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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:53 am
(This post was last modified: March 13, 2016 at 8:54 am by ErGingerbreadMandude.)
(March 13, 2016 at 8:35 am)robvalue Wrote: What?
You're screwing with me. You must be. What is "statistical reasons"? Yes, the probability is in your favour if you switch. I don't know how much clearer I can make it.
Haha. OK fine, I'll leave you be then.
Tell you what rob, I'll explain it to you since you were patient enough to explain things to me before. But I suggest you read carefully.
Consider 5 tries.
First try you get car.
Second try you get car.
Third try you get car.
Fourth try you get goat.
Fifth try you get car.
Yes, the probability of you getting the car is higher in 'x' tries.
Now you decide to switch next time also because you got car most times when you decided to switch.
You choose to do this because of statistical reasons not because of probabilistic reasons. Because you look at the results and you understand that it is beneficial to switch in 'x' tries(emphasis on 'x' tries).
Now get this rob, you are in a show, you are given only 1 chance. How can you say that your chances of winning a car increases if you switch?
Yes, if there is 5 tries you get a car most times if you switch. But what if there is only 1 try?
Believing that you will get a car because you switched all the 5 times and will therefore get a car the next time is analogous to the situation where you get heads 1 million times after flipping a coin and believing that you will get a heads next time around too.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:54 am
(March 13, 2016 at 8:51 am)robvalue Wrote: I don't think we're even talking about the same game.
You pick a door, the guy opens a door (which is not your door, or the prize door. Important!!!)you stick or change, he reveals the winning door. Game over.
I'm saying if you repeat this game N times, with a genuinely random placement of the prize between the 3 doors each time, on average you will win 2/3 of the time by changing.
It's the mean of a probability distribution. Like I say, if you don't know what that is, you need to learn about it.
I have a first class maths degree, I know what I'm talking about here.
If by genuinely random placement you mean each door will be picked the same amount of times through n iterations of the same game, then yes.
You can either address this or not, but whatever degree you have is irrelevant.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:55 am
(This post was last modified: March 13, 2016 at 8:56 am by robvalue.)
It's a probability calculation. If you assume the prize was genuinely randomly placed, you have a 2/3 chance to win if you switch. It's that simple. I don't know what the confusion is. It's not based on observations, it's based on exact statistical analysis.
It's like saying you can pick 1-2 on a dice, or 3-6. Which would you choose? You pick 3-6, because it's more likely to win if someone isn't screwing with the dice.
EP: address what? It's meant to be randomly placed initially. That's the point. If it isn't, I'm not talking about the same game as you.
The door he opens for you, however, is not random.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:57 am
(This post was last modified: March 13, 2016 at 8:58 am by Excited Penguin.)
Edit ^^
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:58 am
(March 13, 2016 at 8:55 am)robvalue Wrote: It's a probability calculation. If you assume the prize was genuinely randomly placed, you have a 2/3 chance to win if you switch. It's that simple. I don't know what the confusion is. It's not based on observations, it's based on exact statistical analysis.
It's like saying you can pick 1-2 on a dice, or 3-6. Which would you choose? You pick 3-6, because it's more likely to win if someone isn't screwing with the dice.
EP: address what? It's meant to be randomly placed initially. That's the point. If it isn't, I'm not talking about the same game as you.
The door he opens for you, however, is not random.
What does random even mean?
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:58 am
(This post was last modified: March 13, 2016 at 9:03 am by robvalue.)
No wait. It doesn't even matter if it's random or not. As long as I choose my first door randomly, he can't psych me out.
What does random mean?
Edit, sorry being rude.
Random means it has the same chance of being in every slot.
You guys are either messing with me, or you're seriously missing how the host works out which door to open. It's not random.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:01 am
(This post was last modified: March 13, 2016 at 9:02 am by Excited Penguin.)
(March 13, 2016 at 8:58 am)robvalue Wrote: No wait. It doesn't even matter if it's random or not. As long as I choose my first door randomly, he can't psych me out.
Whether it's random or not is irrelevant. In order for you to know that you truly have a 2/3 chance to win if you switch you would have to also know that no matter how many times the game is played, each door will be chosen the same amount of times. For example, if they play it 999 times, then door 1 will be picked 333 times, door 2 333 times and door 3 333 times. Otherwise, door 1 could be picked 999 times and your switch strategy means shit if that's the door you're switching to and there's a goat behind it.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:03 am
(March 13, 2016 at 8:58 am)robvalue Wrote: No wait. It doesn't even matter if it's random or not. As long as I choose my first door randomly, he can't psych me out.
What does random mean?
Edit:
Random means it has the same chance of being in every slot.
You guys are either messing with me, or you're seriously missing how the host works out which door to open. It's not random.
What does "has the same chance of being in every slot" mean?
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:04 am
(This post was last modified: March 13, 2016 at 9:06 am by robvalue.)
If I randomly choose a number from 1 to 3, I have the same chance of picking each one.
Whether any real process is actually random is another matter.
No. I don't need to know that. If I choose randomly, then I can't be psyched out. It doesn't matter where he puts it, I have the same chance of being initially right. All that matters is whether I'm initially right, or not.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:08 am
(This post was last modified: March 13, 2016 at 9:10 am by ErGingerbreadMandude.)
Roby, you win most times if you have more than one try and you decide to switch.
The conclusion is :
Therefore it is beneficial to switch given that you have more than a single try because you get more cars and less goats.
The conclusion is not :
Therefore it is beneficial to switch.
You don't have multiple tries in the question.
Are you seeing the fault in your logic?
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