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Can someone explain to me why?

Loserboi
Posts: 1,232
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1/21/2011 7:14:15 PM
Posted: 5 years ago
Ok, at my job there is an exhibit where it requires two people.

There is a money bar and 4 doors. A person chooses a door and I lift up two doors that does not have the money bar. The person now has a 75% of getting it right in finding the money bar if he switches his choice, why is that?

This was also in the movie 21 but it had 3 options.
How come switching now gives you a 75% chance probability of getting it right?
SuperRobotWars
Posts: 3,906
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1/21/2011 7:23:42 PM
Posted: 5 years ago
Wha ? ! ? ! ? !
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Loserboi
Posts: 1,232
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1/21/2011 7:45:34 PM
Posted: 5 years ago
At 1/21/2011 7:23:42 PM, SuperRobotWars wrote:
Wha ? ! ? ! ? !

door 1 nothing
door 2 nothing
door 3Money
door 4 nothing

I know its in door 3

You choose door 2

I reveal door 1 and 4 to show it has nothing

You are left with door 2 and 3

Why is it that switching your answer from door 2 to door 3 gives you a 75% chance of being right?
Loserboi
Posts: 1,232
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1/21/2011 7:47:14 PM
Posted: 5 years ago
At 1/21/2011 7:31:18 PM, tvellalott wrote:
http://en.wikipedia.org...

i dont really understand it

So if there was a 75% of being wrong and i showed you two wrong and the one you did not choose gets the 50% chance from the wrongs i revealed?

Aren't your chances still either your right or wrong? 50%?
FREEDO
Posts: 21,057
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1/21/2011 8:50:18 PM
Posted: 5 years ago
It's not mathematical. It's a mind trick. One that is often used on game shows. When you are asked if you want to change it is highly likely you won't(unless you know the trick) and the person who is trying to trick you knows this.
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lovelife
Posts: 14,629
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1/21/2011 9:00:17 PM
Posted: 5 years ago
At 1/21/2011 8:50:18 PM, FREEDO wrote:
It's not mathematical. It's a mind trick. One that is often used on game shows. When you are asked if you want to change it is highly likely you won't(unless you know the trick) and the person who is trying to trick you knows this.

But then they wouldn't ask, but then they would know you would know they wouldn't ask since they knew that you knew the trick so they would ask, but they would take that into consideration.......

basically try to read the guy. People tend to give signs if their bluffing
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tvellalott
Posts: 10,864
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1/21/2011 9:27:36 PM
Posted: 5 years ago
At 1/21/2011 7:47:14 PM, Loserboi wrote:
At 1/21/2011 7:31:18 PM, tvellalott wrote:
http://en.wikipedia.org...

i dont really understand it

So if there was a 75% of being wrong and i showed you two wrong and the one you did not choose gets the 50% chance from the wrongs i revealed?

Aren't your chances still either your right or wrong? 50%?

No.
In the case of the monty hall problem, it works like this.
You start off picking a door.
One door has a new car and the other two doors have goats.
You have a 1/3 chance of selecting the right door.
The host comes along and opens a door with a goat, since he knows which is the winning door.
Now, you're given the option to stick with your original choice or switch.
This is where the confusion is.
It SEEMS like you have a 1/2 chance now of winning.
However, you don't.
If you stick with your original choice, the odds are 1/3
If you SWITCH your choice to the other case, the odds are 2/3

The explaination is that by switching, you're actually getting two cases, one of which you know if wrong.

Consider it like this...
CAR - GOAT - GOAT
GOAT - CAR - GOAT
GOAT - GOAT - CAR

The first column is your first choice. You see that there is a 1/3 chance.
However, consider the fact by switching you're actually getting two cases with one eliminated; by switching you will win in two out the three scenarios.
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tvellalott
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1/21/2011 9:31:12 PM
Posted: 5 years ago
So in your example...

EMPTY - EMPTY - EMPTY - MONEY BAR
EMPTY - EMPTY - MONEY BAR - EMPTY
EMPTY - MONEY BAR - EMPTY - EMPTY
MONEY BAR - EMPTY - EMPTY - EMPTY

The first row shows the chance of getting it right...

By eliminating two wrong choices from the equation in the second round, you're giving them three of the original four choices.
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Ragnar_Rahl
Posts: 19,297
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1/21/2011 10:36:24 PM
Posted: 5 years ago
The easiest explanation, as wiki noted, is to take the notion that it's half to an absurd conclusion. There are a million doors now, one prize. You pick one door. The host gets rid of 999998 doors. Is it more likely that you picked the one right one the first time, or one of the 999999 wrong ones?

Is it more likely that you picked the one right one the first time in the original, or one of the two wrong ones?
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rogue
Posts: 2,325
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1/22/2011 1:22:38 AM
Posted: 5 years ago
At 1/21/2011 7:14:15 PM, Loserboi wrote:
Ok, at my job there is an exhibit where it requires two people.

There is a money bar and 4 doors. A person chooses a door and I lift up two doors that does not have the money bar. The person now has a 75% of getting it right in finding the money bar if he switches his choice, why is that?

This was also in the movie 21 but it had 3 options.
How come switching now gives you a 75% chance probability of getting it right?

Don't worry it took a long time for me to understand this too.
bluesteel
Posts: 12,301
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1/22/2011 1:50:41 AM
Posted: 5 years ago
At 1/21/2011 9:31:12 PM, tvellalott wrote:
So in your example...

EMPTY - EMPTY - EMPTY - MONEY BAR
EMPTY - EMPTY - MONEY BAR - EMPTY
EMPTY - MONEY BAR - EMPTY - EMPTY
MONEY BAR - EMPTY - EMPTY - EMPTY

The first row shows the chance of getting it right...

By eliminating two wrong choices from the equation in the second round, you're giving them three of the original four choices.

Using tv's chart, let's say your strategy is to "choose door 1 first and then switch to whichever door is left after eliminating two"

Line 1: you get the money bar (eliminate door 2 and 3)
Line 2:you get the money bar (eliminate door 2 and 4)
Line 3: you get the money bar (eliminate door 3 and 4)
Line 4: you lose

that's 75% odds
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