Honestly, that’s the only way I could finally understand it. The host knows which door the prize is behind, and he opens one of the two doors that he knows have no prize. Without his knowledge, with random probability, I couldn’t grasp it.
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The knowledge of the guesser has nothing to do with it. (Edit: This sentence is completely wrong.)
The probability is fixed at the moment the prize is hidden. With three doors, the chance is 1/3 per door. If you then choose one, you have a 1/3 chance of being correct. Someone revealing (with prior knowledge) one of the other incorrect doors doesn’t change the fact that there was a 2/3 chance your guess was wrong. Bringing in a new contestant doesn’t change the fact that of the two remaining doors, one has a 1/3 chance and the other a 2/3 of containing the prize.
An easy way to picture it is to say the host gives you a chance to stay with your 1/3 chance door or switch to BOTH the other doors. Then it is obvious you improve your chances to 2/3. By revealing an incorrect door the host is just concentrating that two thirds into a single choice because he never reveals the correct door (if he did, he would awkwardly reveal the prize one third of the time, and leave you with a 50/50 shot the remaining 2/3 of the time)
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The knowledge of the guesser does matter in that it’s the illustration of the problem. The guesser has to have knowledge of what door was originally picked, otherwise it would be a 50/50 guess regardless of whether there were three or a million doors to start.
No, that’s the knowledge of the host, not the guesser.
I don’t think that really makes a difference. What you KNOW is that he didn’t show you the door with the prize. Whether that was chance or intention what you KNOW NOW is that the prize is not behind the door that he opened. What you know now is that the prize is behind one of the two remaining doors. What if instead of makiing a choice, you told him to open one of the doors, and no prize was behind the door that you told him to open. Your knowledge would be exactly the same as if you had chosen a door hoping that the prize was behind it. You would now know that the prize is behind one of the two remaining doors. The odds of it being behind one of them is 50/50.
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I think this is part of where my brain hangs up. The opening of the doors isn’t fully random since only a door without a prize is opened. Once that is done the distribution behind the doors is now weighted? And we can’t assume two doors means a 50/50 probability? So a new person with no prior knowledge would falsely assume a random distribution between the remaining doors and that they are equally possible, but they are not?
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Bringing in a new guesser doesn’t change the fact that one door is twice as likely to have the prize as the other.
However he has no way of distinguishing, so it is correct, his odds would be 50/50, unless he knew the prior situation and what door the first guesser chose.
The knowledge of the guesser is EVERYTHING. Once the guesser has made a choice, the actual probability that he has picked the right door is either 100% or 0%. He has either picked the prize or not. There IS NO CHANCE. The fact that we ascribe a probability to the outcome is COMPLETELY due to the fact that the guesser and the audience have incomplete knowledge of the system.
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That’s exactly the knowledge I’m talking about. If you don’t know what door was originally picked (i.e., have that knowledge) then you’re making random guess. It’s the knowledge of what door was originally picked that increases the odds.
It’s both. The contestant has to know what door was originally picked, and the host has to know where the prize actually is. That’s what the MHP is. If doors are randomized after the elimination or there is a new guesser who doesn’t have the knowledge, then it’s a random selection of 1/x.
You are correct. I understand the probability, I shouldn’t have said the guesser’s knowledge was not important. It is.
What I was thinking when I said it was that the guesser’s knowledge doesn’t impact the chance a particular door holds the prize. That was decided before the guesser even showed up. The guesser’s knowledge definitely impact the probability of his guess consisting of the correct door, however.
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You can do a Monte Carlo simulation in e.g. Python and demonstrate that it is not 50/50. What you’re feeling is exactly the problem’s noteworthy characteristic: its counter-intuitiveness. Give it some more thought - you’ll get it. (It took me a long time, too, but now seems obvious)
ETA code: https://science-emergence.com/Codes/Monty-Hall-problem-simulation-with-python/
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Any confusion was probably brought about by my wholly non-mathematician and likely sloppy way of illustrating my point. Cheers!
I can see no logic behind the assertion that Monty Hall revealing a no prize door only changes the odds that the door that the contestant has not picked has the prize. Why do the odds THAT door has the prize change to 2/3 while the door that has been chosen stay 1/3?
What if there are two contestents, each picked a door? should they BOTH change.
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Not trying to be flip: Did you watch the Numberphile video in the original post? She makes it very clear what’s going on.
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AnthonyC,
The Monty Haul problem is a completely different one only faced by Dungeon Masters.
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Concentration of probability.
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simonize,
The scenario you describe with two contestants won’t work with three doors, if you think about it for a bit. What if they both guess wrong? Which door does the host reveal?
Goats are also cute with their happy dance. <3
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because door 2’s original probability (1/3) now includes door 3’s (assuming you picked door 1)
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