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someone explain the monty hall problem to me please. i just read this

http://mathforum.org/dr.math/faq/faq.monty.hall.html

and it SORT of made sense, but at the bottom the guy gives an example of a 1000-door problem, saying that if monty shows 998 goats, your chance of finding the car out of the final 2 doors is 999/1000 if you switch. this just seems like it cannot be right. doesn't that assume that monty has no knowledge of where the car is?

wait i think i get it now

Monty does know.

it's not a 50/50 at the end because you still have the 1/1000 chance of guessing right originally. since the host shows you 998 goats, the remaining door HAS to have the car if you didn't guess right at the beginning. and the odds of you not guessing right at the beginning are 999/1000

that was fun

You got it

I just tried to write out a game-theoretic interpretation, and then I tried to apply Bayes' theorem, but I got stuck on both of those.

but yeah rly the answer is "because Bayes' theorem is true"

you got it anyway though, hooray!

monty hall is deep and subtle

this, basically http://formalisedthinking.wordpress.com/2010/10/05/bayes-theorem-and-the-monty-hall-problem/

Also, philosophically speaking, the Monty Hall problem concerns an observation selection effect, which gets into all kinds of wooly things like the doomsday problem, I read part of this guy's book. http://www.anthropic-principle.com/

the weirdos over on LessWrong probably eat that stuff up

Really the best way to understand it is to increase the number of doors like you did. Some people use a deck of cards- you pick one, then the dealer flips over fifty of the remaining fifty-one.

Assumptions underlying rules of probability are subtle- recently read an article called "Mistakes in Probability," I think, about how the founding fathers of the subject- mathematical giants!- pretty much all made various errors in one fashion or another.

"Errors of Probability in Historical Context," by Prakash Gorroochurn.
I found it in The Best Writing on Mathematics 2013, which series I recommend to you, any year you can find.

He wrote a book too. Looks nice. Expensive though.

doesn't that assume that monty has no knowledge of where the car is?

I think it's the opposite. It seems to me there's an implicit assumption that Monty does know where the car is, and it's part of his regular routine to open a door and show a goat, regardless of which door you pick. For instance if it was Monty's policy to only show you a goat if you had picked the correct door on your first try, then it would be a bad idea to switch, right? Or if some random guy walks into the studio with no knowledge of anything and opens a random door, and there's a goat behind it, there isn't any advantage of switching either, right?

yeah Monty and his knowledge are a red herring, what matters is that you see a goat behind a door, allowing to update your prior probabilities about which door hid what.

That's not what the link you linked upthread said:

the information is given by the fact that he cannot open the winning door

i.e., he must know which door is the winning one in order to avoid opening it.

practically he has to know which door is the winning one or else the whole point is lost

Oh crap I lied. Yeah this whole thing is not intuitive at all. And not really in my wheelhouse, I should leave it to others.

Yeah, Monty's knowledge is crucial. The thing the problem forces you to realize is that the, um, equipartition of probability one might consider "intuitive" is actually an assumption you make that should be overridden if and when you have new or other relevant data. That's why they always specify a fair die in certain problems, whereas in real life, you might want to consider the probably that the die has been rigged, but then that's too complicated for a homework problem.

A related implicit assumption you might need to be aware of is whether two events are to be considered independent or not. Or to go way, out there, they way in statistical mechanics particles are considered to be indistinguishable in such a way that changes the way one would have thought the distribution should be calculated.

Do u see?

It seems like the Monty Hall problem is a good example of how cultural context gets smuggled into something that is superficially a pure math problem. It shows how there could be cultural bias in something as apparently objective as the math section of a standardized test.

Recent NY Times article about Bayesian statistics featured the Monty Hall problem.

the weirdos over on LessWrong probably eat that stuff up

Yes - Bostrom's big projects – the Oxford future of humanity/Martin school set-up, the book on Global Catastrophic risks he edited – have contributions from Eliezer Yudkowsky who founded LW.

so here's a problem i'm struggling with

a man has three coins. two are made of gold and one is from a foreign land. if we choose one at random, what is the probability that it is a gold coin from a foreign land.

one might say: 2/3 * 1/3 = 2/9

or one might say: either 1/3 or 0 but definitely not 2/9.

who is right?

anyone?

Recent NY Times article about Bayesian statistics featured the Monty Hall problem.

― You Better Go Ahn (James Redd and the Blecchs), Monday, October 6, 2014 6:59 AM (3 days ago)

yeah this is where i got it from

so here's a problem i'm struggling with

a man has three coins. two are made of gold and one is from a foreign land. if we choose one at random, what is the probability that it is a gold coin from a foreign land.

one might say: 2/3 * 1/3 = 2/9

or one might say: either 1/3 or 0 but definitely not 2/9.

who is right?

― the late great, Thursday, October 9, 2014 3:19 AM (9 hours ago)

i guess if you accept that the foreign coin either is or is not gold, then it's either 1/3 or 0

but not knowing that i guess it's 2/9. both could be right depending on how much info you have?

i am probably wrong about this, but here's how i'm thinking about it:

you have two gold coins and one non-gold coin. the chance that the foreign coin is also a gold coin is therefore 2/3. you then pick one of the three coins. so 2/3 * 1/3 = 2/9.

"1/3 or 0" yes but it's not a 50-50 chance between those two, is the thing.

i am leaning strongly toward ⅓ or 0

problem i have is with this

the chance that the foreign coin is also a gold coin is therefore 2/3

not sure about this

i am a realist, and i "know" (?) that if you did this experiment IRL three million times you would get a gold foreign coin either 0 times or approximately a million times, but not approximately 444,000 times

surely it is possible to aggregate those permutations?

enumerate all ways of choosing the two gold coins and the one foreign coin (there are 9) and compute the probability (of gold & foreign) conditional on that configuration (in 6 of them it is 1/3, and in the rest 0). assuming all configurations are equally likely (uniform prior) then the law of total probability says the probability of choosing a gold and foreign coin is 6/9*1/3 = 1/3

would be different for a different prior of course

sorry 6/9*1/3 = 2/9...

6/9 * ⅓ = 2/9

:)

i'm unclear why you would do this:

enumerate all ways of choosing the two gold coins and the one foreign coin

actually i'm unclear what you mean by this in general

I think the answer here is something like, for a frequentist, you'd write "not enough information to tell a priori", whereas a bayesian would be perfectly happy to answer 2/9, and update their priors after each experiment.

If a given coin has the property "is gold" with probability 2/3, and "is foreign" with probability 1/3, and you have no other information, and you assume that those probabilities are totally indpendent, then 2/9 is totally the right answer to the question "What is the probability that you draw a foreign gold coin?" for a single trial. If you draw a non-gold non-foreign coin or a gold foreign coin, you know that the foreign coin is gold, and your probability of drawing a foreign gold coin on a subsequent trial can be updated to 1/3 (the foreign coin is definitely gold). If you draw a gold non-foreign coin, I don't think you can update your priors, b/c you knew that one non-foreign coin was gold before you started (by the pigeonhole principle), so you've not learned anything.

i think the answer might be 2/9 if we imagined an infinite number of people with three coins in their pocket such that ⅔ of them have a gold non-foreign coin and ⅓ have a gold foreign coin

if we summed an infinite number of experiment over all of those people then I have no doubt that 2/9 of the time a foreign gold coin would be drawn

but i think for one person it collapses down to ⅓ or 0

by enumerate i just mean write down all possibilities for the ways the coins can exist (in terms of gold and/or foreign) then assume they are equally likely. it is an assumption though

The thing is that you can't answer a probability question with "x or y" - if there are two possible scenarios like that you have to determine the probability that each scenario will occur. It's not a real world problem so you have to accept that the coins are different every time you pick one. The way the problem is written implies that either you have scenario A, in which there are two non-foreign gold coins and one non-gold foreign coin (I.e the case in which the probability is 0), or scenario B, in which you have one foreign gold coin, one non-foreign gold coin, and one regular coin (I.e the case in which the probability is 1/3). The probability space breaks down so that - given that these coins are assigned their states at random each time you make a selection - scenario A happens 1/3 of the time and scenario B happens 2/3 of the time, as I explained above. So the total probability that you will pick a foreign gold coin is (1/3 * 0) + (2/3 * 1/3) = 2/9.

conditional on the state of the coins it is either 1/3 or zero yeah xp

given that these coins are assigned their states at random each time you make a selection

imo this is the heart of the 2/9 fallacy

coins don't do that

they definitely could have picked something better than coins, like glasses of water that are randomly colored with blue + yellow food coloring & you have to determine the probability of picking a glass with green water

Uh oh is this going to turn into some kind of plane-on-a-treadmill situation

I agree that if you have 9 zillion cups of water, and you randomly add blue food coloring to ⅓ of them and yellow food coloring to ⅔ of them, you will have about two zillion cups of green water and a 2:9 chance of picking a cup of green water if you randomly select a cup from among those nine zillion

but to me that sounds like a fundamentally different problem - equivalent perhaps to my infinite ensemble of men, each with three coins in their pockets, 1:3 of which are foreign and 2:3 of which are gold, evenly / randomly distributed