Intersting probability situation - NOW with script

[Looney]

Yeah, i dont think you're going to get it. It can't get any simpler than what we've explained in here.

 

[FuzzyBee]

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.

The key word here is picked. I agree with your statement. However, you have another opportunity to pick, and obly two doors to choose between.

 

[TallBill]

Originally posted by: fuzzy bee

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.

The key word here is picked. I agree with your statement. However, you have another opportunity to pick, and obly two doors to choose between.

Wrong again.

 

[FuzzyBee]

Originally posted by: Moralpanic
why would the first door you pick all of a sudden have a higher percentage of being right? It wouldn't.

you're right - it wouldn't, based upon the original selection. Now, however, you have a new selection to make, and only two equally-weighted options to choose from.

 

[FuzzyBee]

Originally posted by: TallBill

Originally posted by: fuzzy bee

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.

The key word here is picked. I agree with your statement. However, you have another opportunity to pick, and obly two doors to choose between.

Wrong again.

WHAT?!?!?!? How is it that you have more than two doors to choose between? How do you not have another opportunity to pick?

 

[TallBill]

Originally posted by: fuzzy bee

Originally posted by: Moralpanic
why would the first door you pick all of a sudden have a higher percentage of being right? It wouldn't.

you're right - it wouldn't, based upon the original selection. Now, however, you have a new selection to make, and only two equally-weighted options to choose from.

No you dont. I'll make a wager on it too.

 

[TallBill]

You do have an opportunity to pick, but as described above MANY MANY times, the correct door will be the other door 2/3's of the time.

 

[Looney]

Originally posted by: fuzzy bee

Originally posted by: Moralpanic
why would the first door you pick all of a sudden have a higher percentage of being right? It wouldn't.

you're right - it wouldn't, based upon the original selection. Now, however, you have a new selection to make, and only two equally-weighted options to choose from.

How can they be equally weighted? It's not a completely new, independent event. It's an event based on previous actions... that the host eliminated some doors for you. That's where your problem is... you think prior information doesn't affect current event, but it does.

 

[FuzzyBee]

Originally posted by: conjur
1/3 - right
1/3 - wrong
1/3 - wrong


One of the 1/3 is made known which leaves (2) 1/3 choices

there's a 1/3 chance the remaining unpicked door is right and there's a 1/3 chance the door you picked is right.

Where is the third third (it's repeating day!) coming from? What door are you using to apply it to?

If you have 2 doors, and one of them is going to have the prize behind it, and youknow that it is either door #1 or door #2, i don't see where you would have any benefit of choosing one over the other.

 

[kranky]

fuzzy bee, you clearly have some training in statistics, so I know you can grasp this. It's cool when you get that Aha! moment finally. I took forever to finally understand it which is why I really like this problem.

 

[FuzzyBee]

Originally posted by: Moralpanic

Originally posted by: fuzzy bee

Originally posted by: Moralpanic
why would the first door you pick all of a sudden have a higher percentage of being right? It wouldn't.

you're right - it wouldn't, based upon the original selection. Now, however, you have a new selection to make, and only two equally-weighted options to choose from.

How can they be equally weighted? It's not a completely new, independent event. It's an event based on previous actions... that the host eliminated some doors for you. That's where your problem is... you think prior information doesn't affect current event, but it does.

It is a completely new event! There are two doors, and one of them has the prize.

I think what is confusing here is that there are people looking at this is the wrong context. I *think* people are trying to predict which of the last two doors you are going to choose between {i]from the beginning[/i]. I am dtermining the probability of which door you will be choosing after the host has already eliminated 98 of the 100 (or 1 of the 3). Heck, he could eliminate 4x10^5475984398579848 for all I care - if there are still two doors left, only one of them can have the prize, and they each have equal probability of having it.

 

[TallBill]

I'll post it again... because people obviously cant read up!

The 2 out of 3 explained

Case 1:
(Odds 1:3)
You pick the correct door to begin with.
1 Empty door is revealed.
1 Empty door is left.
Switching would be bad in this case which happens 1/3 of the time.

Case 2
(Odds 2:3)
You pick an empty door to begin with.
1 Empty door is revealed
1 Correct door is left.
Switching would be good in this case which happens 2/3's of the time

 

[FuzzyBee]

Originally posted by: coder1
Here is a script that test this theory 200,00 times

LINK

If at all possible, could I see the code for this script? Maybe it'd help me understand.

 

[gopunk]

It is a completely new event! There are two doors, and one of them has the prize.

I think what is confusing here is that there are people looking at this is the wrong context. I *think* people are trying to predict which of the last two doors you are going to choose between {i]from the beginning[/i]. I am dtermining the probability of which door you will be choosing after the host has already eliminated 98 of the 100 (or 1 of the 3). Heck, he could eliminate 4x10^5475984398579848 for all I care - if there are still two doors left, only one of them can have the prize, and they each have equal probability of having it.

people are looking at this in the context as described in the original post. choose one of two doors with equal probability is not the scenario described in the original post.

 

[Yax]

I've been convinced that changing doors would give you 2/3 chance of winning. I can see the logic now

Assuming you don't change doors.

Case 1:
Prize in door #1
a. You pick door 1, door 2 or door 3 revealed - you win.
b. You pick door 2, door 3 revealed - you lose.
c. You pick door 3, door 2 revealed - you lose.

Case 2:
Prize in door #2
a. You pick door 1, door 3 revealed - you lose
b. You pick door 2, door 1 or door 3 revealed - you win.
c. You pick door 3, door 1 revealed - you lose.

Case 3:
Prize in door #3
a. You pick door 1, door 2 revealed - you lose
b. You pick door 2, door 1 revealed - you lose.
c. You pick door 3, door 1 or door 3 revealed - you win

By sticking to your door you only win in oneof the 3 cases or 1/3 chance of winning. This means that by switching, you will have the 2/3 chance of winning. This result holds true only because the HOST knows which door to open. It would not hold true if the host just randomly opened one of the doors you did not pick. If he randomly opened a door then asked you whether or not to switch doors after you observed the contents of the open door, then your chances would be 50/50.

 

[Looney]

Originally posted by: fuzzy bee

Originally posted by: Moralpanic

Originally posted by: fuzzy bee

Originally posted by: Moralpanic
why would the first door you pick all of a sudden have a higher percentage of being right? It wouldn't.

you're right - it wouldn't, based upon the original selection. Now, however, you have a new selection to make, and only two equally-weighted options to choose from.

How can they be equally weighted? It's not a completely new, independent event. It's an event based on previous actions... that the host eliminated some doors for you. That's where your problem is... you think prior information doesn't affect current event, but it does.

It is a completely new event! There are two doors, and one of them has the prize.

I think what is confusing here is that there are people looking at this is the wrong context. I *think* people are trying to predict which of the last two doors you are going to choose between {i]from the beginning[/i]. I am dtermining the probability of which door you will be choosing after the host has already eliminated 98 of the 100 (or 1 of the 3). Heck, he could eliminate 4x10^5475984398579848 for all I care - if there are still two doors left, only one of them can have the prize, and they each have equal probability of having it.

Ok, i'll repost what i said earlier, and you answer me:

Lets try this again with the 100 door example.

You pick one door... 1/100

I get to pick 99 doors... so my odds are 99/100 (this represents the doors you didn't pick)

The host opens 98 of my doors, so that there's 2 doors left... yours (1/100) and mine (99/100).

Now you have an option to switch door with me... would you do it?

Would you stay with your original choice, or would you switch doors?

 

[ElFenix]

you have a 2/3 chance of being wrong. the fact that the guy showed you an empty door doesn't change the fact that your decision is 2/3 wrong.

 

[Syringer]

Originally posted by: Syringer
Yup, really interesting stuff.

Although I think the last time this was posted it went on to about 300 posts or so..

And so history repeats itself 🙂

 

[conjur]

Originally posted by: fuzzy bee

Originally posted by: conjur
1/3 - right
1/3 - wrong
1/3 - wrong


One of the 1/3 is made known which leaves (2) 1/3 choices

there's a 1/3 chance the remaining unpicked door is right and there's a 1/3 chance the door you picked is right.

Where is the third third (it's repeating day!) coming from? What door are you using to apply it to?

If you have 2 doors, and one of them is going to have the prize behind it, and youknow that it is either door #1 or door #2, i don't see where you would have any benefit of choosing one over the other.

The door that's already been revealed. It's still a valid choice. A stupid choice, yes, but a valid choice.

 

[stonecold3169]

Alright, you walk into a store and there's a sign that says "pre-scratched lotto tickets, all losers but one a winner! $.05 each, one per customer!". You pick one out without looking at it. The manager of the store walks up to you and says "I'm sick of this promotion... if you give me back the ticket I just gave you, you can take all the other tickets for the $.05". Any resonable person would give him the ticket back and take the other, say 99 (lets assume there were 100 tickets to start, doesn't matter), because you just went from having a 1 in 100 chance to a 99 in 100, right? It's the same thing. His showing you the empty door means nothing... who cares if 98 of the remaining 99 are bad, thats a given, right?

 

[Hector13]

Originally posted by: deftron
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

I'm too lazy to read this entire post to see if someone already pointed out your error here, so here goes:

In your case, what happens if you pick #1, your friend picks #2, and the real prize was in #3?
The host can't open #3 anymore and have it be empty.

The whole point of the problem is that the probabilities are "conditional". On your first pick, all the doors have the same chance of being the prize. But as soon as the host opens up one of the doors (after you have picked one), then the conditional probability of the remaining door being right (conditional on the fact that the door the host opened was wrong), is 2/3.

 

[Hector13]

Originally posted by: conjur


Seems to be a bit of semantics. I still see the remaining, unchosen door as having the same, original odds (1/100 or 1/3...depending on # of original doors).

how could it have the same 1/100 chance? You are missing the new information the host gave you.

The host doesn't leave one door closed at random. Instead, he has to make sure he deliberately leaves the one door that has the prize in it closed (in 99/100 cases, assuming you didn't pick the right door to begin with). Thus, you know that he will open the 98 doors that for sure don't have the prize in them.

 

[Yossarian]

This problem blew my mind! And then it blew it again when I finally understood!!

 

[Looney]

Originally posted by: PipBoy
This problem blew my mind! And then it blew it again when I finally understood!!

Yeah, it's sweet once you understand it... i've explained it to a dozen people at least, and they've always understood it... at least once you take it to the extreme with 100 or 1 million doors.

 

[Venix]

Assume that the prize is behind door 3. Now take a look at what happens depending on which door you choose:

Without switching:
You choose 1. Loss.
You choose 2. Loss.
You choose 3. Win.

With switching:
You choose 1. Win.
You choose 2. Win.
You choose 3. Loss.

When you switch doors, you obviously only win if your original choice was incorrect. 2/3's of the time you'll originally choose an incorrect door, so switching has a higher probability of resulting in a win.