Wednesday, April 14, 2010

Stupid probability question

I got conned into doing this Eat Well, Live Well Challenge where I work, and one of the things you do is try to walk 10,000 steps a day. So I've been walking to a place around the corner for lunch, which is about 5,000 steps round trip.

On my walk today, I was thinking about probabilities. Although it blew my mind at the time, I've come to accept that a probability of zero only means "almost impossible", not actually "impossible". The classic example is that if you pick a random point inside a unit square (0<=x<=1, 0<=y<=1), the probability of any given point being chosen is, necessarily, zero -- but that doesn't mean it's impossible.

Okay, that's great. Of course, you can take any continuous segment of the unit square and give the probability of that, e.g. the odds of landing in (0<=x<=0.5, 0<=y<=0.5) would be 1/4... right? Interestingly, the odds of landing in (0<=x<0.5, 0<=y<0.5) are also 1/4, I think, but that's okay, because the odds of landing exactly on the lines x=0.5 or y=0.5 are that whole "almost impossible" zero again.

So now let's extend it to, pick any random point on an infinite 2-dimensional grid. Again, the probability of choosing any given point is zero. No problem. But now, the odds of choosing any finite continuous segment are also zero... which I guess I can handle.

Okay, but now what are the odds that the point chosen will be in the infinite continuous segment x>0? I don't know how to show the math (I know it involves integration, but I can't seem to quite figure it out), but it's got to be 50%, right?

Okay... so what are the odds that the point chosen will be in the infinite continuous segment x>1? It's also gotta be 50%, right? I think? Which I guess is okay, because the interval 0<=x<=1 has our good old friend "almost impossible" probability of zero.

Can anybody who knows probability theory tell me if that's correct? And if so, how do you sleep at night? This will piss me off almost as much as the Banach-Tarski Paradox...

Edit: I'm guessing this is how I do the math, but I have to get off the computer now so many I can figure this one out tomorrow...


  1. Atheists,

    but you have NO ANSWER TO DEATH... therefore you FAIL...






    Repent and turn to God.

  2. I normally delete Markuze's comments as soon as they pop up, but I am going to leave this one just because I think it's fucking hilarious that he copy-pasted yet another of his macros onto this post of all things.

    Yep, asking questions about the modern interpretation of continuous probability distribution is definitely going to earn me a place in hell.

    Wasn't this worthless fuck supposed to be arrested for making death threats against PZ? And doesn't he live in Canada, where they have laws against hate speech? In what way does this not qualify????

  3. I know this is nearly a month after you posted, but: There's *no such thing* as a uniform probability distribution on the whole of the two-dimensional plane (or, for that matter the whole of the real line, or all the integers).

    Here's why: probability distributions are supposed to be "countably additive", so (working in one dimension for convenience) the sum ... + Pr(-2 <= x < -1) + Pr(-1 <= x < 0) + Pr(0 <= x < 1) + Pr(1 <= x < 2) + ... has to equal 1. But for a uniform distribution, all the terms must be equal, and there's no number such that adding together infinitely many copies of it gives 1.

    There are, of course, plenty of non-uniform probability distributions on R, or R^2, or whatever. Typically they will make Pr(x>0) and Pr(x>1) different, and Pr(0<x<=1) nonzero.

    No weird paradoxical stuff required :-).

  4. Thank you, this was really helpful! It also solves another subsequent mystery I hadn't mentioned -- namely, when I tried to define the probability function and then do an integral on it, it seemed easy if I made the probability distribution non-uniform, but I couldn't even seem to write it for a uniform probability. I guess that makes sense :D

    It still seems counter-intuitive that it is mathematically meaningless to say, "Pick a random integer, with all integers having equal probability." I keep wanting to say, "But the Axiom of Choice!", but of course that only means you can pick from an infinite set, not that you can pick from an infinite set with all elements having equal probability.

    Thanks for the answer!

  5. Yup, the Axiom of Choice is powerless to help you here. But it's intimately related to the Banach-Tarski paradox and other related ideas, which show that some *other* kinds of probabilities don't exist -- e.g., if AC is true then there's no way of assigning a probability to *every* subset of the real numbers so as to satisfy the usual axioms.

    Why? Say that two numbers are "similar" iff their difference is rational; this is an equivalence relation; if AC is true then you can find a set A that contains exactly one element from each equivalence class: that is, no two elements of A are similar, and everything in R is similar to something in A. Now define B to consist of everything of the form {element of A + integer}. I claim that there's no possible choice for the probability of B, for the following reason. Consider all sets of the form B+c where 0 <= c < 1 and c is rational. These are all disjoint (else two elements of B differ by a rational number, contradiction) and their union is all of R (else we can find an element of R that isn't similar to anything in A, contradiction). Since they're all translations of one another, their probabilities should be equal. But now we've got a countable union of disjoint things with the same probability, whose total probability is supposed to be 1, and that won't do for the exact same reason as before: if P(B)=0 then we get P(R)=0, and if P(B)>0 then we get P(R)=infinity. So there's no possible value for P(B), and we lose.

    This shows that there's no way to assign a probability to every subset of R in such a way that probabilities are (1) invariant under translation and (2) countably additive. In fact, it shows that we can't do this even for the periodic-with-period-1 subsets, which can be turned into a proof that we can't assign a positive number to every subset of R with properties 1 and 2, even if we allow P(R) to be infinite after all.

    (That was rather condensed. I hope it makes sense.)

  6. It does make some sense, yeah. In fact, it reminds me quite a bit of my limited lay understanding of the B-T paradox... You could ask a similar question in regards to the probability of a random point within the original unit sphere winding up in one of the 5 "sections"... right? Since you can translate one of the sections to get the union of two other sections, and since the probability is supposed to be invariant under translation, this results in a paradox... if P(A) != 0, and A=B+C, and B and C are disjoint, and therefore P(A)=P(B)+P(C)... Well, my wife is calling me and I'm sure I already said something wrong. But this is starting to make some sense to me. A little bit. I think. :)

  7. Yup, B-T is all about non-measurable sets (just like B in the example I gave) and depends essentially on using the Axiom of Choice to pick one element from each equivalence class of something. Of course the somethings are a bit more complicated, and there's a bunch of group theory and geometry and stuff too, but it's the same basic flavour.