing this window, so that there are again 2 windows, the odds remain 50/50. They do not change to 1/3 and 2/3.Are you saying that the odds do not change whether you stay or switch?

Sorry, but that's complete bollocks. For 3, the odds of getting the car are 1 in 3, for 2, the odds of getting the car are 1 in 2. You can't transfer/compound odds in that way.

IF you have 2 windows and are told that the prize is behind one of them, there is a 50% chance that each has the prize. If another window is added, where you are told the prize is definitely not behind it, the odds remain 50/50. Therefore, by removing this window, so that there are again 2 windows, the odds remain 50/50. They do not change to 1/3 and 2/3.
the factor you arent considering is that monty will NEVER show you the box that has the good thing in it. so the odds of the 2nd "switch" box are compounded. that so the 2nd switch box carries the odds of both switch boxes.

Ok, none of that makes any sense. If I go to door X and they show that door Z contains nothing, then my chances have gone from 1 in 3 to 1 in 2. I still have a 50/50 chance, no matter what door I pick. Because even if I picked the right door, monty would still show me one of the wrong doors. It makes absolutely no difference if I switch or if I stay, because I have a 50/50 chance either way.

Imagine there are a thousand doors, and you are told the car is behind one. You pick door seven. All the doors except seven (which you chose) and fifty-three are taken away. Do you switch? Yeah you do. Unless you want a 1/1000 chance of winning. It's the same with the Monty Hall problem, except 1/1000 gets knocked down to 1/3.

Um, no... because It's two different choices, the one I made before, which was 1/1000, and the one I made after, which was 1/2.

Um, no... because It's two different choices, the one I made before, which was 1/1000, and the one I made after, which was 1/2.

Um, no... because It's two different choices, the one I made before, which was 1/1000, and the one I made after, which was 1/2.

But it's still the same 1000 doors. You picked a door, and the odds were strongly in the favour of it being behind one of the other doors. They've essentially told you which door it's behind.


As a side note, this appeared in one of Nick Hornby's books (the curious incident of the dog in the night time) and he explained it properly there. If you test the origional Monty Hall problem properly, you'll find out that you have more chance of winning if you swap.

Imagine there are a thousand doors, and you are told the car is behind one. You pick door seven. All the doors except seven (which you chose) and fifty-three are taken away. Do you switch? Yeah you do. Unless you want a 1/1000 chance of winning. It's the same with the Monty Hall problem, except 1/1000 gets knocked down to 1/3.

My God, I hope you're joking. In your scenario, once the total door choice has been reduced, your chances are now 1 in 54 - regardless of whether or not you switch doors.

Ok... I think I'm getting it now...
So the chances that the door you picked has nothing behind it is 999/1000.
so 999 times out of a thousand, you will have already picked the wrong door.

My God, I hope you're joking. In your scenario, once the total door choice has been reduced, your chances are now 1 in 54 - regardless of whether or not you switch doors.

I don't think you understood it.

Sorry, but that's complete bollocks. For 3, the odds of getting the car are 1 in 3, for 2, the odds of getting the car are 1 in 2. You can't transfer/compound odds in that way.

IF you have 2 windows and are told that the prize is behind one of them, there is a 50% chance that each has the prize. If another window is added, where you are told the prize is definitely not behind it, the odds remain 50/50. Therefore, by removing this window, so that there are again 2 windows, the odds remain 50/50. They do not change to 1/3 and 2/3.
We did this in AI class last semester. We had a computer run the simulatin about 100K times with switching and without. We came up with nearly an 80% win rate when we switched doors. The key word is not problem but paradox.

Ok... I think I'm getting it now...
So the chances that the door you picked has nothing behind it is 999/1000.
so 999 times out of a thousand, you will have already picked the wrong door.

Right. There is a 999/1000 chance that it is behind any of the other doors. By taking away 998 of these doors, they are basically telling you which door it is behind, unless you were somehow lucky (or unlucky in this case) enough to pick the right door the first time. In the problem mentioned in the first post, there are only 3 doors so you have only already picked the wrong door 2/3 times. I'm not very good at explaining things, but I think I have.

I don't think you understood it.

Fine. Given the rules of the game, explain to me how in your scenario the chance for any door can be anything other than 1 in 54.

Pertinent rules from initial post:

-The prize was in place behind one door before you picked door X and it hasn't been moved since.
-Every single time you pick any door at first, you are guaranteed to be shown an empty door and given an option to switch to the remaining closed door.

3 doors, A,B,C
Prize is behind A

Pick A
Switch-Lose
Stay-Win

Pick B
Switch-Win
Stay-Lose

Pick C
Switch-Win
Stay-Lose

You win 2/3 if you switch, 1/3 if you stay

Fine. Given the rules of the game, explain to me how in your scenario the chance for any door can be anything other than 1 in 54.

Pertinent rules from initial post:

-The prize was in place behind one door before you picked door X and it hasn't been moved since.


-Every single time you pick any door at first, you are guaranteed to be shown an empty door and given an option to switch to the remaining closed door.

the first point is right, the second point isn't. You aren't shown an empty door, 998 of the wrong doors are simply taken away.

The odds are the same for switching and staying. Your just not looking at all the posibilities.

Lets say that the prize is still behind door A. You have two choices to make.

if you first picked door A:
they can open door B. the correct choice is to stay with A.
they can open door C. the correct choice is to stay with A.

if you first picked door B:
they can open door C. the correct choice is to switch to A.

if you first picked door C:
they can open door B. the correct choice is to switch to A.

50% say stay. 50% say switch.



----
A similar problem would be to find the probability of having the "ace of spades" on the top of a deck of 52 cards. But before telling me, I remove a card from the deck without showing you.

I was wrong. If you go into the contest with the strategy of switching - you have a 2/3 chance of winning. With your intitial choice, you only have a 1/3 chance of winning - by switching you effectively get to choose two doors.

It's actually a "force".

They don't always eliminate your first choice. It depends upon how they feel about you.

I was wrong. If you go into the contest with the strategy of switching - you have a 2/3 chance of winning. With your intitial choice, you only have a 1/3 chance of winning - by switching you effectively get to choose two doors.

*nods* You've just been through what everyone who understands the problem went through before they got it.

Lets say that the prize is still behind door A. You have two choices to make.

if you first picked door A:
they can open door B. the correct choice is to stay with A. (x 50%)
they can open door C. the correct choice is to stay with A. (x 50%)

if you first picked door B:
they can open door C. the correct choice is to switch to A. (x 100%)

if you first picked door C:
they can open door B. the correct choice is to switch to A. (x 100%)

1 say stay. 2 say switch.

We did this in AI class last semester. We had a computer run the simulatin about 100K times with switching and without. We came up with nearly an 80% win rate when we switched doors. The key word is not problem but paradox.
80%? It sounds like you were using 5 doors instead of 3. Either that, or your random number generator was very non-random.

It's a force people, you get what they want you to get. There are no probablities. :rolleyes:

Sorry, but that's complete bollocks. For 3, the odds of getting the car are 1 in 3, for 2, the odds of getting the car are 1 in 2. You can't transfer/compound odds in that way.

IF you have 2 windows and are told that the prize is behind one of them, there is a 50% chance that each has the prize. If another window is added, where you are told the prize is definitely not behind it, the odds remain 50/50. Therefore, by removing this window, so that there are again 2 windows, the odds remain 50/50. They do not change to 1/3 and 2/3.But that's not what is happening here. In this problem, the door that is opened depends on which door you pick. Let say:

Door 1 = Nothing
Door 2 = Prize
Door 3 = Nothing


If you pick door 1, door 3 will be opened.
If you pick door 3, door 1 will be opened.
If you pick door 2, either 1 or 3 is opened.

In order for the odds to be 50/50, you'd need to know which door is going to be opened first. You don't. In your scenario you would know.


To begin with, you see 3 doors.
The probability that Door 1 has the prize is 1/3. Doors 2 and 3 are also each 1/3.
If you picked Door 1, the probabilty that you picked right is 1/3.
The probability that the prize is behind door 2 or 3 is 2/3.
When door 3 is opened, the probability that you picked right is still 1/3, and the probability that the prize is behind door 2 or 3 is still 2/3 except you can eliminate door 3 because you can see that there's no prize behind it.

It's a force people, you get what they want you to get. There are no probablities. :rolleyes:
Oh yeah? Well check this out:
The following are given for the purpose of this problem:
...
-Every single time you pick any door at first, you are guaranteed to be shown an empty door and given an option to switch to the remaining closed door.:rolleyes: :rolleyes:

Perhaps I could've phrased it better. I'm saying that, to properly work this problem, you have to 'pretend' that you will definitely be shown a loser door after your initial pick.

Oh yeah? Well check this out:
:rolleyes: :rolleyes:

Perhaps I could've phrased it better. I'm saying that, to properly work this problem, you have to 'pretend' that you will definitely be shown a loser door after your initial pick.

How can they guarantee that your first pick will be a loser door then?

How can they guarantee that your first pick will be a loser door then?

They don't. You have a one third chance of picking the winning door. If you do, they simply one one of the other two at random, showing it to be empty. This is the circumstance under which you lose if you switch, but it only applies 33.3% ofthe time. 66.6% of the time you win if you switch (All decimals recurring for the pedants) as you did not pick the winning door to start with.

They don't. You have a one third chance of picking the winning door. If you do, they simply one one of the other two at random, showing it to be empty. This is the circumstance under which you lose if you switch, but it only applies 33.3% ofthe time. 66.6% of the time you win if you switch (All decimals recurring for the pedants) as you did not pick the winning door to start with.

Yeah, i know that is the point. But that is not what they said. Now, do you get shown what is behind the door that you first pick, or is it simply thrown away.


Actually, I still always get a 1/3 answer, for whatever strategy. That's prolly right, but I can't be sure. (Damn red stripe).

Yeah, i know that is the point. But that is not what they said. Now, do you get shown what is behind the door that you first pick, or is it simply thrown away.


Actually, I still always get a 1/3 answer, for whatever strategy. That's prolly right, but I can't be sure. (Damn red stripe).

No, you do not get shown what is behind the door you first pick, unless you stick with it. But you will know anyway.

Three doors. A, B, C. Pick one (B) for sake of argument.
They open one (A), showing it to be empty.
You choose to swap or to stick. Swap you discover what is behind (C). If it is a prize, then (B) was empty. If it was empty then (B) was a prize. (The game is honest in that there is a prize to be won). See you know, at the end, what was behind all ther doors.

Door............A..........B........C
Chance.......1/3.......1/3......1/3
Select..........x
Accumulate..1/3.......-->2/3<--
Eliminate.......A.........x.........C
Resolve........1/3.......0........2/3

Now decide. Do you stick or switch? I switch.

Okay, this is where I get hung up.

There is only one prize, right? So it is purely fifty/fifty whatever strategy you persue.

Okay, this is where I get hung up.

There is only one prize, right? So it is purely fifty/fifty whatever strategy you persue.

This isn't true.
When you start out, you have a 1/3 chance; if you switch, you have a 50-50 chance.

100 posts, yay!

This isn't true.
When you start out, you have a 1/3 chance; if you switch, you have a 50-50 chance.

100 posts, yay!

No, because they eliminate a losing choice at your first pick. So your only option, whether you have chosen a winning door or not is, either to stay, or to switch. As they do not remove the winning door, there are only two choices left to you: Winning door, or losing door. And what you pick is determined by whether you stay or switch. So essentially you have two options. Stay or switch, one of which is the winning choice. Thus it is 50/50. (Ie. your choice can be assigned the probablity of 1, since it happened, and there are two possible outcomes, right or wrong).

Okay, this is where I get hung up.

There is only one prize, right? So it is purely fifty/fifty whatever strategy you persue.

Nope. 1/3 if you stick. 2/3 if you change.

Actually, I see what's happening... The problem is in the way it is written, it completely disregards the chance that the "removed" box has any chance of being the one with the prize... So the question is not asked in a realistic way. As written, yes, by switching, you do acutally get the effect of having picked 2 boxes - Since you are pre-selecting the removed box as the loser of the 2 unpicked options.

If this was real, though, you don't get that effect. That removed box has a 33% chance of being the winner, which is a lose scenario whether you switch or not - As written, you are atificially adding that to the win % for switching.

If you re-run tests of this, and allow for the chance for the removed box to be a winner (a "lose" scenario), I bet you end up with a 50-50 win rate whether you switch or not. So, in the real world, it doesn't matter if you switch.

If this was real, though, you don't get that effect. That removed box has a 33% chance of being the winner, which is a lose scenario whether you switch or not - As written, you are atificially adding that to the win % for switching.


No, because the quizmaster knows where the prize is, and whatever door you have chosen, he is able to remove a door without a prize.

Nope. 1/3 if you stick. 2/3 if you change.

Don't see it. I must be misunderstanding the rules.

If the quiz master always removes a losing door, then you are left with a win/no win choice. It's fifty/fifty.

No, because the quizmaster knows where the prize is, and whatever door you have chosen, he is able to remove a door without a prize.

Then we are not talking about a truly random scenario - If the rules are that the quizmaster is removing a "loser" option, then I agree you are better off switching.

But then again, I don't Monty Haul ever removed a losing option on LMAD.

Okay, this is where I get hung up.

There is only one prize, right? So it is purely fifty/fifty whatever strategy you persue.

No it isn't. I know it seems like it should be, but there is the act of reducing two options to one, that adds the odds for those two options together and places them on that one.

A, B and C again.

The prize is either at A or at B or at C. OK so far. This means that the chance of the prize being at any given location is 1/3. Yes?

Now you select any one location. What is your chance of being right? 1/3: true.

Now I open another door. Does this have any effect on the chance that the prize is behind the door that you selected? This is where people usually go wrong. NO, it does not. The chance that the prize is behind the door you selected is still 1/3. When you selected the door, you automatically left at least one door without the prize behind it unselected. There is a 2/3rds chance that you left one door without the prize and a 1/3rd chance that you left two such doors. Opening such a door does not change the odds of it existing. What it does do is change the odds of the prize being behind the other unopened door.

It is not a matter of if you select A they open B otherwise they open A. It is that they open the the door without the prize that you did not select.

Door........A..........B........C
Prize.............................X.........For actions: s = Select, o = Open
action......s..........o...................stick = lose, switch = win
action......o..........s...................stick = lose, switch = win
action......o...................s..........stick = win, switch = lose
result.......................................stick = 1, switch = 2
The fact that they could have opened door b in the last example instead of door a is irrelevant as what determines the case is your initial selection, not which of the two doors they choose.
Note, they never open the door with the prize behind it. This is what changes the odds.

You people need to bookmark Wolfram Maths World for easy future reference to these sorts of maths problems:
http://mathworld.wolfram.com/MontyHallProblem.html

There's lots of interesting things there, like did you know the numbers on the roulette wheel add up to 666 (the number of the beast)?
Also the squares of the first seven primes, when added equal 666.
Also, phi (the golden ratio) can be expressed as:
-[sin(666°) + cos(6*6*6°)]

Don't see it. I must be misunderstanding the rules.

If the quiz master always removes a losing door, then you are left with a win/no win choice. It's fifty/fifty.

It's tricky because you have to let go of some basic statistics because the removed door isn't selected at random.

If you pick door one - the removed door predefined as a loser of 2 & 3. So it either 2 OR 3 was the winning door it becomes the remaining door.

No it isn't. I know it seems like it should be, but there is the act of reducing two options to one, that adds the odds for those two options together and places them on that one.

A, B and C again.

The prize is either at A or at B or at C. OK so far. This means that the chance of the prize being at any given location is 1/3. Yes?

Now you select any one location. What is your chance of being right? 1/3: true.

Now I open another door. Does this have any effect on the chance that the prize is behind the door that you selected? This is where people usually go wrong. NO, it does not. The chance that the prize is behind the door you selected is still 1/3. When you selected the door, you automatically left at least one door without the prize behind it unselected. There is a 2/3rds chance that you left one door without the prize and a 1/3rd chance that you left two such doors. Opening such a door does not change the odds of it existing. What it does do is change the odds of the prize being behind the other unopened door.

It is not a matter of if you select A they open B otherwise they open A. It is that they open the the door without the prize that you did not select.

Door........A..........B........C
Prize.............................X.........For actions: s = Select, o = Open
action......s..........o...................stick = lose, switch = win
action......o..........s...................stick = lose, switch = win
action......o...................s..........stick = win, switch = lose
result.......................................stick = 1, switch = 2
The fact that they could have opened door b in the last example instead of door a is irrelevant as what determines the case is your initial selection, not which of the two doors they choose.
Note, they never open the door with the prize behind it. This is what changes the odds.

Okay, I see the problem here. It's not two thirds, because that is not how it would work.

You pick a door - and becuase there are three doors with two that are losers - there is always a door that the quizmaster can open with nothing behind it.

Ok. So whatever your first choice is, it is irrelevant. It does not matter, you are always going to be reduced to a choice of two final doors, one which is the winner, and one which is the loser.

Run through the problem any way you want, you are always left with the choice between two doors, one which wins, one which looses.

Hence 50/50. Like tossing a coin.

Or am I misunderstanding the rules? Can the winning door be eliminated in the first choice. (Which would make it 1/3).

Then we are not talking about a truly random scenario - If the rules are that the quizmaster is removing a "loser" option, then I agree you are better off switching.

Correct. We aren't talking about a random scenario.

Read. (http://en.wikipedia.org/wiki/Monty_Hall_problem)

Ok. So whatever your first choice is, it is irrelevant. It does not matter, you are always going to be reduced to a choice of two final doors, one which is the winner, and one which is the loser.

Run through the problem any way you want, you are always left with the choice between two doors, one which wins, one which looses.

Hence 50/50. Like tossing a coin.

Or am I misunderstanding the rules? Can the winning door be eliminated in the first choice. (Which would make it 1/3).

The winning door is not opened until the very end of the game.

Does this help?

One third of the time:
If you pick A and the prize is behind B, then the quizmaster opens door C.

One third of the time:
If you pick A and the prize is behind C, then the quizmaster opens door B.

One third of the time:
If you pick A and the prize is behind A, then the quizmaster opens either door B or C. (one sixth of the time he will open B, one sixth of the time he will open C).

look, you pick a door, this door has 1/3 chanceof winning.
If they open another door, you still have 1/3 chance of winning with your original pick.
If you switch, you get 2/3 chances of winning.

Imagine 1000 doors (like someone else posted allready)
Your original door has a chance of 1/1000 of winning.
All the other doors combined have a 999/1000 chance.
remove 998 certain losers of these doors, and the remaining one has this 999/1000 chance of being the right door.

edit: thanks for pointing out a possibly confusing mistake.

look, you pick a door, this door has 1/3 chanceof winning.
If they open another door, you still have 1/3 chance of winning with your original pick.
If you switch, you get 2/3 chances of winning.

Imagine 1000 doors (like someone else posted allready)
Your original door has a chance of 1/1000 of winning.
All the other doors combined have a 999/1000 chance.
remove 998 of these doors, and the remaining one has this 999/1000 chance of being the right door.

Clarification: you are removing the 998 loser doors out of the 999 leaving the 1 out of 999 that would be the winner - Whichever one out of the 999 that would be.

Okay, I see the problem here. It's not two thirds, because that is not how it would work.

You pick a door - and becuase there are three doors with two that are losers - there is always a door that the quizmaster can open with nothing behind it.

Ok. So whatever your first choice is, it is irrelevant. It does not matter, you are always going to be reduced to a choice of two final doors, one which is the winner, and one which is the loser.

Run through the problem any way you want, you are always left with the choice between two doors, one which wins, one which looses.

Hence 50/50. Like tossing a coin.

Or am I misunderstanding the rules? Can the winning door be eliminated in the first choice. (Which would make it 1/3).

There are three doors, with two that are losers. Youy pick one, you have a 1;/3rd chance of being right. Whatecer is done now, after you have picked that door, to the other doors, can not and does not change that fact that your chance of being right is 1/3rd. Your chance of this initial pick being wrong is 2/3rds. So the chance of one of the other doors being right is 2/3rds. Hence if all but one of the other doors is now eliminated, the chance that it is the one that is left is 2/3rds. The prize does not move when a door is opened, the odds are not reset. All that happens is one chance of being wrong is eliminated if you switch.

Think of the same situation but with 10 doors. This time they will open 8 empty doors after you picked your one. the chance that you picked the door with the prize behind it to start with is 10%. It stays at 10% as the prize still has nine other doors that it could be behind. there is a 90% chance it is behind one of the other doors. This chance stays at 90%, even when there is only one other door left to be opened. After they open 1 door there is a 90% chance that it is behind one of the remaining eight, then seven etc.

I know, that if you look just at the final situation; two closed doors with a prize behind one of them, it seems to have to be a 50/50 choice, but how you get to that situation matters. If there were just two doors to start with, then your chance of picking the right door would be 50%, if there are a hundred, then it is 1%. After it has been chosen, this chance can not change, all that is happening is that the remaining percaentage is being concentrated in fewer and fewer posibilities.

Clarification: you are removing the 998 loser doors out of the 999 leaving the 1 out of 999 that would be the winner - Whichever one out of the 999 that would be.

Yes. Only losing doors are removed.

The winning door is not opened until the very end of the game.

Does this help?

One third of the time:
If you pick A and the prize is behind B, then the quiz master opens door C.

One third of the time:
If you pick A and the prize is behind C, then the quiz master opens door B.

One third of the time:
If you pick A and the prize is behind A, then the quiz master opens either door B or C. (one sixth of the time he will open B, one sixth of the time he will open C).


Okay, I see where people are confused. It's a poisson ( -sp I, don't really know and I am not about to look it up) function. But, whatever happens in the first choice, has no bearing on the second choice. In every case, you are left with a choice between a winning door, and a losing door.

You cannot name a single contingency that turns out with a choice between to losing doors, hence, whatever happened before is irrelevant. Thus it is a simple win lose decision. Thus 50/50. Diagram it if you don't believe me.

Say A is the winning door out of ABC:

Choose A - the winner, you are left with the choice between A (and B or C)

Choose A - the winner, you are left with the choice between A ( and B or C)

^^you have to say this twice, because the quiz master has two options. (Maybe this is where you fouled up)

Choose B - The Loser, you are left with a choice between A and B, ( C being thrown out as the losing door)

Choose C- - The Loser, you are left with a choice between A and C, ( B being thrown out as the losing door)

So, under any decision tree where A is the winner, you are left with the outcomes: (A, (B or C*)), (A,( B or C*)), (A or B), (A or C).

Thus A, is in fifty percent of the outcomes, and by inspection, it can be seen to apply to any other variable that is assigned as winning.

Does that help.


*(B or C) were used, in this case, to represent the indeterminate choice that the quiz master could have made in the event that the winning door was chosen. But it is irrelevant anyway, as you are always left with a win lose choice.

Okay, I see where people are confused. It's a poisson ( -sp I, don't really know and I am not about to look it up) function. But, whatever happens in the first choice, has no bearing on the second choice. In every case, you are left with a choice between a winning door, and a losing door.

You cannot name a single contingency that turns out with a choice between to losing doors, hence, whatever happened before is irrelevant. Thus it is a simple win lose decision. Thus 50/50. Diagram it if you don't believe me.

Say A is the winning door out of ABC:

Choose A - the winner, you are left with the choice between A (and B or C)

Choose A - the winner, you are left with the choice between A ( and B or C)

^^you have to say this twice, because the quiz master has two options. (Maybe this is where you fouled up)

Choose B - The Loser, you are left with a choice between A and B, ( C being thrown out as the losing door)

Choose C- - The Loser, you are left with a choice between A and C, ( B being thrown out as the losing door)

So, under any decision tree where A is the winner, you are left with the outcomes: (A, (B or C*)), (A,( B or C*)), (A or B), (A or C).

Thus A, is in fifty percent of the outcomes, and by inspection, it can be seen to apply to any other variable that is assigned as winning.

Does that help.


*(B or C) were used, in this case, to represent the indeterminate choice that the quiz master could have made in the event that the winning door was chosen. But it is irrelevant anyway, as you are always left with a win lose choice.

The option to change is not independent of the option of the door you first selected. Now go back and redo your math using dependent probabilities rather thsan independent probabilities. It is not a Poisson, it is Bayles Theorem.

The option to change is not independent of the option of the door you first selected. Now go back and redo your math using dependent probabilities rather thsan independent probabilities. It is not a Poisson, it is Bayles Theorem.

Yah, it is, because they always remove a losing door, never a winning door. So it is totally independant.

Yah, it is, because they always remove a losing door, never a winning door. So it is totally independant.

No it is not. The situation before you choose is not the same as the situation after. Their action depends, totaly, upon your initial action. No independence here whatsoewer, except in the case where you have picked the winning door. If you pick a losing door, your acton constrains theirs. How is that independant?

No it is not. The situation before you choose is not the same as the situation after. Their action depends, totaly, upon your initial action. No independence here whatsoewer, except in the case where you have picked the winning door. If you pick a losing door, your acton constrains theirs. How is that independant?

Because if you pick the winning door, they have two options, don't they? So it all squares away in the end. So if you want to look a dependency of previous choice, you have to count picking the winning door twice in the final outcome, because there are two different choices the quizmaster could make.

But that is all irrelevant anyway. Name me one circumstance in which you are not left with a simple choice between a winning door and a losing door. Any, and all, choices you make will lead to this outcome, hence it is not conditional.

Can I just point out that IT IS ACEPTED THAT SWITCHING IS THE BEST STRATEGY. It's not an argument about the best strategy, its about trying to understand.

I've only encountered it as a demonstration of the alarming limits of human reasoning (which all the confusion demonstrates). I wouldn't call it a paradox, just a surprise that makes you kick yourself when you understand.

So calm down and have a think about it.

Because if you pick the winning door, they have two options, don't they? So it all squares away in the end.Let's actually try this out.

Suppose that we're doing this with one million doors. So 1 door has a car behind it, the other 999,999 have nothing. In the role of Monty, I know which door has the car behind it.

So, pick a number at random between 1 and 1,000,000.

Done?

Now I'm going to open every door except the one you chose (whatever it might be), and door 287,148.

Are you going to stick with your original choice, or switch to 287,148?

But switching doesn't give you an advantage. Given the 3 door problem, A is picked and C is opened. Switch?

Both doors have an equal chance of hiding the prize, because the quizmaster isn't going to open the door you picked. The odds change because the situation changes. The situation is dynamic, based on the whim of the quizmaster. As a probability problem, only the second set of choices is important, the first set doesn't as all choices lead to the second set.

Thus, you have 2 choices. Changing your choice to B makes no difference, as both doors have an equal 1 in 2 chance.

The same applies to the 1000 door problem.

Let's actually try this out.

Suppose that we're doing this with one million doors. So 1 door has a car behind it, the other 999,999 have nothing. In the role of Monty, I know which door has the car behind it.

So, pick a number at random between 1 and 1,000,000.

Done?

Now I'm going to open every door except the one you chose (whatever it might be), and door 287,148.

Are you going to stick with your original choice, or switch to 287,148?

Different problem :headbang:

But switching doesn't give you an advantage. Given the 3 door problem, A is picked and C is opened. Switch?

Both doors have an equal chance of hiding the prize, because the quizmaster isn't going to open the door you picked. The odds change because the situation changes. The situation is dynamic, based on the whim of the quizmaster. As a probability problem, only the second set of choices is important, the first set doesn't as all choices lead to the second set.

Thus, you have 2 choices. Changing your choice to B makes no difference, as both doors have an equal 1 in 2 chance.

The same applies to the 1000 door problem.

Only the second choice is relevant if the probabilities are independant. In this problem they are not. Which door is revealed to be empty is dependent, 3/3rds of the time on your original choice. The first set of choices do affect the second. It is called Bayesian probability, and is very well understood mathematicaly. It is not as a previous poster said, a question about which is right. It is a question about whether you can understand that switching does improve your chances. It doesa. that is not in dispute. (N. B. Sorry for the misnomer of Bayles instead of Bayes earlier, a memory failure on my part.)

Different problem :headbang:

Same problem. In what way is it topologically or structurally different?

Different problem :headbang:Which problem are you discussing, then? If your only objection is the quantity of doors we can play the same game with three, though the demonstration will take rather longer.

Same problem. In what way is it topologically or structurally different?

It lead to a lose lose choice most of the time.

It lead to a lose lose choice most of the time.

So does the three door problem, unles you think 2/3 is not most of the time.

It lead to a lose lose choice most of the time.It'll never lead to a lose/lose choice - remember that the host knows where the car is. He's clearly never going to start by opening the door that has the car behind it, since that would destroy the point of the game entirely.

So does the three door problem, unles you think 2/3 is not most of the time.

Yah, I am a dickhead :headbang: . i am going to go away and sober up and think about this.

Only the second choice is relevant if the probabilities are independant. In this problem they are not. Which door is revealed to be empty is dependent, 3/3rds of the time on your original choice. The first set of choices do affect the second. It is called Bayesian probability, and is very well understood mathematicaly. It is not as a previous poster said, a question about which is right. It is a question about whether you can understand that switching does improve your chances. It doesa. that is not in dispute. (N. B. Sorry for the misnomer of Bayles instead of Bayes earlier, a memory failure on my part.)

So does the door opened have anything to do with the whim of the quizzer? If it does, then it isn't a Bayesian probability. And the probability is dependent on the one examining the problem. It could be a frequency probability, in which case I am correct. The problem can be solved with many "degrees" of accuracy in mind, all of which will change the results slightly.

This confused me a lot when I first heard about it.

There are still mathematicians who cannot believe it's true, but it's obvious when you think about it.

You just have to consider why each of the boxes have made it to the final two - one is there just because you chose it initially. The other is there because it either is the prize (i.e. you were wrong in your first guess) or it is empty (i.e. you chose correctly first time).

Essentially, by switching to the other door, you are taking the odds that you were wrong in your initial guess.

This confused me a lot when I first heard about it.

There are still mathematicians who cannot believe it's true, but it's obvious when you think about it.

You just have to consider why each of the boxes have made it to the final two - one is there just because you chose it initially. The other is there because it either is the prize (i.e. you were wrong in your first guess) or it is empty (i.e. you chose correctly first time).

Essentially, by switching to the other door, you are taking the odds that you were wrong in your initial guess.

Which is why it isn't obvious. You aren't going to gain any odds because behind the door is either nothing, or a car. So just thinking you are wrong, doesn't make you wrong.

Yes, it is either nothing or a car. But the probability of it being a car is higher than the probability of it being nothing.

i cant believe y'all are still arguing about this. why doesnt the simple working out of the problem convince you?

Yes, it is either nothing or a car. But the probability of it being a car is higher than the probability of it being nothing.

And how, praytell, is that? 2 possibilities. It's a car, or it isn't. The car is behind door A or door B. Sounds like a 1 in 2 chance to me.

Just because there's two options, doesn't mean they're both equally likely.

If you were right in the first place, then you'd win if you stuck.

If you were wrong in the first place, you'd win if you switched.

The probability of you being right in the first place is only 1 in 3.

And then one of the choices is removed, leaving a 1 in 2 chance.

i cant believe y'all are still arguing about this. why doesnt the simple working out of the problem convince you?

It begins as a 1 in 3 shot of being correct. A second choice after one choice is removed leaves a 1 in 2 shot of being correct. The suggestions of Bayesian probability say that you were wrong to begin with and so should switch because you have a 2 in 6 chance of winning if you switch, and only 1 in 6 chance of winning if you stay.

By saying this, the assumption is that if you choose to switch, you increase your chances because you have chosen more options. If this was the case, you couldn't lose! The chance is 1 in 3 you are correct to begin with. If you switch when given the second choice, you have a 2 in 3 chance of being correct. But since one of the 3 has been taken away, you have to win, right?

I'm saying wrong.

Imagine there are a thousand doors, and you are told the car is behind one. You pick door seven. All the doors except seven (which you chose) and fifty-three are taken away. Do you switch? Yeah you do. Unless you want a 1/1000 chance of winning. It's the same with the Monty Hall problem, except 1/1000 gets knocked down to 1/3.
Nope. The odds of door seven have are no longer 1/1000 they've improved to 1/2.

A practical test of this principle. You will need someone to help you.

Take a pack of cards and spread them out on the floor face down without looking at the faces at all. Now select one card which you think is the ace of spades. Keep your finger on that card.

Now your assistant looks at the other cards one at a time, removing the card if it is not the ace of spades until there is only one card left that you do not have your finger on.

Now is the chance that you have your finger on the ace of spades 1 in 52 (presuming no jokers) or 1 in 2? If someone were toi give you a car if you could select which of the two remaining cards is the ace of spades, which one would you choose?

This isn't true.
When you start out, you have a 1/3 chance; if you switch, you have a 50-50 chance.

100 posts, yay!
Yeah, but if you stay you are still making a choice between two options. it's still 50/50, no?

It's tricky because you have to let go of some basic statistics because the removed door isn't selected at random.

If you pick door one - the removed door predefined as a loser of 2 & 3. So it either 2 OR 3 was the winning door it becomes the remaining door.
I don't get it, but I'll concede. It seems people who know alot more about math than me beleive it's better to switch. Who am I to argue?

And then one of the choices is removed, leaving a 1 in 2 chance.

This would only be true if the removed choice was randomly selected from the initial 3. In this case, the removed choice is predetermined to be a wrong choice out of the remaining 2 choices. It is not random! The removed door is only selected after the winning door is determined (but not revealed to the choser). The act of removing the door has no statistical bearing on the 1st choice being right. In fact it has no bearing on anything - You don't even have to remove it - We already know that one of the two unchosen doors has to be a loser.

So that means by switching, you are getting the benefits of BOTH doors 2 & 3.

As the end result, your options would be:

1) Your initial choice (if you stay) = 33% of winning

or

2) The better of the other TWO choices (if you switch) = 33% + 33% chance of winning.

The confusion lies in the fact that it seems like the situation is: one is the prize and one isn't, so it's 50:50. But it isn't like that because one is more likely to be the prize than the other.

By switching, you only lose if your first guess was correct.

When Monty's removing a box:

-If you picked the prize (1 in 3 chance), he removes a random wrong answer and if you swap, you lose.

-BUT if you picked a wrong one (2 in 3 chance), Monty is FORCED to remove a wrong answer, and if you swap, you get the prize.

Since the latter scenario is most likely, it's best to swap.

And how, praytell, is that? 2 possibilities. It's a car, or it isn't. The car is behind door A or door B. Sounds like a 1 in 2 chance to me.I bought a lottery ticket. It's either a winner or a loser. Does that make it 1 in 2?

I bought a lottery ticket. It's either a winner or a loser. Does that make it 1 in 2?
You bought a lottery ticket. So did your brother. You are told that ONLY you or your brother have the winning ticket. One of you HAS to win. Does that make it 1 in 2?

You bought a lottery ticket. So did your brother. You are told that ONLY you or your brother have the winning ticket. One of you HAS to win. Does that make it 1 in 2?
i dont know but im glad my brother sleeps in on sunday, ill be at the lottery office claiming the prize before he's even out of the shower!

depends on how they decided that one of us 2 is the winner.

It's like that, but if your brother has two tickets and you have one.

id switch
my brother has always had better luck than i do

but should my brother switch too?

You bought a lottery ticket. So did your brother. You are told that ONLY you or your brother have the winning ticket. One of you HAS to win. Does that make it 1 in 2?You buy a lottery ticket. Your brother buys all the other lottery tickets. Your brother scratches all the other lottery tickets. Your brother shreds all of his scratched-off lottery tickets except one. He did not shred the losing ticket. One of you has the winning ticket.

If this experiment is repeated for many lotteries, who do you think will end up with more winning tickets? Will it be you, who bought one ticket per lottery, or will it be your brother, who bought all the other tickets?

You buy a lottery ticket. Your brother buys all the other lottery tickets. Your brother scratches all the other lottery tickets. Your brother shreds all of his scratched-off lottery tickets except one. He did not shred the losing ticket. One of you has the winning ticket.

If this experiment is repeated for many lotteries, who do you think will end up with more winning tickets? Will it be you, who bought one ticket per lottery, or will it be your brother, who bought all the other tickets?if you're gonna do it like that, than you also have to give me the chance to switch tickets with my brother ;)