You should be able to solve this.

Create a method to choose a random integer between 0 and infinity such that no integer is more likely to be chosen than any other.

Yes it is possible, I don't care what your probability book says (Notice I didn't even use the word "probability").  You may assume the axiom of choice.

2?

x = anything, where x is the random integer

Another one

x/0=y, where y is the random integer

anything divided by 0 can be anything because any number of 0s fits into 0

>>3
>>4
missed the point completely. OP said <i>choose</i>. you haven't come up with an actual method, just different ways of expressing a random number.

>>5
<i>choose</i>

aww.. fuck me :P

Let F be the set of all continuous functions from [0,inf] to [0,1]. Let f be an biyective function from c to F.
Consider g a (continuous) function from [0,inf] to [0,1] defined by:
g(x) = sum from i=i to w of (1/2^i)f(i)(x). Let y = inf { x in R : g(x) is maximum }.

Choose n = min { m in N : m > y}

Is that what you wanted OP?

Flip a coin forever so that heads is 1 and tails is 0.  You will generate your number in binary.  You won't ever finish, but that's ok because you math weirdos defined real numbers as limits of inifinite sums anyway.

>>8
>You won't ever finish

The axiom of choice says you can ignore that little detail; it just says it is possible to get infinitely many coin flips, or flipping an infinite number of coins. (You can see now why the axiom is controversial.) Other than that you have the solution.

>>1
Isn't this basically a straightforward application of the axiom of choice? You have an infinite set of bins, each containing the digits 0-9. Select one from each bin. Concatenated together, the digits form a number. More formally, value = sum{n=0}{inf}(b_n * 10^n) where b_n is the digit selected out of the nth bin. This is the same solution as >>8 with a different radix.

This is a possible algorithm, though I realized I haven't really proven that the probability is equal. My set theory is a little rusty.

>>7
I don't know if I understand.  In the third line, what's w or "f(i)(x)".  F is going to have higher cardinality than the reals, so if the "c" in the first line is the complex numbers, there is no such bijection.

>>8
>>9
You can't do the coin flip thing, since you're not guaranteed to ever get a finite number.  At some point, you'd have to have chosen a finite number of digits, and get nothing but zeros from then on.



I'll post the answer later today.  If you want a hint, look at this: http://en.wikipedia.org/wiki/Vitali_set

>>9
lol quote failure.

w is the first infinite ordinal (The natural numbers). c is the first ordinal with the same cardinality as the real numbers. f is a function defined from c to F, so f(i) is a continuous function. F has the same cardinality as the reals, there is a easy injection from F to Q^Q, where Q is the set of rational numbers.

>>12

>>10

>>12
>F has the same cardinality as the reals
I'm not sure about that, but I'll go with it for now.

So let me see if I get this.  Of the uncountably infinite set of functions from [0,\infty) \rightarrow [0,1], you're selecting a sequence f_i := f(i) (how? just any sequence?), and setting g(x) := \sum_{i=1}^{\infty} f_i(x).  Then g is a uniform limit of continuous functions, and so is continuous, and the output is the next integer above the smallest local maximum.

How do you know g has a local max at all?  Also, where's the "random" part of it?  Are you selecting the f's at random?  How are you doing it?

>>10
well, it's later today.  how do you manage to finish choosing a number, then, if each is truly equally likely to be chosen.

>>15
By using the standard construction of unmeasurable sets as in the wikipedia article (or, better yet, this version: http://everything2.com/title/unmeasurable%2520set), you can partition the unit interval I=[0,1] into a countable union of disjoint subsets, each of which is identical to all the others, only translated left or right by some real number, so that they're all the same size (even though they don't really have a "size", strictly speaking, since they're unmeasurable).  Number these sets V_1, V_2, ....

Now just select a random number x \in [0,1] (which can be done with the coinflip thing, since an infinite number of non-zero digits *after* the decimal point is OK), and let your random integer be the index i of the set V_i that x happens to be in.

Since you can't have a probability space that contains all the V_i, you can't talk about the "probability" of choosing 7567865 (or whatever), but since the chance of choosing 7567865 only depends on the size and shape of V_{7567865}, which is the same as all the other V_i's, no number is more likely to be chosen than any other.

How's that any better than the flipping the coin thing?  We imagine that all infinity digits are chosen at once anyhow, not simply from left to right as though we were reading them.

>>17
It's obviously better, because it always converges, the other way doesn't.

In fact the other diverges with probability 1.

>>18
I don't see that it does, for any accepted meaning of convergence.

it's impossible to do

>>19
You flip a coin an infinite amount of times to decide a, for the sake of simplicity, binary decimal expansion.

So You have numbers X_i

X_1 = 0.a_1
X_2 = 0.a_1a_2
X_3 = 0.a_1a_2a_3

etc

We define the number we pick to be X = Lim X_i as i-> inf

Now obviously X_i is a cauchy sequence and |X_i - X_j| is at most 1/2^i for all j > i

So the sequence converges. This is all fine with the axiom of choice.


The other way though, if we pick digits going to the right, you can easily see that the sequence you get is not cauchy and doesn't converges.


As a very simple example what value do you give to the number if all your coin flips are heads?

>>19
There's nothing that needs to converge in the first place.  You just pick a random number in [0,1], and then pick the set that number is in.

throw a ball out into space, and count how long it takes to get back to you in seconds.
i guarantee you, every number from 0 to infinity is equally as unlikely to be the number of seconds it takes for the ball to get back to you.

>>23
oh right, uh, just to cover a technicality, round to the nearest second

>>23
<i guarantee you, every number from 0 to infinity is equally as unlikely to be the number of seconds it takes for the ball to get back to you.
<i guarantee you,

Prove it.

>>14
You're right, g doesn't necessary have a max; instead restrict F to the sets of functions with at least one maximum and just take f(0) as g. The random part is at the moment of taking f, which i dont have a flipcoin/throw a ball in the universe argument (And maybe will never have it, as c doesn't have a physicall representation)

PD: F being of size c is a consecuence of the fact that two continuous functions are equal iff they are equal in a dense set, in this case, the rational numbers. So c <= |F| <= |R^Q| = c^w = c

>>21
Where does Axiom of Choice actually come in? AC implies "there exists" such an infinite sequence, but you're already defining one hence don't need it.

>>21
That part obviously converges, but that's just an intermediate result. There's no convergence after doing the mapping.

>>22
Right, so no convergence. It's all or nothing.

>>16
Assuming that "countable union of disjoint subsets" refers to the method on the everything2 page, you can't say that they have the same size and shape after you've cut them to the [0, 1] range.

I wondered if you could choose the elements in X so as to all have a difference of more than 1, but then you'd have covered a real interval with no more than one element from each of an allegedly countable amount of sets, which can't be good.

Meh, too hot right now, will think more about it later.

>>28 All the W_i have the same external measure (In fact, they are constructed as translations modulo 1 IIRC). In that way you have the 'same probability' of being chosen, just as the OP said.

>>29

*translations by a rational number modulo 1 of some set

>>9 here...

>>10
You can't do the coin flip thing, since you're not guaranteed to ever get a finite number.  At some point, you'd have to have chosen a finite number of digits, and get nothing but zeros from then on.

Your second statement is correct here; after a certain point, you'd need an infinite number of zeros. This is not a problem. This is in fact what the axiom of choice allows us to do.

>>17
>>18
What does this have to do about convergence?

>"method...random"
wut

>>31
>This is not a problem. This is in fact what the axiom of choice allows us to do.

Uhhh, no.

>>27
Might not, I'm never sure, just use it as a cover all.

But I'd say picking the a_i uses choice, cause you're basically defining a choice function, but every A_i is just the set {0,1}, so maybe you're ok.

>>33
Wtf? Of course it does. Are you going to provide any rational explanation for why you think it doesn't?

>>35

Why do you think it does??  You're looking at the (countable) set of all infinite sequences of coinflips which are eventually always zero, and just claiming that AC lets you pick a random element of that set, which is basically the same problem we started with.

http://en.wikipedia.org/wiki/Axiom_of_choice

>>36
You're looking at the (countable) set of all infinite sequences of coinflips which are eventually always zero
Yeah...

and just claiming that AC lets you pick a random element of that set
It does...

which is basically the same problem we started with.
What the fuck? No it's not, it a valid answer to OP's question. "Create a method to..." well there's the fucking method!

>>37
>It does...
You don't need AC to pick an element from one infinite set.  Besides, AC doesn't say anything about the distribution of the element chosen.

>What the fuck? No it's not, it a valid answer to OP's question. "Create a method to..." well there's the fucking method!

The problem was to pick an element uniformly at random from the integers (a countable set) which is the same thing as picking an element at random from the set of sequences of coinflips that are eventually all zero, since the two sets are in bijection.

This is a cute thread, and I like the Vitali set solution, but isn't this cheating?  The notion of "likelihood" just seems like an alternate phrasing of "probability" and if it is indeed supposed to be a different animal, then it hasn't been defined.

I don't feel we can speak about one choice being "more likely" than another in any definite way without resorting to some notion of measure.  The solution OP proposed seems to rely on saying that sets that are congruent modulo translation have the same measure - even though they are nonmeasurable!  For this to work we need to assert that a number randomly selected from the unit interval is as 'likely' to appear in one element in the partition as in any other - but again, this seems to resort to the idea of measure.

imo, this is a clever way of dodging the question rather than a solution to the question as asked.  I'd love for someone to convince me otherwise, though!

>>39 here again, with a more serious objection to the alleged "solution" in
>>16
You assert that if a subset of [0,1] is obtained from another by partitioning it into finitely many pieces and translating them, then if we choose a random number in [0,1] it is "as likely" to appear in either one.  If we apply the same reasoning to [0,1]^3 we run into problems with Banach-Tarski.

Namely, if we have three disjoint balls in [0,1]^3 of the same radius, and we choose a random point in the unit cube, it should be "equally likely" for this point to lie in balls 1, 2, or 3.  But with AC, balls 2 and 3 can be obtained from 1 by partitioning 1 into finitely many pieces, and translating them; so by your reasoning it should be "as likely" for this point to be in ball 1 as in the union of balls 2 and 3.

Anyway, I sort of doubt what OP is suggesting is possible without an utterly trivial notion of "likelihood", but this is still a fun question to think about.