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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.

The question reminds me of Bertrand’s paradox from Calcul des probabilités. If you’re unfamiliar with Kolmogorov axioms, don’t read the Wikipedia article until you’ve worked it:

There’re three boxes.

• contains two gold coins
• contains two silver coins
• contains one gold coin and one silver coin

I love this problem. While it’s similar to Monty Hall, the geometric, probabilistic, and statistical solutions are far more elegant.

Scott Aaronson likes to describe quantum information thy as the extension of the concept of "probability" to allow negative values/amplitudes.

Interesting, but this interoperation marginalizes probability theory—whose usefulness is due to its simplicity and few dependencies. Information theory, on the other hand, requires a vast and complex framework. Fundamental objects in probability, e.g events and random variables, are easily reducible into the simplest mathematical objects. Hell, this includes continuous interpretations, whose fundamental theorem is simply proved with characteristic functions. While Shannon entropy is formally expressed using expected values and random variables, they aren’t equivalent. It’s far more nuanced than just the addition of negative values. Oh, and Aaronson is an asshole. So that’s probably motivating my nitpicking.

2/3, right?

Aaronson is an asshole? Tell me more! This is the kind of scuttlebutt I'm really looking for here.

no wait! tell us more about information theory's vast and complex framework

Aaronson is an asshole? Tell me more! This is the kind of scuttlebutt I'm really looking for here.

He wrongly threw my former advisor under the bus. It still pisses me off.

no wait! tell us more about information theory's vast and complex framework

LOL. This is what happens when I drink a bottle of wine before posting.

i am serious!

ok good to know! I really like his work (& er know someone who has to write about it) so I like to know about who I'm going to laud

i've heard a few explanations of how you can look at quantum stuff not just as a generalization of probability, but even more basically of logic. you just do the right sort of algebraic setup and let your variables "vary" over a different, weirder domain.

Anyone read any of this guy's stuff? http://www.newyorker.com/magazine/2008/03/03/numbers-guy

it's great. i know some of his work with his students on the munduruku (sp?) re their grasp of small numbers and then another category for "bigger than small" rather than more different numbers. So like 1 2 3 4 5 6 7 big

Also the tribe people get a diff number concept just from learning in schools taught in European languages.

Thanks. Wonder if it's worth reading his number sense book, or any of the others.

Number Sense is the famous one. could just look for articles on the net if you want a sample before buying

http://webmuseum.mit.edu/detail.php?t=exhibitions&type=exh&f=&s=3&record=6

Grothendieck is dead.

RIP :(

Thought revive would be for tomorrow's anniversary of Euler's formula V - E + F = 2.

RIP BIG G

Grothendiecks to watch out for

http://inference-review.com/article/a-country-known-only-by-name

http://upload.wikimedia.org/wikipedia/commons/thumb/d/da/Icosian_grid_small_with_labels2.svg/500px-Icosian_grid_small_with_labels2.svg.png

Been reading Four Colors Suffice: How the Map Problem was Solved, by Robin J. Wilson, which is very well done, informative and entertaining. If you don't believe me, there is a rave review by John H. Conway on the back.

Here is a related cheat sheet: http://www.mei.org.uk/files/conference07/A2.pdf

Here is review in the AMS: http://www.ams.org/notices/200402/rev-toft.pdf

Also dipping into related book: Euler's Gem: The Polyhedron Formula and the Birth of Topology
By David S. Richeson