"Anything that can happen, will happen, given enough time." -- Really?

Factual Questions
1

I have previously (and on this board) argued against this statement. To illustrate my reasoning, let’s pretend for a moment that there’s three things that can happen (with equal probability), A, B, and C. A ‘world’ would then consist of a sequence of these events, corresponding to a string of the symbols ABCABCABC… or ABBAACCABA… or anything like that. However, there are evidently possible worlds only consisting of events A and B, like ABABAB… or AABAABAAB… or ABBABBABB… or infinitely many (uncountably many, even) others.

However, intuitively and viscerally, it is absolutely clear to me that if you wait long enough, C will happen; and the probability for it not to happen does surely approach zero if you ‘try’ infinitely often.

So what’s the real deal, here? I can’t find fault with my reasoning in the first paragraph, yet I can’t for the life of me imagine a world ridiculous enough that something with a probability of one third per second, say, doesn’t happen even once if given an infinite amount of time.

As an added bonus question, if A, B, and C all have a probability of one third to happen, yet none must happen at any given point, is there a possible world in which none of the three, and thereby nothing, ever happens?

2

There’s no obvious fault. And the fact is that something with a probability of one third per second won’t happen even once if given an infinite amount of time… somewhere out in sample space

I think perhaps why you are struggling is that you are imagining how you would react if you were placed in such a universe, with the knowledge you have now. But that’s not valid. If you lived in a universe where, for example, dice never rolled twos or threes then you would not have any reason to believe that the odds of rolling a two or three were one in three. You would believe that there is some underlying principle that makes such an event impossible. Because that is what science would tell you: that rolling a two or a three has never been observed to occur.

And maybe we live in such a universe right now. How would you know? We believe for example that matter never spontaneously moves away from matter. We call that effect gravity and we have invented all sorts of convoluted (and mutually contradictory) theories to explain it. But what if in reality there are three equally probable options: mass attracts mass proportional to mass, mass repels mass proportional to mass and mass has no effect on mass? How would we determine that when the world we live in consists only of option A repeated endlessly?

Ridiculous examples of course, but there to illustrate a point. You are seeing this as being ridiculous because you have a retrospective view of the probabilities of the outcomes. But how do we define the probabilities except by retrospective analyses of the outcomes? If you never saw the other two events occur why would it strike you as odd that they never occurred?

If none must happen then they don’t have a probability of one third to happen. Your probabilities have to add up to one for all possible outcomes, and you’ve given us four possible outcomes: A, B, C and null. Either they each have a 1/4 chance to occur or they have probabilities or .3, .3, .3 and .1

I suspect what you are trying to describe is a scenario in which an initial event has a certain chance to occur, and if it does occur then it immediately triggers another probabilistic event which has three equally probable outcomes. For example an isotope decays or it doesn’t, and if it decays it releases a particle that is either re-absorbed, lost or absorbed by a neighbouring atom. In that case then, yes, there is there a possible world in which none of the three occur, with a probability equal to the chance of the initial event not occurring.
But you can’t say that every time you roll the dice you have a 1/3 chance of rolling 1-2, 3-4 and 5-6 and then say that you have a chance of not having rolled the dice at all. You define your probability of outcomes based on all the possible outcomes at the time the event takes place. You can’t say that one possible outcome is “the event never took place”. That’s like asking what colour an invisible four sided triangle is. The question only makes sense linguistically but not logically.

3

This is true as long as they’re mutually exclusive, but if more than one could happen in the same second, that is, if the probabilities are completely independent, then the cumulative chances that none of them occur in a given second are 2/3 * 2/3 * 2/3 or approx 0.2963

Alternatively, the OP may have only meant to indicate that they were equivalent in probability to each other as ‘one-third’, or might not be versed enough in statistics to understand the need for absolute probabilities to build models for this sort of thing. :slight_smile:

4

No, my question relates to a more fundamental issue within the interpretation of probability, I believe. There are things for which I can compute an a priori probability of them happening – for instance, a fair coin lands either heads or tails up with probability 1/2. I can test that by throwing a coin so-and-so often to then, a posteriori, conclude that heads or tails has probability 1/2 contingent on the number of times I’ve thrown it. Both figures are generally taken to be the same; my question essentially amounts to asking whether or not there’s a reason for that (for instance, if the sequences where both heads and tails crop up equally often are far more than those where they don’t), or if things just happen to work out that way.

No. If something has a 1/3 chance of happening, it has a 2/3 chance not to happen. Say I’ve got a button wired up to some generator of true randomness in such a way that if I press it, a red light will come on in 1/3 of all cases. That means I can push it down, and have nothing happen. Now figure I got three of those buttons, labelled A, B, and C. Each of them has a 1/3 chance of switching its light on, but I can just as well push down all three and have nothing happen.

5

I think that we’re touching on a number of interesting probability questions here, but I’m not quite sure how to tie them all in clearly. The first thing I thought about when you were mentioning ‘a priori’ and ‘a posteriori’ odds was the Principle of Indifference, which is about assigning equal odds to different outcomes when you have no other basis for how likely those outcomes are to occur, just based on not knowing anything different about the outcomes except for their names. :smiley:

In terms of the sequence tendencies, you might want to try and find some coin flip bell curves for different sample sizes of flips. I didn’t have much luck with a quick google search, but I remember these from math class back in school - basically the shorter a sequence you’re looking at, the more likely it is to have wild swings towards only heads or only tails. (Because in longer sequences, those will often get countered by opposing swings of the other kind.)

I’m intrigued by your button and light example, but - what happens if you push two of the buttons, say? Is the light supposed to go on if you leave it alone, or if you push it??

6

Perhaps I should reformulate my question. In the grab-bag of possible worlds (the term being used here in the same sense as in the OP, as a sequence of the events A, B, and C), there are uncountably many that do not include C (or any of the others) even once, yet the notion of using statistics to determine probabilities (as with the tossed coin) is predicated on us drawing almost always one where A, B, and C are represented in equal frequency, when I can see no reason why that should happen even particularly often (or at all, actually).

Or consider any of the many, many ways the string could be built with the wrong frequencies for A, B, and C. In fact, for each string with the ‘right’ frequency for C, for instance, it appears to me that I could construct again uncountably many strings where that frequency is off, by interspersing it with each of the uncountably many strings that only consist of As and Bs. So it would seem to me that the sets of ‘correct’ and ‘wrong’ worlds in the grab bag can be of equal cardinality at best, so there’s as much reason to grab a wrong one as there is to grab a right one. Yet that’s clearly counter to our everyday experience.

So basically, if the fact that something can happen doesn’t imply that eventually it will happen (i.e. if there’s no reason that strings with the correct distribution are favoured over those with a wrong one), how can any sort of probabilistic reasoning be considered valid?

7

Why are you only including the uncountably many instances where something doesn’t happen? There’s an uncountable many where it did as well.

Splitting infinities into percentages is a meaningless occupation. That’s why we deal in probabilities.

If something has X chance of happening, then it has X chance of happening. “Probability” doesn’t mean that given an infinite number of tries it will happen, or there’d be no difference between a probability of 99.99999% or 0.00000001%. In an infinite number of tries at either of those probabilities, you’re going to have an infinite number of hits–which is why I say the exercise is meaningless. All probability does mean is that the rules of the system are known to be such that your safest bet will be on a particular ratio of output. Nothing more and nothing less.

Now taking this back to the Titular Question: No, not really, but it would certainly be your safest bet.

8

That’s the question I’m asking – do I? Didn’t you also argue that I might get zero hits (if it’s not the case that everything that can happen, will happen)? And couldn’t I just as well get exactly three hits? If I can get zero hits, what does it mean for something to have a probability of 0.0000001%? If I can’t get zero hits, why?

9

I’m saying that mixing probabilities and infinity is similar to dividing by zero or trying to figure out how a baseball ever arrives according to Zeno. You’re approaching the problem in the wrong way.

10

Addendum, you might as well simplify your question to: Why does 1/infinity == 1 gazillion billion/infinity? Forget the probability portion of it, that’s a red herring.

11

I agree with this. One-third of infinity divided by infinity equals what? That is a very different answer than one-third of [arbitrarily large number] divided by [arbitrarily large number]. That you’re getting the one-third from something probabilistic confuses the issue.

This is true, but there is no contradiction. The probability and ratios of things when you try an enormous amount of finite times is different that trying infinite times. Offhand, it seems like you have a fairly good grasp on probability, it’s “infinity math” you’re having issues with.

12

All of which has a very well-defined answer in the right framework; a/0 = infinity, hence, a/infinity = 0, on the real projective line, or the Riemann sphere, for example, and the baseball arrives because infinite sums can have finite values.

All of which is pure smartassery, of course, but I think it shows that dismissing such questions out of hand misses the point somewhat.

Besides, I’ve presented a concrete model to which my question applies, the three buttons. If you keep pressing them for eternity, is it actually possible that one of the lights never lights up? (And to satisfy your curiosity, chrisk, what happens if you press two buttons at once is that each light lights up with a probability of 1/3 – you can always chose one to be ‘first’ in any case, since the only difference that makes is one of ordering, which doesn’t change the overall distribution --, and the light only comes on when the button is pushed, and then only with a probability of 1/3. Just so you don’t think I’m ignoring you. ;))
And if it is possible that one light always stays dark, what does it mean to say it has a probability of 1/3 to light up?

13

The odds of it not occuring becomes infinitesimally small as the line extends out into infinity. So that makes the answer to your question, according to official infinite math:

infinity/1 * 1/infinity = 1/1

I take that to mean that the light is guaranteed to light up.

14

Sort of correct, but in “official infinite math” dividing by infinity is undefined, just as dividing by zero is undefined.

Infinite is not a number, it is an endless process that can be represented as a limit. That gives the answer to Half Man, Half Wit’s problem. Although you can have a finite run of C not occurring that is as long as you want, and in fact as many finite runs of C not occurring as you want, the math says that no finite run can ever be infinitely long. That can only occur if C has a 0% chance of occurring, not a finite chance, no matter how small. The limit of any finite occurence occuring over infinite trials is finite, not zero.

I’ll apologize to the mathematicians for the impreciseness of my language.

15

Every mathematician in the world would be very surprised to learn that. I don’t know what you’re getting at with your infinite math, but I think you’ll have a hard time convincing me that it’s at all coherent.

Reasoning about problems like this is fine as long as you stay within the framework of probability theory. As soon as you try to bring intuition into the picture, you will go off the rails and get confused. Keep this in mind as you read the rest of my post.

If you have a sample space consisting of countably long sequences of three events A, B and C and the distribution of the next event is independent of the current one, then the probability of any particular sequence is just the product of the probabilities of each element in the sequence. This implies that the probability of seeing any individual sequence–that is, one where all the elements are fixed–is zero. Not very small, not infinitesimal, but exactly zero.

The reason that this is OK is that it satisfies the Kolmogorov axioms, which is all that we require of some notion of probability. Furthermore, it’s consistent with the case of finite length sequences in the sense that if you compute the probabilities for those sequences and take the limit as the length of the sequence increases without bound, you’ll get this measure of probability.

However, we can allow all but a finite number of the elements in any sequence be A or B or C, which means that the probability of seeing that element be any three of those is one. The product of an infinite number of ones is one, so the probability of drawing a sequence with constraints on only a finite number of its elements is just the product of the probabilities of those elements. So for instance, the probability of seeing a sequence that starts with A is just the probability of A. The probability of seeing a sequence that has A in the second, thirty fourth and 100!-th positions is just the probability of A cubed.

However, the probability of seeing any sequence that has constraints on an infinite number of elements is exactly zero. That includes sequences that don’t contain C, or don’t contain C after a finite number of elements, or anything like that. That doesn’t mean that those events can’t happen! Every single sequence has probability zero, but one of them has to be realized, so it must be the case that an event with probability zero happens. That can’t be the case for countable or finite sample spaces, but when you start looking at uncountable sample spaces, your intuition becomes worthless like I warned above.

16

Which is what I said. And I’m fairly sure that I understood your post and that you didn’t say anything different from what I did, I just probably stated it in terms that aren’t correctly used if one was in the math world.

17

I think the issue is that, when you get into infinite sample spaces, an event can have probability 0 without it being impossible, or have probability 1 without it being impossible for it not to happen. (On preview, I think this is what ultrafilter is saying too.)

Your sample space, if I understand correctly, is the set of all strings, finite or infinite, made up of the letters A, B, and C, and you want the probability that a randomly chosen such string includes no Cs anywhere. This, I think, would be similar to asking the probability that a randomly chosen real number in the interval from 0 to 1 has no 6 digits anywhere in its decimal expansion. Such numbers do clearly exist, but if the set of such numbers has measure 0, the probability that one such would be chosen is 0. (Any mathematicians know whether it does or not? or whether I’m even thinking about this correctly?)

18

I wasn’t aware there was ‘official infinite math’ – as I’ve already pointed to, in several frameworks, division by infinity (and zero) isn’t a problem at all; in others, it is. So, whether or not you have a problem depends on the framework you’re using, nothing else.

Whether or not infinity is a number depends on the concept of number you’re using, nothing else. By the same token, you can say that pi isn’t a number, or 2/3, or 5.

If you’re talking about convergence and such, perhaps. But is the number of elements in an infinite set an ‘endless process’? Is the length of the Koch snowflake?

And that’s again where my problem comes in: you can compute the probability for any given finite run to not include C, no problem. And the way the math for that looks, if you go to infinity, that C occurs becomes a necessity, i.e. anything that can happen, will happen. But then, what exactly prevents worlds of the form ABABAB… (or any of the other forms I have so far outlined that would show a different distribution for C than its having 1/3 probability would lead us to expect) or whatever from being actualized? Under apparently the same rules of probability that state that C most occur eventually, it’s perfectly conceivable that, at any given point, C does not happen.

19

How do you define probability? For that matter, how do you define “possible”?

More to the point, if one lived in a universe where C never came up in an infinite number of trials, is there any definition of “probability” under which one could say that C has a 1/3 probability of coming up? Is there even any definition of “possible” under which one could say that C is even possible?

20

Yes. It’s exactly the same situation, except that you have ten possible events rather than three.