ℝolliℵg M∀th Thr∑a∂

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

maybe i'm too subscribed to the bayesian viewpoint but i don't see it as a fallacy. by choosing a different prior distribution on the state of the coins (e.g. one of the states has probability 1 and the rest zero) you can get the 'one person' answer also.

Well if it helps, consider that before you draw a coin out of your pocket, you are effectively one of that infinite ensemble of men, and you don't know which one. Once you draw a coin, you might know more. (Don't worry about the idea of replacing the coin and them changing, we're only worrying about the first draw.)

as silby said earlier it's a v nice example of the difference between the frequentist and bayesian inference

yeah i think i've got my bayesian vs frequentist lecture down now

the other problem i have is that by the 2/9 interpretation you can conclude that he has a 2/9 chance of drawing a gold foreign coin, a 1/9 chance of drawing a foreign nongold coin, a 2/9 chance of drawing a domestic nongold coin and a 4/9 chance of drawing a domestic gold coin.

so if he has three coins, then he has

⅔ of a foreign gold coin
⅓ of a foreign nongold coin
⅔ of a domestic nongold coin
4/3 of a domestic gold coin

but this is impossible! he can only have 0, 1 or 2 of each of those types of coin

all we have calculated is the average number of each type of coin among a large group of coinholders

So 2:9 only holds if you're picking one coin from one man out of a large group of men

sorry for beating this into the ground

If this is making you disgruntled then you'll hate quantum physics

Scott Aaronson likes to describe quantum information thy as the extension of the concept of "probability" to allow negative values/amplitudes. There's no classical physical interpretation of a negative probability, just like there's no common-sense interpretation of 4/3 of a coin. But we do the math that way because we get the right answer.

Full marks to crüt for his science-dropping explanation.

whoever came up with this problem is a jerk

Indeed.

so if he has three coins, then he has

⅔ of a foreign gold coin
⅓ of a foreign nongold coin
⅔ of a domestic nongold coin
4/3 of a domestic gold coin

but this is impossible! he can only have 0, 1 or 2 of each of those types of coin

you lost me a bit here. if you are choosing 3 coins from the same guy then you need to consider that you are picking the coins without replacement and the probabilities should be different - the prior should change each time. if you are choosing 3 coins from different guys then you could absolutely get 3 coins of the same type!

i majored in quantum mechanics!

sorry I was calculating expectation values for # of each type of coin

ok my flight just got cancelled so i have far too much time on my hands. i am thinking abt what happens if we simply take all 3 coins from a single guys pocket (without placement) where the coins in his pocket satisfy the rules of the initial problem (2 gold, 1 foreign). there are 4 possible states for each of the coins in his pocket. writing a coin's state as:

(0,0) = not gold, not foreign
(0,1) = not gold, foreign
(1,0) = gold, not foreign
(1,1) = gold, foreign

then the 3 coins in his pocket have 3 possible states:

state 1: (1,0),(1,0),(0,1) - prob: 1/3 (uniform prior)
state 2: (1,0),(1,1),(0,0) - prob: 1/3
state 3: (1,1),(1,0),(0,0) - prob: 1/3

then after we take all 3 coins from the guys pocket the probability that we have at least one coin of one of the 4 states is:

Pr(we have at least 1 (0,0) coin) = 2/3
Pr(we have at least 1 (0,1) coin) = 1/3
Pr(we have at least 1 (1,0) coin) = 1
Pr(we have at least 1 (1,1) coin) = 2/3

state 2 and state 3 are equivalent, no?

oh i see why you did that, nm

i majored in quantum mechanics!

― the late great, Thursday, October 9, 2014 1:32 PM (1 hour ago) Bookmark Flag Post Permalink

sheepish emoji face

‘I do not even know what a matrix is’, Heisenberg had lamented when told of the origins of the strange multiplication rule that lay at the heart of his new physics. It was a reaction widely shared among physicists when they were presented with his matrix mechanics.