ArtificialAngel

gabriel (2 inches taller)

@ArtificialAngel

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Alastar Gabriel (but you can call me anything). I'm an ex-professional software developer, now I make weird art and music :p I will give you bug facts unprompted

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ArtificialAngel
@ArtificialAngel
eniko
@eniko
eniko
@eniko

I hate it when people bring up the Monty Hall problem because I can never remember why it works and I have to go look it up and work through it in my head and it's all very unpleasant.

Anyway having just done that, here's my explanation. Start from the assumption that the door you picked at random isn't the right door. That means one of the two other doors has the car. Monty cannot open the door with the car, and your initial choice was incorrect, so he must open the door in the remaining pair of doors that doesn't have the car.

But if your initial guess was wrong, then Monty is effectively showing you where the car is. There are three doors, you picked one that had a goat. He showed the other goat. The car is behind the door he didn't open and you didn't pick.

Said another way, if your initial guess was wrong, switching will give you the car. Your chance of being wrong initially is 67%, so you have a 67% chance of winning by switching vs the 33% chance of your initial guess being right.

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in reply to @eniko's post:

modulusshift
@modulusshift

I love the trick this plays on psychology. people naturally want to commit to a choice they've made, even with no information, just because they made it. Monty Hall can literally offer to give the game away 2/3 times, ruining the house odds, because the average person isn't ever gonna suss this out.

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sol-hsa
@sol-hsa

An easy way to figure it out is to scale it up. Monty shows you a million doors. You pick one. Then he opens up 999 998 doors revealing goats. Your pick and another door remain. Do you feel lucky, punk?

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lexyeevee
@lexyeevee

it may help to envision this concretely: you pick door 1, he goes down the line and opens every single other door except for 71936. would that not seem a bit conspicuous

(it'll be a different door for each game, of course, but he always skips one in particular out of a million — and you know he knows where the car is)

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lexyeevee
@lexyeevee

this is why i liked the explanation i came up with, haha — there are no cases at all, just a reframing of the question. you can either open door A, or effectively open both doors B and C (except that monty opens a dud from among them for you). taking both B and C naturally gives you twice the odds

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sol-hsa
@sol-hsa

There's two sets. Set A has 3 items, set B has 2 items. Your choice of "keep what I picked" selects set A, and "switch" selects set B. The only reason why A feels better is that it was your choice to begin with. If you ignore that, it's more clear. If that doesn't do it, let's say set A has a million items and set B has two items. If you still feel sticking to set A makes more sense (where your odds are 1:1000000) than switching to set B (odds 1:2), well.. I tried. =)

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JoelAtkinson
@JoelAtkinson

You are not picking the option that would be fifty-fifty versus the option that was one in a million, as that doesn’t add up to 1 like all good probabilities need to. You’re picking between the option you’d previously selected that was 1 in a million versus one which must therefore be 999,999 in a million.

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fizbin
@fizbin

I assume you saw one of my posts on the problem, likely through
@lexyeevee's re-chost which gave the second one reach.

In those posts I present three different ways to think about the problem; I should probably see if there are more different approaches because I think that the key here is that what explanation works for a particular person is a highly individual thing, and so what you really need to do is just come up with different angles approaching the problem and just cover the conceptual space with genuinely different approaches rather than trying to clearly explain one particular approach more and more deeply.

For the record, the three approaches I used in those posts are:

  • Multiverse counting
  • Comparing the relative likelihoods of two histories
  • Bayesian inference using Bayes' Theorem

The last of those approaches is almost certainly not a good way to develop intuition around the problem for anyone; however, Bayesian inference can often be a useful way to do probability calculations and for someone who already "gets" the Monty Hall problem looking at it with Bayesian inference could be a useful introduction for how to do Bayesian inference.

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