Common sense is useless in a discussion of scientific topics.

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posted on Aug, 31 2013 @ 06:36 AM
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Barcs,

re: " Obviously if you are allowed to choose 2 doors instead of 1, your odds go up substantially. "

And my question remains: What's the difference if you personally open #3 or the host opens #3 for you? Either way you get to look at both #2 and #3.




posted on Aug, 31 2013 @ 10:51 AM
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reply to post by rstrats
 




As I asked previously, if after your initial pick of curtain #1, and before the host opens any curtains, I told you that you could stay with #1 or switch to BOTH #2 and #3, what would you do? My guess is that you would switch so that you could look behind both curtains #2 and #3 in spite of the fact that you KNOW that at least one of the curtains does not have a car behind it. Lets say that the first curtain that you open is #3 and you find that there is no car behind it. You now get to open curtain # 2, giving you 2 chances. What is the difference if you personally open curtain #3 or the host opens curtain #3 for you?


The difference is that it is two different games. Using your Area A and Area B scenario, you are exactly right Area A has 1/3 and Area B has 2/3, and it doesn't make any difference who opens the curtain.

But as soon as you are given a new choice between you original curtain and the remaining 'Area B' curtain, Area A and Area B DISAPPEAR and the original probabilities collapse.

Instead you now have Area X and Area Y, and you are free to choose between the two of them. The fact that one of them is your original choice is irrelevant, you are free to choose between either curtain. You have a binary decision: Area X or Area Y. Two curtains, one car, one goat.

This decision is entirely INDEPENDENT of the previous decision. It is an entirely new game, with an entirely new probability distribution.



posted on Aug, 31 2013 @ 12:17 PM
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Originally posted by rstrats



Barcs,

re: " Obviously if you are allowed to choose 2 doors instead of 1, your odds go up substantially. "

And my question remains: What's the difference if you personally open #3 or the host opens #3 for you? Either way you get to look at both #2 and #3.


I already answered that. There is no difference, except the host knows the outcome, but you are thinking of a different game. You do not get to look at both 2 and 3 until the game is over, and odds are irrelevant at that point. You keep diverting the topic from the simple fact that choosing between 2 doors = 50% choice at that moment. You are asking non sequitars irrelevant to the game we are discussing.
edit on 31-8-2013 by Barcs because: (no reason given)



posted on Sep, 1 2013 @ 05:24 AM
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Barcs,

re: "There is no difference..."

So if there is no difference, why wouldn't you jump at the chance to get into area "B" where there is a 2/3 chance that that is where the car is located?



re: "You keep diverting the topic from the simple fact that choosing between 2 doors = 50% choice at that moment."

In reality you'd be choosing between area "A" and area "B". There is 1/3rd chance the car is in area "A" and a 2/3rd chance it's in area "B".



posted on Sep, 1 2013 @ 05:33 AM
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rnaa,

re: "But as soon as you are given a new choice between you original curtain and the remaining 'Area B' curtain, Area A and Area B DISAPPEAR and the original probabilities collapse."

That is incorrect. Area "A" still exists with one unopened curtain in it and area "B" still exists with one opened and one unopened curtain in it.



posted on Sep, 1 2013 @ 12:11 PM
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Originally posted by rstrats


Barcs,

re: "There is no difference..."

So if there is no difference, why wouldn't you jump at the chance to get into area "B" where there is a 2/3 chance that that is where the car is located?



re: "You keep diverting the topic from the simple fact that choosing between 2 doors = 50% choice at that moment."

In reality you'd be choosing between area "A" and area "B". There is 1/3rd chance the car is in area "A" and a 2/3rd chance it's in area "B".


Nope. One door gets eliminated and the odds change. It's 50/50 every time. Area A vs Area B is relative and no longer exists once a door is eliminated.



posted on Sep, 1 2013 @ 05:03 PM
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Barcs,

Since it's apparent that I'm not able to convince you that you are incorrect with your reasoning, you might want to type "Monty Hall Problem" in your browser's search block. A number of web sites will come up that explain why you should always switch because that will double your chance of picking the curtain with the car.



posted on Sep, 1 2013 @ 05:08 PM
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Common sense has nothing to do with science and factual information.

For example, to me it makes sense that mountaintops would be far hotter than the surrounding land- Mountain Tops should be like deserts (in my common sense) because they are closer to the Sun.

The Earth looks flat to me- Common sense tells me if it were round, at some point I would fall off into the sky.

-Now granted- I totally understand that what I just said is not true- But if I didnt have some scientific foundation (which trumps common sense) I would argue these points to the end of time.

Nothing to do with the video , I think- But the premise is true. Too much in science trumps what my common sense tells me SHOULD be true...Like why we dont fly off the Earth if its spinning so fast (I still dont get this...) because if I put some ants on a baseball and spin it fast enough, they would fly off-



posted on Sep, 2 2013 @ 01:29 PM
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reply to post by rstrats
 


You know what. I think I am actually wrong. I ran a simulation with all possible choices, and in 2 of 3 scenarios the car is chosen by switching, which means by switching you do indeed have a 66% chance. It doesn't seem logical, but based on simulations it seems accurate.

Here are all possible arrangements of the game.

GAME ONE:
D1__D2__D3
_C__G__G

GAME TWO:
D1__D2__D3
_G__C__G

GAME THREE:
D1__D2__D3
_G__G__C


In Game one, you pick door 1. You switch you lose
In Game one, you pick door 2. You switch you win
In Game one, you pick door 3. You switch you win

In Game two, you pick door 1. You switch you win
In Game two, you pick door 2. You switch you lose
In Game two, you pick door 3. You switch you win

In Game three, you pick door 1. You switch you win
In Game three, you pick door 2. You switch you win
In Game three, you pick door 3. You switch you lose

You only lose by switching 1/3 of the time, and that is the 1/3 of the time when you luckily pick the car.

I would still argue that at the time of choice, it's 50/50, but in the grand scheme of the game, switching wins more than it loses so it would be a viable strategy.
edit on 2-9-2013 by Barcs because: (no reason given)





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