Intersting probability situation - NOW with script

[AntaresVI]

Originally posted by: 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.

::sigh:: the probability is not 50/50 before the first pick.

ok. Pick number 1. You have a 33% chance of being right, 66% chance of being wrong.

One door is taken away. You now must choose between two doors that remain. You know that one is correct and the other is incorrect.

All that the "50% people" are saying here is that, AT THIS POINT IN THE GAME after the choice has been reduced to two, the chance is now 50/50 between the two doors. This is obvious and irrefutable; there are two choices and it does not matter that one has survived the "test" of the last round. To say that one of the choices is more likely to be right is to say that some mystic intuition led the chooser to pick his door initially.

 

[TallBill]

Originally posted by: LeRocks

Originally posted by: 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.

::sigh:: the probability is not 50/50 before the first pick.

ok. Pick number 1. You have a 33% chance of being right, 66% chance of being wrong.

One door is taken away. You now must choose between two doors that remain. You know that one is correct and the other is incorrect.

All that the "50% people" are saying here is that, AT THIS POINT IN THE GAME after the choice has been reduced to two, the chance is now 50/50 between the two doors. This is obvious and irrefutable; there are two choices and it does not matter that one has survived the "test" of the last round. To say that one of the choices is more likely to be right is to say that some mystic intuition led the chooser to pick his door initially.

Come on.. read the thread a little

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

 

[AntaresVI]

Originally posted by: 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.

and THIS is why you don;t understand the problem. How is it not a redeal if you are not given extra information about the door you picked? THe odds to start were 33/33/33, now that there's only two doors left, those odds MUST be redistributed to 50/50. Just because you picked the door in the first round does not make it any more likely to be the right door than the one that you did not choose yet made it to the second round anyway.

 

[Haircut]

LeRocks You just don't get it do you.
I will post again when I posted above, if you think it is wrong can you please show me where the flaw is.

If you had picked the correct door at first (1/100 probability) then the host could remove any 98 of the remaining 99 empty 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.

The other situation is that you pick the incorrect door at first (99/100) probability. The host then can remove all 98 other empty doors. 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.

 

[coder1]

I threw an ASP script together at LINK

This script runs through 200,000 attempts of this scenario. In the script choice 1 and 2 are bad choices or no-wins. Choice 3 is the winner. I had the computer randomly choose a door each time. After the choice it located the other bad choice door and then switched to the only closed door. You can now see that in 200,000 attempts you have on an average 66% chance of getting the prize by switching. Not 50/50

 

[AntaresVI]

Originally posted by: TallBill

Originally posted by: LeRocks

Originally posted by: 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.

::sigh:: the probability is not 50/50 before the first pick.

ok. Pick number 1. You have a 33% chance of being right, 66% chance of being wrong.

One door is taken away. You now must choose between two doors that remain. You know that one is correct and the other is incorrect.

All that the "50% people" are saying here is that, AT THIS POINT IN THE GAME after the choice has been reduced to two, the chance is now 50/50 between the two doors. This is obvious and irrefutable; there are two choices and it does not matter that one has survived the "test" of the last round. To say that one of the choices is more likely to be right is to say that some mystic intuition led the chooser to pick his door initially.

Come on.. read the thread a little

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 HOWEVER YOU DONT KNOW THIS

Case 2
(Odds 2:3)
You pick an empty door
1 Empty door is revealed
1 Correct door is leftHOWEVER YOU DONT KNOW THIS

I completely agree to that. However, it only addresses the first round of the game.

 

[DrPizza]

For whoever it was who still didn't get the 100 doors thing (and I can't see *HOW* anyone couldn't realize they're wrong), look at it from yet a similar situation.

100 doors. 1 million behind one of the doors. You *DON'T* get to change your mind.

If you win, which will be 1 out of 100 times, the host is going to open 98 losing doors for the suspense.
If you lose, which will be 99 out of 100 times, the host is going to open 98 losing doors for the suspense.

1 out of 100 times, you won, and there's still one unopened losing door.
99 out of 100 times, you lost, and there's an unopened winning door.

See it yet???!!

Oh, and to make it 50/50, the host would have to say "now turn around, we're going to reshuffle what's behind the door you chose and the remaining unopened door"

 

[Looney]

Originally posted by: LeRocks

Originally posted by: 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.

and THIS is why you don;t understand the problem. How is it not a redeal if you are not given extra information about the door you picked? THe odds to start were 33/33/33, now that there's only two doors left, those odds MUST be redistributed to 50/50. Just because you picked the door in the first round does not make it any more likely to be the right door than the one that you did not choose yet made it to the second round anyway.

Those odds are redistributed... but how does your first pick odds increase just because another door is opened? IT'S NOT! It stays the same. So the other door must inherit the extra distribution.

 

[TallBill]

LOL, Lerocks, you suck at math. It does not get any easier then my explanation.

 

[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.

 

[DrPizza]

Originally posted by: TallBill
LOL, Lerocks, you suck at math. It does not get any easier then my explanation.

TallBill... what you have to experience is having a class of students, explaining it the way you did *right before* the test... the first 2 minutes of class. It's the sixth time you've done that exact same problem. That's question 3 on the test, and half the students get it wrong anyway.

 

[Looney]

LOL damn, i still can't believe some of you guys aren't getting this.

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?

 

[TallBill]

Originally posted by: DrPizza

Originally posted by: TallBill
LOL, Lerocks, you suck at math. It does not get any easier then my explanation.

TallBill... what you have to experience is having a class of students, explaining it the way you did *right before* the test... the first 2 minutes of class. It's the sixth time you've done that exact same problem. That's question 3 on the test, and half the students get it wrong anyway.

hehe, i could only imagine. im not gonna lie and say i understood it right off the bat, but using a little logic (hey, i got an A in that class) it was pretty damn easy.

 

[DrPizza]

I've got a whole pile of similar such problems where the correct answer goes against common sense... If I get time when I get home, I'll post a couple.

 

[TallBill]

Originally posted by: DrPizza
I've got a whole pile of similar such problems where the correct answer goes against common sense... If I get time when I get home, I'll post a couple.

Looking forward to the problems.. not looking forward to explaining them

 

[conjur]

Originally posted by: Moralpanic
LOL damn, i still can't believe some of you guys aren't getting this.

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?

But you don't have 99 picks (or two picks in the case of 3 doors). You have one pick. So your remaining door also has a 1/100 chance. Each door has an assigned probability (1/100...1/3)...that doesn't change.

 

[Looney]

Originally posted by: conjur

Originally posted by: Moralpanic
LOL damn, i still can't believe some of you guys aren't getting this.

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?

But you don't have 99 picks (or two picks in the case of 3 doors). You have one pick. So your remaining door also has a 1/100 chance. Each door has an assigned probability (1/100...1/3)...that doesn't change.

That was an example... *I* represented the doors you didn't pick.

 

[TallBill]

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

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

Come on folks!

 

[Looney]

Each door has an assigned probability (1/100...1/3)...that doesn't change.

E X A C T L Y!!!! The door you pick initially, stays at 1/100. But the doors you DIDN'T pick, changes, because the host tells you which do not contain the prize... so in essence, he's saying that out of those 99 doors that you didn't pick, these 98 doors do not contain the prize... he reduces 99 doors to 1 door!

 

[conjur]

Originally posted by: Moralpanic

Each door has an assigned probability (1/100...1/3)...that doesn't change.

E X A C T L Y!!!! The door you pick initially, stays at 1/100. But the doors you DIDN'T pick, changes, because the host tells you which do not contain the prize... so in essence, he's saying that out of those 99 doors that you didn't pick, these 98 doors do not contain the prize... he reduces 99 doors to 1 door!

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).

 

[Looney]

Originally posted by: conjur

Originally posted by: Moralpanic

Each door has an assigned probability (1/100...1/3)...that doesn't change.

E X A C T L Y!!!! The door you pick initially, stays at 1/100. But the doors you DIDN'T pick, changes, because the host tells you which do not contain the prize... so in essence, he's saying that out of those 99 doors that you didn't pick, these 98 doors do not contain the prize... he reduces 99 doors to 1 door!

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).

It's not semantics if you think out of 100 doors, and 98 of them are opened, that both doors still only have a 1/100 each of having the prize....

 

[DrPizza]

Actually, Conjur's right.... the remaining door still has a 1/3 chance of being correct...
IF and ONLY IF the host picks a door at random. He didn't pick a door at random though. The deck has been stacked.

Conjur, see my post above about 100 doors and NOT getting to change your mind.

 

[conjur]

Originally posted by: Moralpanic

Originally posted by: conjur

Originally posted by: Moralpanic

Each door has an assigned probability (1/100...1/3)...that doesn't change.

E X A C T L Y!!!! The door you pick initially, stays at 1/100. But the doors you DIDN'T pick, changes, because the host tells you which do not contain the prize... so in essence, he's saying that out of those 99 doors that you didn't pick, these 98 doors do not contain the prize... he reduces 99 doors to 1 door!

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).

It's not semantics if you think out of 100 doors, and 98 of them are opened, that both doors still only have a 1/100 each of having the prize....

That's right...each door only has a 1/100 chance of being right.

If you take into account new information (the opened doors), that leaves 2 doors and, thus, a new equation with 50/50 odds.

 

[TallBill]

Read my post above. Its not gonna get more basic then that. And if you dont get it, you wont get it.

 

[Looney]

Originally posted by: conjur

Originally posted by: Moralpanic

Originally posted by: conjur

Originally posted by: Moralpanic

Each door has an assigned probability (1/100...1/3)...that doesn't change.

E X A C T L Y!!!! The door you pick initially, stays at 1/100. But the doors you DIDN'T pick, changes, because the host tells you which do not contain the prize... so in essence, he's saying that out of those 99 doors that you didn't pick, these 98 doors do not contain the prize... he reduces 99 doors to 1 door!

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).

It's not semantics if you think out of 100 doors, and 98 of them are opened, that both doors still only have a 1/100 each of having the prize....

That's right...each door only has a 1/100 chance of being right.

If you take into account new information (the opened doors), that leaves 2 doors and, thus, a new equation with 50/50 odds.

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