I think that what’s tripping you (and, in a different fashion, snailboy) up is the Gambler’s Fallacy.
Let’s illustrate it with something a bit less abstruse than floating bowling balls: Tossing a fair coin (one with exactly equal chances of landing on either side).
If you toss the coin once, the probability of it coming up heads is 0.5.
Toss the coin a second time. The probability of any random pair of tosses of the coin giving two “heads” results is 0.25. Does that mean that if your first toss came up heads, the second toss only has a 0.25 probability of heads? Of course not. The outcome of the first toss doesn’t affect the outcome of the second in any way. The probability of it coming up heads is still 0.5.
Toss the coin 24 times. It is possible for it to come up heads all 24 times–the chance of it doing so is 1 in 16,777,216. Once you’ve managed to hit that one-in-millions chance, however, the probability of the next flip coming up heads is still 0.5. There is no mystical force that will turn the coin over because a “tails” result is “due”.
Now the floating bowling ball has a substantially lower probability than a fair coin landing heads-up. However, since we’re assuming it’s floating because of the random motion of particles in the ball, there is no causal link between the levitation of one ball and the levitation of another. The chance of finding two bowling balls levitating is half the chance of finding one, yes…but the chance of finding a second one after finding the first is the same as the chance of finding only one. And if you’ve found two, well, the third one has the same chance. At no point does finding another floating ball become impossible just because you’ve already found others. (The same reasoning can be applied to the probability of the ball continuing to float.)
**Balance ** (great name, btw), my objection to the bowling ball situation goes beyond the gambler’s fallacy. The situation is that our physics is descriptive. The so-called laws of physics do not rule reality, they just model it for our understanding. We say that things fall to the ground, because we have seen them fall to the ground. If we throw any number of balls to the ground, here on earth, we will observe that they all fall to the ground. If we observed one that didn’t, we would chalk it up to scientific error and carry on.
Now, if we have this other place, where bowling balls consistently and predictably hover at one inch from the ground, that observation doesn’t hold our laws of physics. Either gravity works differently in that place (and knocks down a good chunk of our understanding of the universe), or there is another phenomenom at work there, that explains why bowling balls hover in a way that would be repeatable elsewhere, if that phenomenom were also at work in that other place.
As for **Pochacco’s ** problem with the ABC’s, let’s see it this way. ANY particular combination of letters has probability zero. Be it …ABBAABBAABBA… , …ACABBCAACBCA… or …CCCCCCCCC… Whatever it is you are observing, has the same zero probability, and yet it happened.
Similarly, in the lottery, the 00000 has the same chance of coming out than any other particular combination. People might hesitate to play it because they compare that probability with the probability of any non-repeating number coming out. But consider any number that has come out and see how many times that one number has come out before. All combinations are just as [un]likely.
I’m clearly out of my depth on this question, yet I will forge ahead anyhow.
We all agree there is some chance that any bowling ball will hover for a second, right? Let’s be extremely extremely generous and assign that a probability of 1/26th. You might model it by picking a letter of the alphabet, say Z.
So in our normal emprical planet Earth, we see ABBKCKDLKASFAJLKJELKE, and hardly ever a Z to be found, although there is nothing forbidding Zs from appearing.
Let’s examine an infinite series of letters, symbolizing all the bowling balls in the universe, across some time period. Could there be a sequence of eight thousand Zs in a row? It would be very unlikely, a very low probability, and yet it could happen. What about eight million Zs? Eight billion? Same answer, very very unlikely, maybe even zero probability (if I understand the other answers) but clearly possible.
Now if you experience a bunch of Zs in a row, what’s the probability that you’ll keep on getting more Zs? Very very low, but clearly possible. In the same way, if you momentarily experience a bunch of levitating bowling balls, the odds that they will continue to do so by chance are very very low, but clearly possible.
Do the laws of probability work differently in that sequence? No, they just came up with a very unlikely roll. Do the laws of physics work differently on that planet? No, it’s just a very unlikely outcome. Might a scientist on that planet deduce different laws of physics? Almost certainly, but he would be wrong.
Or maybe we are the misguided scientists living in a strange corner of the universe where a billion Z’s have occurred in a row, and the true “laws” of physics actually make it ridiculously unlikely that bowling balls will fall to the ground and not hover eternally.
Muhahahaha!
(Now is the time to discuss the Anthropic Principle… Except my atrophied mind lost its ability to do so about 15 years ago.)
Well we’re talking about the physical world here. The original post asked about an infinite universe. Personally, I don’t believe the physical universe can possibly be infinite because of the fallocies we’ve talked about along with others, but we’re assuming it is possible. Mathematics really has no clean way of dealing with infinities. There is an infinite number of sets that you can create with only A’s and B’s. However, for every single combination, there is an infinite number of combinations with some number of C’s. As proof, for any combination of A’s and B’s you can come up with (if it was possible for you to create an infinite list), I could come up with any number of C’s and add them into the list between various other letters. Thus I could come up with an infinite number of lists with the same pattern of A’s and B’s (if you took out the C’s) just by adding C’s, again if it was possible for humans to make infinite lists. It’s the same concept as how there are an infinite number of integers, but an infinite number of real numbers between each integer. I think you get what I’m saying.
So with that said, what are the odds of getting a list without C’s? There’s an infinite number of lists as well as an infinite number of lists without C’s, but infinity divided by infinity is as meaningless an operation as you can hope for. But like I pointed out in the previous paragraph, for every list without C’s, there’s an infinite list with C’s. So it forms a 1:infinity ratio. If it was a 1:9 ratio, we’d say there was a 1/10 chance, or 10% chance. Well I think it’s obvious that infinity plus 1 is still infinity. There are an infinite number of positive integers. Add 0 to that list (the countable numbers) and there are still an infinite number of them. So we could say that there’s a 1/infinity chance of getting a list without C’s.
Now what is 1/infinity? In mathematics, I’m pretty sure that’s undefined. There are a few real world examples though. Density is mass divided by volume. Just the same, volume is mass divided by density. A singularity has a positive mass (we’ll say 1 for convenience) and zero volume. How much density does it have? Infinity. That’s what people say anyway. So from this, we can conclude that infinity = 1 / 0, and that 0 = 1 / infinity. Need more examples? How about speed? An object moves a distance of 1 in a time of 0. What is it’s speed? I’m sure there are plenty of other examples, but I need to sleep so I’m going to wrap this up.
So what are the odds of randomly getting a list without C’s? 0. It’s not some incredibly-close-to-0 number. It’s 0. Yes, you can make a list without C’s, but that wouldn’t be random. Some people have said the odds can be 0 and it still be possible, but it can’t, not by random chance. It’s impossible to continue a lucky streak until infinity without cheating the system somehow.
It’s been pointed out that any combination that could pop up would have had a 0 chance of occuring, and I agree with that. There is a 0 chance of any given point in the universe being the location of a singularity, and yet there are points that are singularities. But any individual list has a chance of 0 of occuring. On the other hand, the chance of having 0 C’s is 0, while the chance of having infinity C’s is 1. If every possibility has a 0 chance of happening, yet one of them will, you can bet that whatever happens will have had a 0 chance of happening. But if something has a 0 chance of happening and something else has a 1 chance of happening, I’m betting on the latter.
Technically, yes, you can’t have a random letter generator working for an infinite time. We may already have something equivalent to that though. Earlier, people were talking about the half-life of protons. There’s a 1/2 chance that a proton will decay before it reaches that age. There’s a 3/4 chance it’ll decay before it’s twice that old, etc. Heat death is a theoretical end of the universe that ends with all matter decaying into photons. As you go forward in time, the odds of a proton having decayed increases, but there’s no time that you can choose and say it will have definitely decayed by then. Does that mean it’s possible for the universe to follow the heat death route, yet have at least one proton survive infinitely? As with the letter generator, I’ll have to say no. The last proton can survive for any real number of years, but not infinity. It can’t get lucky forever. That is assuming there are a finite number of protons though. If there are infinite protons due to an infinite universe, all bets are off.
Maybe so, but since I don’t really know how to deal with infinity in the physical world (do you?), I have to go with an idealized mathematical formulation.
What do you mean by “clean”? I personally believe that the way mathematics deals with the infinite is quite elegant. But then again I’m already sold to the beauty of math.
Well, that’s not the way you do things in probability theory (1/infinity is pretty meaningless), but yeah, sure, if you want.
Sure, I agree. I’ve said so many times.
:dubious: And what do you mean by that? Don’t you agree that getting, say, the sequence {A, A, A, …} as a result of your random letter generator is a valid output? You have 1/3 probability of getting the first A, then if this happens, 1/3 probability of getting the second A, and so on. Of course, the probability of the sequence {A, A, A, …} being the output of your random letter generator is 0 (and exactly 0). But the point is that, in an uncountable probability space, events with probability 0 are not necessarily invalid events.
After all, the probability of any single sequence occurring is 0, and you agree with that. But naturally if you start your infinite sequence generator (in an idealized world, of course, since infinite sequence generators don’t exist in the physical world), it will return a sequence. You can then point at it and say that this particular sequence had a probability of 0 of being the one returned, and yet this is exactly what happened.
See above. Any result you obtain (by random chance) had an a priori 0 probability of occurring.
Well, yeah, I’m not disagreeing with that.
I see that what you’re using some sort of probabilistic exponential decay model. I should mention that if there’s 1/2 chance that a proton will have decayed when we reach the half-life of the proton, and a 3/4 chance it will have decayed when we reach twice this time, the only time we can say that there is a probability 1 that all protons will have decayed is when we “reach” infinity. In other words, if we’re in the real world, we never reach that point. Is it possible for a proton to survive indefinitely long? I say why not? After all, do we even know how to accurately describe the decay of particles when we have a very small number of them? When we reach these points, your model isn’t very useful.
Snailboy, in your post you offered what you took to be a proof that the probability of a non-C sequence is zero. You said this means it is certain that the sequence will have a C in it somewhere.
Think of a dartboard. How many points are on it? Now pick one point. What is the probability that a dart thrown at that board will be centered exactly on the point you picked?
Since the point, being a point, has area of zero, it seems the probability must be zero.
So: The probability is zero, yet, the dart can surely hit that point.
I don’t know if the probability actually is zero, but my argument that it is zero is analogous to your own argument that the non-C probability is zero.
I missed this paragraph on my first readthrough. I see you already understand the point I made in my previous post.
And here at the end, you say what I think is the correct thing to say: that we should bet on there being a C somewhere.
But this does not support your conclusion that it is absolutely certain there will be a C somewhere. In fact, the paragraph above lends support for the opposite conclusion. The probability is zero, but it could still happen. It is, therefore, not absolutely certain that it won’t happen.
In the first sentence, you say nothing can happen that has a probability of zero. In the second, you say that certain things happen that have a probability of zero.
