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The Monty Hall Problem (math riddle)
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2019-03-31 at 9:17 PM UTC
Originally posted by whoami No because when the game is set up there's only 1 out of 3 configurations where your initial choice is the prize, that still holds true after Monty selectively opens one of the doors, meaning the other 2/3 is accounted for by the only remaining option.
But he's always going to remove a door with nothing behind it. It's always going to come down to 2 doors, one of which has a thing and the other doesn't. The fact that he eliminates an irrelevant door doesn't change anything. Math is retarded. Reeeeee.
You can't fool me with your jedi kabbalah maths magic. -
2019-03-31 at 9:58 PM UTC
Originally posted by whoami You double the odds of winning from 1/3 to 2/3 by switching. Most people get this wrong because they treat it as a standard probability problem and don't account for the fact that Monty Hall knows which door has the prize behind it and isn't opening them at random.
There are 3 possible configurations for the objects behind the doors: prize behind 1, goats behind 2 and 3; prize behind 2, goats behind 1 and 3; prize behind 3, goats behind 1 and 2. Assume you pick door 1. In the first configuration, he knows the prize is behind 1 so he will open one of the others. If you switch, you lose, if you stay, you win. In the second configuration, he knows the prize is behind 2 so he'll open 3. If you switch, you win, if you stay, you lose. In the third configuration, he knows the prize is behind 3 so he'll open 2. If you switch, you win, if you stay, you lose. You win 2/3 games by switching and 1/3 by choosing to stay. Simple.
Ok I was thinking in terms of increasing the odds after the goat door is open in which case it's still 50/50. I guess the wording in the OP had me thinking of calculating after its determined to be a binary choice. Thank you -
2019-03-31 at 10:06 PM UTCYou niggers got zonked that was trick question
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2019-03-31 at 10:20 PM UTC
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2019-03-31 at 10:39 PM UTCOld
.The following users say it would be alright if the author of this post didn't die in a fire! -
2019-03-31 at 11 PM UTCi have to be honest he explained it well but i still see it as 50/50The following users say it would be alright if the author of this post didn't die in a fire!
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2019-03-31 at 11:24 PM UTC
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2019-04-01 at 7:10 AM UTC
Originally posted by HTS But he's always going to remove a door with nothing behind it. It's always going to come down to 2 doors, one of which has a thing and the other doesn't. The fact that he eliminates an irrelevant door doesn't change anything. Math is retarded. Reeeeee.
You can't fool me with your jedi kabbalah maths magic.
Right, that's the point. If you think about the situation before the elimination there was a 2/3 chance of the prize being behind one of the doors you didn't pick. There remains a 2/3 chance that you were wrong in your initial pick, even after one of the doors was eliminated, since the other door is the only place the prize could be if it isn't the one you picked, then there's a 2/3 chance of it being there.
It's weird, I didn't believe it at first. I remember thinking about it for a couple of days and being like "no way dude", I wrote a simple program to test it many many times and it turns out the 2/3 chance of winning thing is borne out empirically.
My favorite "riddle" is this one my friend told me where it's basically set up to sound like a reframing of the monty hall problem, but like it's set up differently so the intuitive line of thinking holds (switching does nothing). So smart asses who have heard of the monty hall problem will be like "you gotta switch" without actually thinking through the scenario and be wrong. Something about exploiting people's willingness to recognize the similar setup and immediately assume it's monty hall cracks me up, it's like some kind of brain trap. -
2019-04-01 at 1:25 PM UTC
Originally posted by Lanny Right, that's the point. If you think about the situation before the elimination there was a 2/3 chance of the prize being behind one of the doors you didn't pick. There remains a 2/3 chance that you were wrong in your initial pick, even after one of the doors was eliminated, since the other door is the only place the prize could be if it isn't the one you picked, then there's a 2/3 chance of it being there.
It's weird, I didn't believe it at first. I remember thinking about it for a couple of days and being like "no way dude", I wrote a simple program to test it many many times and it turns out the 2/3 chance of winning thing is borne out empirically.
My favorite "riddle" is this one my friend told me where it's basically set up to sound like a reframing of the monty hall problem, but like it's set up differently so the intuitive line of thinking holds (switching does nothing). So smart asses who have heard of the monty hall problem will be like "you gotta switch" without actually thinking through the scenario and be wrong. Something about exploiting people's willingness to recognize the similar setup and immediately assume it's monty hall cracks me up, it's like some kind of brain trap.
Mossad changed the code to your simple program so that you'd get that result and start to believe their jedi lies. How can you trust empiricism if you are capable of being fooled by jedi trickery? -
2019-04-01 at 1:32 PM UTC
Originally posted by yabbadabbadindunuthin The Monty Hall Problem gets its name from the TV game show, Let's Make A Deal (which is hosted by Monty Hall 1). The scenario is such: you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide “goats” (or some other such “non-prize”
Um goats are a great prize. -
2019-04-01 at 3:01 PM UTC
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2019-04-01 at 3:06 PM UTC
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2019-04-01 at 4:03 PM UTC
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2019-04-01 at 4:06 PM UTCYup, used to own 2 fine animals until the neighbors dog killed them along with a bunch of chickens and quail...so I killed it
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2019-04-01 at 4:08 PM UTC^ didnt you 2 faggits get enough of eachother at cedar creek
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2019-04-01 at 4:33 PM UTC
