Is probability meaningful in cases of infinity?

Is it meaningful to speak of probability in cases of infinity?

For instance, consider me having an infinite line of balls arranged in the manner: -

Red, Green, Blue, Red, Green, Blue, Red.......

Now, I'm picking a ball randomly from this line. Am I allowed to ask the question, "What is the probability that the ball you picked is Green?" And if an answer exists, what would it be?

My gut feeling is that the probability doesn't change, i.e. it still remains $1/3$ (just as in the case of $3$ balls of different color). But I'm not able to justify this mathematically. It's because I thought that in this case, the number of red, green and blue balls would all be infinitely many, and so the probability computation would essentially involve infinity, making me doubt the question's validity.

I mean, in this case, by definition,

$P$ (picking a green ball) = (No of green balls)/ (Total no of all the balls) = $∞ / ∞$, which is undefined.

Or am I going about it the wrong way?

• How would you characterize 'picking a ball randomly from this line'? – copper.hat Dec 12 '13 at 19:39
• @copper.hat: Sorry, but I'm not sure I understand your question. – Train Heartnet Dec 12 '13 at 19:43
• Well, to some extent, the issue is how to characterize picking something randomly from a countable set. You cannot have a uniform probability in this case. However, in your example, since the pattern repeats, you are really only selecting one number from $1,2,3$, so you can assign a uniform probability here since there are only three outcomes (or at least any outcome is equivalent to one of three canonical outcomes). – copper.hat Dec 12 '13 at 19:55

The closest thing is the concept of natural density As you take larger and larger finite sets starting at $1$, the fraction of red balls converges toward $\frac 13$ If it converges, you can reasonably say the probability of a red ball is $\lim_{n \to \infty}\frac {\text {number of red balls in }[1,n]}n=\frac 13$
Using the intuitive definition of probability leads to issues here, but the measure-theory based probability was formulated to deal with such things. There indeed you can define the probability measure in such a way that event of drawing a green ball has a measure of $1/3$...
• Well, we can indeed find measures that give green probability $\frac{1}{3}$. But "most" measures won't. – André Nicolas Dec 12 '13 at 19:52
• Perhaps the sample space should be $\Omega = \{ \{ 1,4,7,10,...\}, \{2,5,8,11,...\},\{3,6,9,12,...\} \}$ instead? – copper.hat Dec 12 '13 at 20:39