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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 9:16 am by Mr.wizard.)
(March 13, 2016 at 8:51 am)Excited Penguin Wrote: (March 13, 2016 at 8:49 am)Mr.wizard Wrote: The probability does not change, if you use a pick and hold strategy you will win 1/3 of the time, if you use a pick and switch strategy you will win 2/3 of the time, it's that simple, no it does not guarantee you a car but it is in your best interest to switch mathematically.
I repeat myself, that would only be true if you knew for sure that no matter how many times the game is played, the number of times each door is picked will be the same, for instance 999999 times each.
I have no idea what you are trying to say, the probability is right, the math is right, I can't figure out what your problem is.
The number of times each door is opened is irrelevant, your choosing between a random choice of 1/3 and curated filtered choice at 2/3, every time the game is played. Somebody gave the example of 1 out of 100 doors, Monty opens 98 goats and re-offers you the choice, do you stick with your random 1/100 pick or switch to the filtered pick?
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:09 am
(This post was last modified: March 13, 2016 at 9:12 am by robvalue.)
What?
Apply that logic to rolling a die. Will you pick 1-2, or 3-6?
You pick the thing most likely to win. Changing doors is most likely to win.
You don't seem to understand probability at all.
The average holds even if it's one attempt.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:12 am
(This post was last modified: March 13, 2016 at 9:12 am by ErGingerbreadMandude.)
Yes,the math is right. The answer is wrong.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:13 am
You don't understand probability, that's all this is showing.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:25 am
(This post was last modified: March 13, 2016 at 9:26 am by robvalue.)
OK. I'll give it one last try. I'm still not convinced you guys aren't messing with us.
Let's start at the beginning. The host can put the car wherever he wants. Behind any of the 3 doors, I don't care.
I decide randomly which door I'll pick first. Say I roll a dice. 1-2 I pick door 1, 3-4 I pick door 2, 5-6 I pick door 3.
Do you agree that I have a 1/3 chance of initially picking the right door?
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:25 am
(This post was last modified: March 13, 2016 at 9:35 am by Mr.wizard.)
(March 13, 2016 at 9:08 am)pool the great Wrote: 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?
Wrong, the math is still the same, you initially pick with a 1/3 random chance and then are re-offerd a filtered choice, mathematically it's in your best interest to switch whether you play once or a hundred times. Your always starting out at a guess from 1/3, meaning your chance is always 33% if you stay with your door.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:39 am
(This post was last modified: March 13, 2016 at 9:46 am by robvalue.)
Yes, they agree. If they don't, we're really in trouble.
I have probability 1/3 that I'm in front of the right door. It doesn't matter if monty is a sneaky cunt because I've chosen randomly, he can't psych me out. Take that Derren Brown.
So I have a 2/3 probability I chose wrong.
Now, he must open one of the other two doors. So if I chose wrong initially, I'm now guaranteed to win if I switch. The prize can only be behind the remaining door.
This situation has probability 2/3. The probability I switch and lose only happens when I chose right initially, probability 1/3.
In other words, I'm planning to choose the two doors I don't pick initially. Monty picks one for me, I pick the other. If it's behind either of those, I win. If it's behind the one I pick initially, I lose. As I have 2 chances to win, it's 2/3.
Here's the crucial part: him opening a door gives me no more information about whether I picked right initially, or not. He is making a forced choice (although it's random between the two if I chose right). It looks the same to me, either way.
Now, if monty could also open my door, the game is up. I have no advantage. It's because he can't that I get an instant win 2/3 of the time.
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:50 am
If it is beneficial to switch rob, why did you return with a goat last time we played this game?
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:50 am
Did....
Did it work?
Am I free.....?
Freeeddoooommmmmmmmm!
http://youtu.be/2mc13fBlhOg
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RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 9:52 am
(This post was last modified: March 13, 2016 at 9:53 am by robvalue.)
(March 13, 2016 at 9:50 am)pool the great Wrote: If it is beneficial to switch rob, why did you return with a goat last time we played this game?
Because you hadn't put the prize anywhere. You were just going to say I lost whatever I picked.
Monty places the prize, before I pick a door. That's the game. You didn't do that, or if you did, I got unlucky.
If you weren't cheating, I had a 2/3 chance.
I don't know why you're so insistent on trying to cheat the game.
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