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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 9:45 pm
(This post was last modified: March 8, 2016 at 9:52 pm by Jenny A.)
If it's arrogant to think you are doing the math wrong than I'm arrogant. But I'm sure enough to give you over two to one odds that I am right. Come play. I've always wanted to visit Budapest. AND I'd be happy to go on your dime.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 9:53 pm
I feel sorry for entering this childish debate. Grow up, come up with a good argument and we'll talk.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 9:58 pm
(This post was last modified: March 8, 2016 at 10:01 pm by Jenny A.)
The proof is here. For me it would be like betting that two plus two is four. http://betterexplained.com/articles/unde...l-problem/ Well not quite, a ridiculous string of bad luck could lose it for me, but in a thousand tries the odds of that would be extraordinarily low.
What exactly is is it that you find wrong about the math? You can test it. Test, or give it up.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:01 pm
That's not proof. I don't know how that game works.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:03 pm
(This post was last modified: March 8, 2016 at 10:08 pm by Excited Penguin.)
You can find me any proof you like, it still doesn't make sense that probability helps you in that scenario. I will say it's all luck since you can't explain it.
As I already said, the fact that the moderator takes out all the bad variants but for one doesn't tell you anything about whether you made the right decision or not. Just the fact that he takes out all the other doors, doesn't mean that he made the unchosen left variant any more plausible than the one you already chose. And that's because no matter if you were wrong or right at first he was going to take out the same number of doors after you made your choice.
In some cases people refer to the 2/3 logic. That's bad logic. It's not 2/3 anymore because there's no three left. There are only two options left and so it's 50/50, not 25/75.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:08 pm
(This post was last modified: March 8, 2016 at 10:08 pm by Jenny A.)
(March 8, 2016 at 10:03 pm)Excited Penguin Wrote: You can find me any proof you like, it still doesn't make sense that probability helps you in that scenario. I will say it's all luck since you can't explain it.
As I already said, the fact that the moderator takes out all the bad variants but for one doesn't tell you anything about whether you made the right decision or not. Just the fact that he takes out all the other doors, doesn't mean that he made the unchosen left variant any more plausible than the one you already chose. And that's because no matter if you were wrong or right at first he was going to take out the same number of doors after you made your choice.
Translation: there is no proof I will accept. Now that's childish.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:12 pm
(This post was last modified: March 8, 2016 at 10:13 pm by Excited Penguin.)
The math doesn't make any sense. As I already said, I will accept proof but it won't be proof of what you think it is. It will merely be proof of a particular pattern, which I already explained is how probability works in the first place. But having no previous knowledge of any patterns and facing this game at first you have no good reason to either switch or not. It only comes down to preference, and since you have no reason to switch and you've already made your choice, I'm simply assuming you won't change your mind for no reason at all when I vote no in the poll.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:24 pm
(This post was last modified: March 8, 2016 at 10:36 pm by Jenny A.)
There are no patterns to consider. One door always has the prize, and the others don't. Choose one door and you have a one third chance of winning. Choose two doors and you have a two thirds chance of winning. After Monty's reveal switching gives you effectively two rather than one doors. Staying with the first door leaves you with a one third chance.
EDIT: Look around it this way. Suppose you had the choice of choosing one door or choosing two doors initially with the understanding that Monty would show you a goat behind one of the two doors you choose and you'd get the remaining door of your two door choice. That's what's actually happening if you switch.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 10:30 pm
(This post was last modified: March 8, 2016 at 10:36 pm by Excited Penguin.)
You have a third chance of winning no matter what you do. Revealing a door's prize doesn't change the door you chose at first. It doesn't say that you chose the wrong one or the right one, it merely tells you that you don't know that you lost yet. If he revealed the right door you would know that you lost.
To better illustrate this consider that you chose the right door first and then, after the reveal you switch and choose the wrong one.
Now consider that you chose the wrong one first and then after the reveal you chose the right one.
Either way, he will still reveal a false option and that won't tell you anything about how likely is your option to be true or not. It all comes down to what door you choose first if you're going with the switch method. If you choose the right one more often than not, and there's nothing to say that you won't, because there's nothing to say that you're more likely to choose a wrong door than a right one, then you'll lose the game more often than not(because of your switch method).
The math may very well make sense theoretically, but it doesn't apply in the real world at all in this case. It's an illusion, albeit a very strong one, by the looks of it.
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RE: The role of probability in solving the Monty Hall problem
March 8, 2016 at 11:28 pm
(This post was last modified: March 8, 2016 at 11:38 pm by Excited Penguin.)
(March 8, 2016 at 10:24 pm)Jenny A Wrote: There are no patterns to consider. One door always has the prize, and the others don't. Choose one door and you have a one third chance of winning. Choose two doors and you have a two thirds chance of winning. After Monty's reveal switching gives you effectively two rather than one doors. Staying with the first door leaves you with a one third chance.
EDIT: Look around it this way. Suppose you had the choice of choosing one door or choosing two doors initially with the understanding that Monty would show you a goat behind one of the two doors you choose and you'd get the remaining door of your two door choice. That's what's actually happening if you switch. Then the choice has been 50/50 all along, and it doesn't change in the least. You're just either in on it or not, that's all that changes. The chances of you being right, however, are the same throughout.
At the start of the game, running the original experiment, from your perspective you have a 1/3 chance of getting right. From the moderators' perspective and the one that is actually right since he has all the information you have a 50/50 chance. So this is not so much a puzzle as it is a trick. No mystery whatsoever here, the chances stay the same from the beginning, only your understanding of them improves, not so the actual chances.
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