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I'm not a logician, so I apologize if what follows translates to nonsense. I would like to try to define a different theory of random choice. I hesitate to call it probability theory because I do not expect it to follow the usual rules of probability. I will however refer to it as a warped theory of probability.

For the sake of simplicity take $\mathbb{Z}$ or $[-\infty,\infty]$. It is an easy fact that one CANNOT define a discrete uniform distribution on either of these spaces in the usual sense. One of two things necessarly goes wrong: normalization or countable additivity. However, according to the axiom of choice, I can pick an element of $\mathbb{Z}$ or $[-\infty,\infty]$. Even though these sets are well ordered, I would still like to use the axiom of choice as explained below.

In fact, if I have infinitely many copies of $\mathbb{Z}$ or $[-\infty,\infty]$ then I can pick elements from each copy turning it into a product space. Since there is no real recipe for how the axiom of choice picks elements, I would like to think that if I want to define a "uniform" distribution on $\mathbb{Z}$ or $[-\infty,\infty]$, then I would invoke the axiom of choice to generate an element. Can any of this be formalized into a useful theory? In short, I would like to pick an element of each set with the underlying notion that I have no preference to the choice I make. I'm invoking the axiom of choice as a means to do so. That is, I am defining the notion of a uniform distribution through the process of saying "by the axiom of choice I can pick an element." Literally, I am thinking of the axiom of choice as a black box into which I feed an arbitrary collection of sets from which it spits out picked elements.

Of course this would not be in line with usual probability. In particular there would be huge issues with sets such as $[-1,1]$ and $[-\infty,\infty]$ where in the usual theory of probability I CAN define a uniform distribution on $[-1,1]$ but cannot on the latter, even though the two sets have a bijection. I am not sure how to reconcile this.

My experience with the axiom of choice has been mostly in proofs that say, involve some collection of equivalence classes so I can pick representatives from each one. The point is in this case I don't care which representatives I pick. So if I were to suddenly care, is there some warped notion of probability that one can invoke to make "inferences?"

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    You do not need the axiom of choice to pick an element from a well-ordered set, just take the smallest element. In fact, this is how the proof that the axiom of choice is equivalent to the well ordering theorem goes.2012-04-29
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    You don't even need the axiom of choice to pick finitely many elements from finitely many non-empty sets. Only when choosing from infinitely many elements we begin to thread on the grounds of the axiom of choice.2012-04-29
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    Your question in its current form is unclear. It seemed to me that you were asking about uniform distributions on countable sets, and you thought that the axiom of choice is needed in the proof. Please correct your question.2012-04-29
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    I've chosen a very simple example. What if I want to generalize it to countably many copies of $\mathbb{Z}$. As well, what if I want to talk about unordered sets. I mentioned below in one of the answers that I want to think of the axiom of choice as a generator of a random element from such infinite sets with no intent of it being consistant with the usual rules of probability. Of course I can prove the equivalence of axiom of choice by exhibiting a choice function but I would this idea of a uniform distribution to be grounded in the fact that I can choose any choice function2012-04-29
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    I'll try to edit it for clarity2012-04-29
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    You don't need the axiom of choice to choose from countably many copies of the same countable set.2012-04-29
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    The fact that you can define a uniform distribution on $[-1,1]$ means that you can on $[-\infty,\infty]$ via transfer of structure (use a bijection between the two to generate the probability on the latter). It will, however, be incompatible with what you would like to think is the real line (in terms that this is no longer guaranteed to play nice with translation or scaling). Regardless to that, if you simply taking copies of the *same* set you do not need the axiom of choice. Remember that you also **fix** a choice function, so once you fixed that you suddenly care what your elements are.2012-04-29
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    @Asaf: Agreed, the issue of bijections is quite troublesome. I see what you mean about fixing a choice function. Can I still think of my fixed choice function as a random choice? I know what I've said is basically circular.2012-04-29
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    @Sam: Well, I'm not 100% clear about what you are trying to do. However the way I see it there are two options. (1) You are giving each equivalence class a probability determined by that of the representative, so you care about the representative; (2) You give the representative probability based on the equivalence class and then you have reduced to the original case of a uniform probability on $\mathbb N$.2012-04-29
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    You should read [this paper](http://www.dam.brown.edu/people/mumford/Papers/OverviewPapers/DawningAgeStoch.pdf) and the Freiling paper mentioned in it.2012-04-29
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    Cox's Theorem prevents you from distributing "indifference" among infinitely many things. That's not a state of knowledge one can possibly have, and [probability is in the mind](http://lesswrong.com/lw/oj/probability_is_in_the_mind/)2012-04-29
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    @EMS: Thank you for the interesting links!2012-04-29

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It seems to me that the question is based on an unfortunate but quite common double-use of terminology. When people apply the axiom of choice, they often replace formally correct statements like "let $f$ be a choice function, $\dots$, apply $f$ to the set $\dots$" with more colloquial formulations involving words like "pick" or "choose arbitrarily". In probability theory, one also talks about "picking" or "choosing" elements of some space according to some probability distribution. The question asked here seems to be based on the idea that this coincidence of terminology might correspond to a coincidence or at least some close relationship in the underlying mathematics, so that something like the probabilistic "picking" can be obtained from the axiom of choice "picking". As far as I can see, the answer is no; these are two quite different uses of "choose" and "pick".

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    You dig deep! :-)2012-11-10
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    I still tend to see them as related, and I'll give you an example of why. Please let me know what is wrong in my construction.How I believe you can use AC to build a random number generator: Supposse we have $\omega$ well ordered copies of the set N. We fix a choice function that will pick one element of each set. Then put the chosen numbers in the same order than the copies of N, within a black box. Then by definition my "natural number generator" box will generate a sequence of natural numbers. If this were possible to imagine,do I have any expectations about the distribution probability?2014-07-14
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    @julianfernandez: If I were fixing a choice function for $\omega$ copies of the set $N$, it would be the identically zero function. You want to fix a rather different choice function. How, exactly, do you want to do that?2014-07-14
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    Assume the choice function treats all the elements of N as unlabeled, or undiferentiated, so the function looks truly random.2014-07-14
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    Or I am not allowed to "fix" that?2014-07-14
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    How does "the function looks truly random" follow from "the choice function treats all elements of $N$ as unlabeled or undifferentiated"? For that matter, what do "treat" and "undifferentiated" mean in this context?2014-07-14
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    OK, let us say that the sets consists of actually N identical balls. Well you know they are not identical because you can brake them and they have a label (n) inside. But you will only use or know the labels after you put the chosen balls in the Black box. I am abusing AC?2014-07-14
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    @julianfernandez Yes. What you're doing has little to do with AC. All AC tells you in this situation is that there exists a choice function. It doesn't guarantee that this function is uncorrelated with the hidden labels. A choice function is just a set of ordered pairs; it has nothing to do with what *we* know about the entities involved.2014-07-14