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re: Let's discuss the Monty Hall problem (probabilities, odds)
Posted on 8/12/17 at 2:44 am to fightin tigers
Posted on 8/12/17 at 2:44 am to fightin tigers
You have a 1/3 chance of being right from the outset... just because he showed you a goat, doesn't change those odds, you still have a 1/3 chance of being right... 3 doors, and you picked one... the goat increases the odds of the last door to 2/3 by negating the goat's door to zero
He adds a value input to the remaining door, by stripping away the value of the goat's door... your initial door is still at 1/3 because you picked it with no added value input
You have a 1/3 chance of being right from the start, and a 2/3 chance of being right if you switch
They are counting on people being either too arrogant or too uncomfortable to change
He adds a value input to the remaining door, by stripping away the value of the goat's door... your initial door is still at 1/3 because you picked it with no added value input
You have a 1/3 chance of being right from the start, and a 2/3 chance of being right if you switch
They are counting on people being either too arrogant or too uncomfortable to change
3
Posted on 8/12/17 at 2:48 am to TheArrogantCorndog
Say you were in the business of giving away your own money.
Would you offer people to switch off of a goat and onto your own money?
The math part makes sense in theory. Using a gameshow to explain this is bunk.
Would you offer people to switch off of a goat and onto your own money?
The math part makes sense in theory. Using a gameshow to explain this is bunk.
Posted on 8/12/17 at 3:02 am to TheArrogantCorndog
Let's keep it easy and always pick door 1... keep in mind he can't show you door 1 (your pick) and he can't show you the car
1 2 3 stay switch
C G G W L
G C G L W
G G C L W
2/3 odds if switch
1/3 odds if stay
1 2 3 stay switch
C G G W L
G C G L W
G G C L W
2/3 odds if switch
1/3 odds if stay
Posted on 8/12/17 at 12:53 pm to TheArrogantCorndog
quote:
You have a 1/3 chance of being right from the outset... just because he showed you a goat, doesn't change those odds, you still have a 1/3 chance of being right... 3 doors, and you picked one... the goat increases the odds of the last door to 2/3 by negating the goat's door to zero
He adds a value input to the remaining door, by stripping away the value of the goat's door... your initial door is still at 1/3 because you picked it with no added value input
You have a 1/3 chance of being right from the start, and a 2/3 chance of being right if you switch
I get all of this.
Here is my question, though. There are two goats, so no matter what door you pick, he's going to be able to show you a door with a goat behind it.
When he asks you if you want to switch, if you ignore the door you initially picked, you have a 50/50 chance of picking the door with prize behind it at that point. Is that incorrect?
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