do you switch your door?

[ElFenix]

Originally posted by: DAGTA

Originally posted by: cubby1223

Originally posted by: DAGTA
The reality is that the door that is opened is completely eliminated from the decision process. You then have a 50% chance of winning by choose to stay with one door or switch to the other.

The problem is not with your programming skills. The problem is with how you are defining the model.

Just answer this question - what are the odds of initially selecting the 1 correct door out of 3 total doors?

1 out of 3 = 33%

Then a bad door is revealed

Now you have a new decision. You again have to select a door, from 2 total doors.
If you stay with the one you picked the first time, you are still making a choice.

1 out of 2 = 50%

The first decision becomes meaningless and the only odds that matter are the final odds of deciding on one door out of two.

the first decision does not become meaningless because the door monty opens is dependent 2/3 of the time on the door you pick. as he can only pick a door with a goat behind it, if you by sheer chance pick a door with a goat behind it there will only be one door he can open. because 2/3 of the time you pick a door with a goat behind it, 2/3 of the time he will open the only other door with a goat behind it, and 2/3 of the time the door you did not pick and he did not open will have a car behind it. it is NOT a new decision.

 

[DAGTA]

Originally posted by: ElFenix

Originally posted by: DAGTA

Originally posted by: cubby1223

Originally posted by: DAGTA
The reality is that the door that is opened is completely eliminated from the decision process. You then have a 50% chance of winning by choose to stay with one door or switch to the other.

The problem is not with your programming skills. The problem is with how you are defining the model.

Just answer this question - what are the odds of initially selecting the 1 correct door out of 3 total doors?

1 out of 3 = 33%

Then a bad door is revealed

Now you have a new decision. You again have to select a door, from 2 total doors.
If you stay with the one you picked the first time, you are still making a choice.

1 out of 2 = 50%

The first decision becomes meaningless and the only odds that matter are the final odds of deciding on one door out of two.

the first decision does not become meaningless because the door monty opens is dependent 2/3 of the time on the door you pick. as he can only pick a door with a goat behind it, if you by sheer chance pick a door with a goat behind it there will only be one door he can open. because 2/3 of the time you pick a door with a goat behind it, 2/3 of the time he will open the only other door with a goat behind it, and 2/3 of the time the door you did not pick and he did not open will have a car behind it. it is NOT a new decision.

Thank you.

I finally see how information can be revealed using the first decision.

 

[Suspicious-Teach8788]

Originally posted by: Gibson486

Originally posted by: KillyKillall

Originally posted by: joedrake
OMFG
you've been reading too much Da Vinci code or w/e
in no way at all does that make sense
its flucking random, go back to 3rd grade math-class

Um...they don't discuss this type of probability until junior high... I recommend you read up on it.

Also, this is a college porbabilty porblem....

They talk bout this in college. If you take math for computer science they will talk about this. No, you won't learn it in probability in high school nor have I learend it in my stats 1 class...

This is known as the Monty HALL problem. If you dispute it, you're retarded. I can understand if you don't GET IT, but don't tell me its not true, because a computer simulation will show you that it is...

 

[Gibson486]

Originally posted by: ElFenix

Originally posted by: DAGTA

Originally posted by: cubby1223

Originally posted by: DAGTA
The reality is that the door that is opened is completely eliminated from the decision process. You then have a 50% chance of winning by choose to stay with one door or switch to the other.

The problem is not with your programming skills. The problem is with how you are defining the model.

Just answer this question - what are the odds of initially selecting the 1 correct door out of 3 total doors?

1 out of 3 = 33%

Then a bad door is revealed

Now you have a new decision. You again have to select a door, from 2 total doors.
If you stay with the one you picked the first time, you are still making a choice.

1 out of 2 = 50%

The first decision becomes meaningless and the only odds that matter are the final odds of deciding on one door out of two.

the first decision does not become meaningless because the door monty opens is dependent 2/3 of the time on the door you pick. as he can only pick a door with a goat behind it, if you by sheer chance pick a door with a goat behind it there will only be one door he can open. because 2/3 of the time you pick a door with a goat behind it, 2/3 of the time he will open the only other door with a goat behind it, and 2/3 of the time the door you did not pick and he did not open will have a car behind it. it is NOT a new decision.

Correct.....

Here is the math:

You are given three doors to choose from:

P(picking the right door) = 1/3.

After the door is picked, he shows a door that will have no car. You now have 2 doors, so

P(pick right in 2nd try) = 1/2

In probabilty, if you have a given condition to calculate you use

P(A|B) = P(AB) / P(B)

That the probability picking the right door(A), given that the host took away one pick(B), is equal to the intersection (AB [means where they share common results ie picking the right door even after the host takes a door away]) devided by the probaility of the condition (B).

P(picking the right door|pick right on 2nd try) = (1/3)/(1/2) = 2/3.

 

[cubby1223]

Originally posted by: Gibson486
Correct.....

Here is the math:

<snip>

I'd hate to prolong this thread much longer, but your math & reasoning is completely wrong from start to finish.

The end goal is trying to find P(pick right in 2nd try)

What you described was a situation where if the contestant chooses correctly the first time, he wins all the time. And if he chooses wrongly the first time, he gets a second 50/50 chance to win. Which only by pure coincidence comes out to 2/3 chance.

Just read the link above, it explains all.

 

[cKGunslinger]

Originally posted by: DAGTA
1 out of 3 = 33%

Then a bad door is revealed

Now you have a new decision. You again have to select a door, from 2 total doors.
If you stay with the one you picked the first time, you are still making a choice.

1 out of 2 = 50%

The first decision becomes meaningless and the only odds that matter are the final odds of deciding on one door out of two.

Incorrect, the first decision is NOT meaningless. What is meaningless is being shown an empty (or goat-filled) door. Since he will *always* be able to show you an empty door, no matter what which door you pick (100%,) that is the the meaningless items here. So your choice is *not* 1 out of 2, but still 1 out of the original 3 (leaving a 2-out-of-3 cahnce it's one of the other doors you didn't pick,) as him showing you an empty door has *0* effect on your decision. It comes down to "Is the prize behind the 1 door I chose, or behind one of the 2 I didn't choose." Which is a 33% of you winning is you don't switch, and a 66% if you do.

Think about the example with 100 doors. You pick one. That leaves 99. Now what are your odds? 50% or 1%? There's a 1% chance your door is correct, and a 99% it's one of the other doors. If you could switch right now from your door, to ALL 99 other doors, would you? Of course you would. Would knowing that at least 98 of those other doors are empty stop you? No. So being shown those 98 empty doors does nothing to change the original odds. Your orginal choice still has 1% odds. The non-opened door of the remaining 99 still has the 99% chance.