I believe that Cabbage was saying that this isn’t the case. Just because there’s a probability 1 that you’ll get one of those outcomes, doesn’t mean that you’ll certainly get one of those outcomes. In this case, it happens to be true, but it doesn’t follow from the fact that P = 1.
What other interpretation is there?
IIRC. the result of a number divided by infinity is not zero, it is undefined.
The limit of n/x as x approaches infinity is zero, but not n/x itself.
Urban Ranger: P(b) is a well-defined quantity, so it has to have a value. By the axioms of probability, it must be non-negative. If we can show that P(b) is less than any positive number (and I think we can) then P(b) = 0. No dividing of infinities required.
P(b) has a value, with is 1/x where x is the number of real numbers in [0,1]. A value does not have to be numerical, there is nothing that prevents P(b) being a function of some other parameters.
For a discrete random variable X, if P(X=x)=0 then X cannot take the value x. For a continuous random variable X, P(X=x)=0 for every x.
I see what you’re thinking, and that idea works for modeling a number picked at random from finitely many numbers. However, one of the axioms of a probability function is that the range of probabilities must be contained in the real interval [0,1]. Any sort of 1/infinity is not a real number, and generally not defined at all. So, while you may be getting an intuitive feel for what’s going on by thinking of it that way, that’s really not the way the probabilities are defined in any mathematical sense, and probability is known for having some very counter-intuitive results, so relying on your intuition is liable to lead to mistakes.
Lebesgue measure is defined in terms of limits, but the Lebesgue measure of any (Lebesgue measurable) subset of [0,1] is a real number in the interval [0,1]. The Lebesgue measure of a single point is not “approaching” zero in any sense, it is exactly zero. And Lebesgue measure satisfies all of the probability axioms, so it’s a perfectly valid probability model, and one which demonstrates that probability one is not the same as “guaranteed to happen” (GTH).
Having said what I said before about probability functions with a finite sample space, and how you can construct models with such probability functions so that probability one does not mean GTH, I have a strong feeling that you can make refinements to any such model to rectify this situation–simply take the “largest” subset with probability zero, and get rid of it. In other words, for example, take my original sample space {1,2,…,6,something else happens}, and simply reduce it to {1,2,…,6}. Doing this doesn’t significantly change your probability model, it simply discards the events that are not interesting. In fact, I suspect this may also be doable when your sample space is countably infinite, though I haven’t actually tried proving either of these ideas. Keep in mind, however, that this doesn’t demonstrate that, even in the finite case, probability one = GTH; while the mathematical side of probability is rigorous, I am still free to model real world examples in such a way that probability one != GTH.
However, when your sample space is uncountable, there’s no way around it–there will be events with probability one that are not GTH (unless your probability function is simply too trivial).
Actually, scratch that, forget it. I think my mind got off on a tangent somewhere, and I don’t think the above is really relevant to what we’re discussing, after all. Reducing the sample space doesn’t change the fact that something else can still happen to the die–it’s not guaranteed to come up 1-6, even though it has probability one of doing so.
Still, there is a fundamental difference between the finite case and the uncountable case. Maybe what I was trying to say could be rephrased as this: In some cases, such as the above case, probability zero can possibly be interpreted, not as “can’t happen”, but rather, “not interesting”. The “something else happens” is not impossible, rather, it’s just that our attention is focused solely on what happens when the die is rolled the proper way with a proper outcome.
This is distinct from the uncountable case. With the picking a real number/Lebesgue measure example; the probability of any particular number being picked is zero, however, taken as a whole set, there’s probability one that one of them will be picked. That’s a significant and fundamental property of the probability function in this case, the probabilities of zero for any particular number can’t be easily dismissed as in the preceding example, since together they end up forming the whole sample space. Anyway, I hope that makes sense.
Cabbage,
I agree with what you’ve written, but at the same time I think it is misleading.
Suppose you choose a random real number on the interval [0,1]. As you said, the chance that it is rational is 0. But also, the chance that it is representable as a mathematical expression is 0. In other words, mathematical expressions are countable, and the real interval [0,1] is not countable, so the probability of a random point on real [0,1] being representable is 0.
So if you were to actually generate a string of these numbers, you would not be able to represent any of the particular values you generated. The best you could do is say they were greater than a certain representable number, and less than another representable number.
So what I am arguing is that the chance that you randomly generate a rational number is 0, and furthermore, just as you would expect, it couldn’t happen. Furthermore, the chance that you will not generate any particular rational number is also 1, and furthermore, that will always happen. What will always happen is that you generate a number which is not precisely representable by any finite string of mathematical symbols.
jawdirk, I’m afraid I don’t see how that makes what I said misleading. The point I was trying to make is that any particular number has zero probability of coming up. That means any number, whether it’s rational, expressable by some finite slgorithm, or some other number that we can’t even generate with any kind of algorithm. The point being that, whatever number is picked (whether it’s a “nice” number or not), the probability of it being picked was zero, and yet it was still picked. It’s not misleading to say that probability zero does not mean “impossible”, and it’s not misleading to that that probability one does not mean “GTH”, since that’s exactly what happens here, and that’s the only point I’m trying to make.
I think that what jawdirk was saying is that it’s not meaningful to say you’ve “picked” a number if that number is not representable. If you’ve got your random number generator, and I ask you what number it returned, what do you say?
I would say that I’ve picked a number. Yay axiom of choice!
I’d say I picked a number. Yay axiom of choice!
On a more serious note, every real number is representable. What’s at stake is whether we’ve “picked” a number if the number we picked is not finitely representable. I have no problem with allowing that case. After all, the set of finitely representable numbers in [0, 1] is countable, so it has measure zero–therefore, it is almost certain (i.e., probability 1) that we will pick a number which is not finitely representable.
Here’s a question: what is the probability that we’ve picked a rational number less than or equal to 1/2, given that we’ve picked a rational number?
Yes Chronos, that’s the point.
Cabbage, it’s not just that it’s not a nice number, it’s that you can’t even say what number you’ve picked. You can’t differentiate these numbers from other numbers that also have (equal) probability 0 of being picked. Between any two representable numbers there will always be an infinite number of unrepresentable numbers, and you will have no way to say which of these was randomly chosen except by narrowing the bounds again.
I could be wrong about this – I’m just working on a CS Masters not math, but I believe this is territory where the axiom of choice becomes important. I would guess that it is unprovable that it’s ok to assume that a particular number has been chosen. In other words, all the numbers do have probability 0 of being chosen, and “choosing a number at random” doesn’t violate this, because you can’t say which number has been chosen. Unless you embrace the axiom of choice, you can’t even assume that a particular number can be chosen.
Please educate me if I’m misinterpreting the axiom of choice.
For those who don’t know anything about the ‘axiom of choice’, it is an axiom, in that a coherent mathematics can either hold it true or false, as far as anyone knows. When the axiom of choice is held true certain unintuitive results arise, and when it is held false other unintuitive results arise, so there is no concensus on which mathematics is ‘preferable.’ In consequence, mathematicians study mathematics with and without the axiom of choice. In general, denying the axiom of choice limits what can be proven.
Well, we’re obviously talking about a thought experiment in the first place anytime we talk of picking something from any infinite set. The fact is, is that Lebesgue measure does form what can be called a probability function on the interval [0,1]. Is there any way to realize an actual mechanism that will do the picking for us? Of course not, but we can still speak of what would happen if we had such a mechanism, and I still claim that it’s perfectly fine to speak of picking a particular real number at random from [0,1], using the axiom of choice (which, really, is commonly accepted). As far as I’m concerned, all I need to say is that some particular number has been picked; I don’t have to specify what that number is in order to demonstrate my point that probability zero is not the same as impossible–a particular number was picked, and it had probability zero of being picked.
For that matter, you claim that picking a rational number “couldn’t happen”. How so? I agree that it’s extraordinarily unlikely; it has probability zero of occuring, so it’s natural to think of it as being extraordinarily unlikely, but my point all along has been that that doesn’t imply that it’s impossible.
Good question. It’s tempting to say that, intuitively, the probability should be 1/2, but I’m not really satisfied with that. The problem is that, given that a rational number is picked, we’ve really restricted our sample space to a countable set. Unfortunately, it’s impossible to construct a uniform probability function on a countable sample space (which is essentially what the question is asking for), so I’d have to say that it’s pretty much unanswerable (under the standard probability axioms, anyway).
Ugh, I hope the failed post I’m rewriting actually failed:
In response to: what is the probability that we’ve picked a rational number less than or equal to 1/2, given that we’ve picked a rational number?
As Cabbage said, it’s certainly not 1/2. There is a 1-to-1 correspondence between any two countably infinite sets, so the interval [0,1/4] has just as many rational numbers as the interval [0,1/2]. So any argument used to show that the probability was 1/2 could be used to show that the probability was 1/3, or any other value between 0 and 1.
Cabbage wrote:
See that’s the crux of it. In this case my intuition seems to go against the axiom of choice, because as far as I am concerned, probability 0 means impossible. As you said, the axiom of choice is commonly accepted, but also, it is commonly rejected. Mathematicians point out that their proof requires the axiom of choice in order to differentiate what can and can’t be proven without it. It has been proven that it is an axiom – it can’t be proven to be either true or false. As I said, some of its results are generally regarded as intuitive, and some of its results are generally regarded as absurd.
I have heard that with the axiom of choice it can be proven that there exists a way to cut a solid sphere into a finite number of pieces so that they can be reassembled into two solid spheres with volume equal to the original sphere! Importantly, it is not possible to describe the shapes mathematically – but it is necessary that they exist if the axiom of choice is true.
So I’m glad we agree that the axiom of choice is the question here.
Here is a question for you:
Suppose a mechanism randomA outputs a random real number on [0,1]. Then let us define randomB as:
x=0;
while (x is rational) repeat {x=output of randomA}
output x;
The question is, does randomB produce a different output from randomA? Note that with probability 1 the while loop iterates exactly once!
This probably merits a new thread–there sure oughta be a way to do it, and if not, it’s gonna engender some interesting discussion. I’ll open it tonight.
No, because the probability depends on the measure of the set, not the size. The two are very different.
With probability 1, the output of randomB is the output of randomA. That doesn’t mean it’s guaranteed to happen.
Promised spinoff
Cabbage, would you say your interpretation “P(x) = 0 does not imply x cannot happen” is somewhat unorthodox? I like it and all, because it’s so weird, but my probability book defines a sample space as something like, “The set of all possible outcomes”. And I think this definition is not uncommon.
This is orthodox. If probability zero is not impossible, then your sample space can contain non-empty subsets of probability zero.