Intersting probability situation - NOW with script

[gopunk]

Originally posted by: fuzzy bee

Originally posted by: gopunk

that makes no sense to me. every time you have the option of choosing, it is a new case. new weighting must be applied based on the probability at that point, disregarding any previous probabilities. if there are two doors available, and you have to choose one (the situation you would be in at that point), there is a 50-50 chance you will pick the correct door. it doesn't matter how many doors have been opened prior.

no it is not a new case, because you have existing information on the doors. if you were to mix up the two doors such that you didn't know which one was which, then yes, you would have 1/2 chance each. but you know which door is which.

the only existing information you have is this:

there are two doors. one has a prize. one does not. they have equal opportunity to have the prize - it's not like one door has been opened and you have been shown what's behind it.

no, that is incorrect. you have the following information:

the door you picked has 1/3 chance of being right. you can not challenge this...

there is one door left... what probability do you assign to it?

do you really think you are right? why don't you do a little experiment to see... seriously, we're just going to go around in circles here. go ahead and do a test.

 

[conjur]

Originally posted by: Moralpanic

But when you rolled a dice, each event is not dependent on the past... but in this instance, IT IS dependent on what the host does. Because he eliminates a door for you.

Lets take a very extreme example... there are 1 million doors.
You pick one, your chances are 1/1000000.
The host eliminates 999998 other doors until 2 doors are left... one you choose, and another closed one.

You're telling me the one you chose has the same odds as the other one?

Then your odds are 1/2, not 999999/1000000.

 

[FuzzyBee]

Originally posted by: Moralpanic

Originally posted by: fuzzy bee

Originally posted by: gopunk

Originally posted by: conjur

Originally posted by: gopunk

Originally posted by: conjur
Why can't the door picked also have 2/3 chance since 1/3 of the original choices are now gone? That seems a bit flawed to assign 2/3 chance to an unpicked door and not to the picked door.

you had a 2/3 chance you were wrong. this does not change simply because the number of ways you could have been wrong is reduced. it simply increases the probability that any particular remaining way of being wrong was the case.

Ok...but the now known door removes 1/3 from the equation so there's a 1/3 chance you're wrong and a 1/3 chance you're right. You can't throw the other 1/3 back into the equation. It would be more accurate to say your remaining chances are 1/2 but you still have the original 1/3 probability.

no you must throw the 1/3 back, the total probability must sum to 1.

look at it this way... the set of doors you did not choose has 2/3 probability. by removing a door from that set, you don't remove any probability, you just remove the number of elements in that set, meaning that when it comes time to distribute the probability of that set amongst it's elements, each remaining element will get more

but those elements don't matter any more. that's equivalent to saying that you have a lower probability of rolling a six just because you just rolled a six. each case is independent.

But when you rolled a dice, each event is not dependent on the past... but in this instance, IT IS dependent on what the host does. Because he eliminates a door for you.

Lets take a very extreme example... there are 1 million doors.
You pick one, your chances are 1/1000000.
The host eliminates 999998 other doors until 2 doors are left... one you choose, and another closed one.

You're telling me the one you chose has the same odds as the other one?

Yes, that's exactly what I'm saying. All you know is that one of those doors is the winner, and one is a loser. You have to pick one. You have a 50% chance of getting it right.

Have to go home now. If this debate is still up when I get there, I'll jump back in.

 

[TallBill]

The 100 door problem makes it easy. Unless you picked the correct door on yer first chance (1out of 100), the 98 doors that are removed will be empty, because the correct door cant be revealed, and the only door left (other the the first pick) will be the correct one.

 

[gopunk]

Yes, that's exactly what I'm saying. All you know is that one of those doors is the winner, and one is a loser. You have to pick one. You have a 50% chance of getting it right.

Have to go home now. If this debate is still up when I get there, I'll jump back in.

why are you so keen on ignoring existing data?

 

[FuzzyBee]

Originally posted by: kranky

Originally posted by: fuzzy bee

Originally posted by: kranky Can you see how your odds didn't magically grow to 50/50 just because I showed you 98 empty doors? You would be saying that you had a 50/50 chance of being right from the start, and we know that's wrong. You had only a 1 in 100 chance.

correct - you had a 1-in-100 chance. now, you have a 1 in two chance. you can't pick the doors that have already been opened, can you? prior selections don't affect the current selection, once you make it there.

How about this? There are 100 doors, and you pick one. Let me have the other 99. The host shows that 98 of my doors are empty (what do I care? I knew 98 of them would be empty anyway), and now it's down to your door and the one I have left. Do we have an equal chance of being right?

You are being fooled by the fact that some of the doors are shown to you before you get to switch. That changes nothing about your original chances of being right.

Yes it does, because you can't pick those doors. there are 2 doors to choose between, and you have to pick one of the 2 equally weighted doors.

 

[conjur]

Originally posted by: gopunk

no, that is incorrect. you have the following information:

the door you picked has 1/3 chance of being right. you can not challenge this...

there is one door left... what probability do you assign to it?

do you really think you are right? why don't you do a little experiment to see... seriously, we're just going to go around in circles here. go ahead and do a test.

You've changed the equation. You've opened the box with Schroedinger's cat in it.

 

[DrPizza]

Originally posted by: gopunk
what the f*ck is wrong with you people, have you never taken any sort of probability course in your lifetime? this is not flawed logic, you have flawed thinking. do a f*cking test if you still think you're right.

ugh.

I couldn't have said it better!

 

[TallBill]

Im gonna keep repeating my logic, as its prety damn easy to understand.

The 100 door problem makes it easy. Unless you picked the correct door on yer first chance (1out of 100), the 98 doors that are removed will be empty, because the correct door cant be revealed, and the only door left (other the the first pick) will be the correct one.

 

[gopunk]

Yes it does, because you can't pick those doors. there are 2 doors to choose between, and you have to pick one of the 2 equally weighted doors.

except they are not equally weighted. you picked one with 1/100 probability, set it aside... regardless of what you do, that door will always have 1/100 probability.

 

[deftron]

Originally posted by: gopunk

Yes, that's exactly what I'm saying. All you know is that one of those doors is the winner, and one is a loser. You have to pick one. You have a 50% chance of getting it right.

Have to go home now. If this debate is still up when I get there, I'll jump back in.

why are you so keen on ignoring existing data?

Proof there is no "existing data"

3 doors
2 people

You pick door #1
Other person picks #2

Host shows door #3 is empty

If you both switch doors, you'd both have "2/3 chance" .. making it 4/3 total (impossible)

It's 50/50

 

[Looney]

Originally posted by: fuzzy bee

Originally posted by: kranky

Originally posted by: fuzzy bee

Originally posted by: kranky Can you see how your odds didn't magically grow to 50/50 just because I showed you 98 empty doors? You would be saying that you had a 50/50 chance of being right from the start, and we know that's wrong. You had only a 1 in 100 chance.

correct - you had a 1-in-100 chance. now, you have a 1 in two chance. you can't pick the doors that have already been opened, can you? prior selections don't affect the current selection, once you make it there.

How about this? There are 100 doors, and you pick one. Let me have the other 99. The host shows that 98 of my doors are empty (what do I care? I knew 98 of them would be empty anyway), and now it's down to your door and the one I have left. Do we have an equal chance of being right?

You are being fooled by the fact that some of the doors are shown to you before you get to switch. That changes nothing about your original chances of being right.

Yes it does, because you can't pick those doors. there are 2 doors to choose between, and you have to pick one of the 2 equally weighted doors.

WTF?! That's the point, they're not equally weighted! Take Kranky example... you pick 1 door out of 100. He gets to pick 99 doors. The host then decides to open 98 of his doors to show that they don't contain anything. Now you have the option to switch doors with Kranky... would you do it?

 

[CChaos]

Look at it this way:

When you first pick, there's a 1/3 chance you are right and the prize is behind your door.

There's also a 2/3 chance that you are wrong and the prize is not behind your door.

The prize has to be behind a door and it can't be behind the door that's opened.

So, even when there are only 2 doors left, there's a 1/3 chance you picked right and a 2/3 you picked wrong.

It's more likely that you picked wrong to begin with and there's only one other door that can be right.

Switch!

 

[gopunk]

Originally posted by: deftron

Originally posted by: gopunk

Yes, that's exactly what I'm saying. All you know is that one of those doors is the winner, and one is a loser. You have to pick one. You have a 50% chance of getting it right.

Have to go home now. If this debate is still up when I get there, I'll jump back in.

why are you so keen on ignoring existing data?

Proof there is no "existing data"

3 doors
2 people

You pick door #1
Other person picks #2

Host shows door #3 is empty

If you both switch doors, you'd both have "2/3 chance" .. making it 4/3 total (impossible)

It's 50/50

sure, because you're guaranteed that either you or the other person picked the right door. your analogy would only hold if the host picked between #2 and #3, thus allowing for the possibility that #3 is the winnah.

 

[Haircut]

Originally posted by: fuzzy bee

Originally posted by: kranky

Originally posted by: fuzzy bee

Originally posted by: kranky Can you see how your odds didn't magically grow to 50/50 just because I showed you 98 empty doors? You would be saying that you had a 50/50 chance of being right from the start, and we know that's wrong. You had only a 1 in 100 chance.

correct - you had a 1-in-100 chance. now, you have a 1 in two chance. you can't pick the doors that have already been opened, can you? prior selections don't affect the current selection, once you make it there.

How about this? There are 100 doors, and you pick one. Let me have the other 99. The host shows that 98 of my doors are empty (what do I care? I knew 98 of them would be empty anyway), and now it's down to your door and the one I have left. Do we have an equal chance of being right?

You are being fooled by the fact that some of the doors are shown to you before you get to switch. That changes nothing about your original chances of being right.

Yes it does, because you can't pick those doors. there are 2 doors to choose between, and you have to pick one of the 2 equally weighted doors.

No, it doesn't work like that.
If you had picked the correct door at first then the host could remove any 98 of the remaining 99 doors. In this situation you would win by staying on your door instead of moving. This can only happen if you pick the correct door first off, this is 1/100 probability.

The other situation is that you pick the incorrect door at first (99/100) probability. The host then can all 98 other doors that have nothing behind them. Thus in this situation you will win by moving to the other door.

Now 99 times out of 100 you will have to move door to be correct, the other 1 time you will be correct by staying where you are.

End of story.

 

[TallBill]

Jesus christ, im gonna shoot someone!

The 100 door problem makes it easy.

Case 1:
(Odds 1:100)
You pick the correct door.
98 empty doors are revealed
1 empty door is left

Case 2
(Odds 99:100)
You pick an empty door
98 doors are revealed
1 correct door is left.

 

[Ynog]

Those think the answer is 50/50 are looking at it wrong.

Lets work with this case. Four doors.

I pick one of the doors. I have a 1/4 chance I picked the right door. And a 3/4 chance I picked the wrong door.

So If I remove ONE of the three remaining doors leaving 2 left.

Now what are the odds If I stick with the first door, what are the odds i have it right.

This is the way you solve it.
If you guessed it right initially with 1/4 probablilty and wrong with a 3/4 probabilty.
If you were right initially then you won the prize and wrong you don't. The odds don't
change. Cause you didnt' change your answer.

So that 1/4 *1 + 3/4 * 0 = 1/4

now if you switch you have different odds.
If you were right initialy then you lose. if you were wrong then you have a 1/2 probablitly that you guess the right
door the second time.

So thats 1/4 * 0 + 3/4 * 1/2 = 3/8.

Now when you go back to three doors
its 1/3 * 0 + 2/3 * 1 = 2/3. If you guessed wrong the first time 2/3 then you have to be right the second time.

YOU CANNOT IGNORE EXISITING DATA. The previous choices do effect the outcome.

 

[TallBill]

The 2 out of 3 explained

Case 1:
(Odds 1:3)
You pick the correct door.
1 Empty door is revealed.
1 Empty door is left

Case 2
(Odds 2:3)
You pick an empty door
1 Empty door is revealed
1 Correct door is left

 

[CChaos]

Sorry but it can't be 50/50. There are 3 choices when you start and the probability is based on the choice you make from those three.

Don't be fooled by the door being opened. They will open whatever door is empty if you chose wrong and either of the other doors if you chose right. No matter how you chose there will be at least one empty door so don't infer that the odds change based on a door being opened.

All that matters is your first choice has a 66% chance of being wrong!

 

[gopunk]

Originally posted by: TallBill
The 2 out of 3 explained

Case 1:
(Odds 1:3)
You pick the correct door.
1 Empty door is revealed.
1 Empty door is left

Case 2
(Odds 2:3)
You pick an empty door
1 Empty door is revealed
1 Correct door is left

ooh good explanation. hopefully that is clear..

 

[TallBill]

If someone cant understand that, they will die.

 

[CoveX]

Hi, new here. =)

This problem made my head spin when I first encountered it some years ago. Let's try an alternate explanation. There's a 1/3 chance your initial guess was correct, and a 2/3 chance it was wrong. So either:

1/3 chance-- You were Correct, so sticking with your chosen door will guarantee the prize.

2/3 chance-- You were Wrong, and a second wrong door is opened showing it empty. Switching now will guarantee you the prize.

Since you were more likely to be wrong (2/3 chance) than right (1/3 chance), it makes sense to always switch.

Edit: Agh, someone beat me to it. Oh well

 

[Ynog]

This is an old problem. I have seen it more than 5 times through High School and College.
Its amazing how many people don't understand it. If you really believe its 50/50 talk to your
math teacher or professor. He will explain it for you.

 

[DrPizza]

Originally posted by: deftron

Originally posted by: gopunk

Originally posted by: conjur

Originally posted by: gopunk

Originally posted by: fuzzy bee
Thinking back over all of my statistics classes, I don't remember any point where previous random "drawings" affected a current "drawing".

When you get down to 2 doors, there are only 2 options: pick door a, or pick door b. You have a choice between one or the other, not between one out of 100. All that matters is the probability at that point, not the probability of you getting to that point.

sure, but the weight assigned to the door you didn't pick is 2/3, and the weight assigned to the door you did pick is 1/3.

Why can't the door picked also have 2/3 chance since 1/3 of the original choices are now gone? That seems a bit flawed to assign 2/3 chance to an unpicked door and not to the picked door.

you had a 2/3 chance you were wrong. this does not change simply because the number of ways you could have been wrong is reduced. it simply increases the probability that any particular remaining way of being wrong was the case.

also, if both doors had 2/3 chance, you would have 4/3 total probability, which is impossible.

What about cheapbidder's scenario?

3 doors
2 people

You pick door #1
Other person picks #2

Host shows door #3 is empty

If you both switch doors, you'd both have 2/3 chance .. making it 4/3 total

That scenario doesn't work.... The only way the host could show what's behind the 3rd door would be if neither person picked the losing door. In the 1 door picked scenario, the host will ALWAYS be able to show a losing door. Also, in the one door scenario, the rule is that the host has to always show what's behind one of the other doors.

Supposed the host showed the door *only* when a person picked the winning door.... then, if half the people switched, half of them lost.


edit: fixed one word

 

[DrPizza]

Originally posted by: fuzzy bee
Considering it's a "redeal" of opportunities at this point, and you aren't dumb enough to pick a door that has already been opened, you will have a 50/50 chance no matter which door you choose. Any other answer is bunk. None of the prior decisions affect the fact that there are 2 doors, and you have to choose 1 of the two.

THIS is why you don't understand the problem.... it's NOT a "redeal"
A redeal would be if after they showed you a door, they shuffled the other two doors.
If they randomly reassigned the winner to the two remaining doors, then, and only then, would it be a redeal.