Probability Question
I have a metaphysical probability question:
Suppose there are an infinite number of parallel Earths. Alice uses a teleporter to teleport to a random Earth. Bob tries to follow Alice, but he has to guess which Earth she teleported to. What are Bob's chances of getting it right? Is there any way for a teleporter machine to randomly select an Earth out of an infinite number of them in a finite amount of time, or is there always going to be, practically speaking, only a finite amount of Earths for Alice to teleport to because of the limitations of the machine? What if I cheat and say the teleporter pokes a hole into the universe and the universe somehow, through a mysterious process, randomly picks an Earth out of an infinitely large ensemble for Alice to teleport to? Are Bob's chances of teleporting to Alice's world zero?
Suppose there are an infinite number of parallel Earths. Alice uses a teleporter to teleport to a random Earth. Bob tries to follow Alice, but he has to guess which Earth she teleported to. What are Bob's chances of getting it right? Is there any way for a teleporter machine to randomly select an Earth out of an infinite number of them in a finite amount of time, or is there always going to be, practically speaking, only a finite amount of Earths for Alice to teleport to because of the limitations of the machine? What if I cheat and say the teleporter pokes a hole into the universe and the universe somehow, through a mysterious process, randomly picks an Earth out of an infinitely large ensemble for Alice to teleport to? Are Bob's chances of teleporting to Alice's world zero?
Comments (32)
ETA: Bob makes a guess as far as the machine is concerned, but in the second scenario, Bob is at the mercy of the universe, so would it be possible for the universe to send Bob to the same parallel Earth it sent Alice to?
Are there a countable or uncountable infinity of worlds?
Reason I ask is that there is no uniform probability measure on a countably infinite set. In other words if we throw all the positive integers 1, 2, 3, ... into a bag, and reach in and pull one out, there is no sensible way to assign probabilities to that scenario, if we insist that each number has the same chance of being picked as any other. It's literally impossible to do that; and the impossibility is actually very easy to prove.
But there is a uniform probability on an uncountable set.
It turns out (I'm no expert, I saw a Sean Carroll video on this) that the multiverse cosmologists are terribly concerned about this problem. In an infinite multiverse, how many universes are like this, and how many like that?
They call this the measure problem. This is not to be confused with the measurement problem of quantum physics, the question of what is a measurement that makes the wave function collapse. As Wiki says:
"The measure problem in cosmology concerns how to compute the ratios of universes of different types within a multiverse. It typically arises in the context of eternal inflation. The problem arises because different approaches to calculating these ratios yield different results, and it is not clear which approach (if any) is correct."
This is a deep, unsolved problem in multiverse cosmology.
According to Thoralf Skolem's construction, i.e. by injecting infinite cardinalities in the model's structure, which is a countable set of symbols, there is at most a countable number of models of arithmetic.
The strong assumption here is indeed the continuum hypothesis:
In her lecture on the subject, Victoria Gitman confirms this:
If the physical multiverse is somewhat structurally similar to the arithmetical multiverse, it should also have a countable number of physical universes.
If we deny the continuum hypothesis, however, then most of the then uncountable universes would be unreachable because there can still only be a countable number of infinite cardinality symbols to do so.
Whoa, the subject is multiverse cosmology in eternal inflation. Physics theory. Not math.
Quoting Tarskian
Nothing whatever to do with CH, though I do take your point (if this is your point) that if cosmology considers an infinite multiverse, questions of set theory arise.
Quoting Tarskian
I'm not sure what any of this has to to with multiverse cosmology. Nothing, actually, though I'm familiar with Hamkins's student Gitman.
Quoting Tarskian
Wrong thread, wrong topic.
Quoting Tarskian
Multiverse cosmology is not about set theoretic models.
Quoting Tarskian
Non sequitur much?
Here, see this.
https://en.wikipedia.org/wiki/Eternal_inflation
and this.
https://en.wikipedia.org/wiki/Multiverse
I think you have conflated Hamkins's set-theoretic multiverse with the physical multiverse of speculative cosmology. Or maybe cosmetology.
It is.
If the (unknown) theory of universe is not categorical, then the physical universe is part of a larger multiverse.
Therefore, it is a mathematical problem.
In an infinite multiverse, there would be an infinite number of Boltzmann Brain universes, so what are the odds you're in one? 50-50?
The number of models (universes) depends on the spectrum of the theory:
The number of models is not necessarily countable, but according to Gitman, it happens to be countable for arithmetic.
Oh man, you do have a unique take on things.
I'm convinced I am one. Its statistically more likely than that I just showed up 13 billion years after the big bang.
The technical point is that if there are countably many universes, we can't put a probability number on it at all. Not 50-50, not 1 in a zillion, not anything. There's no way to assign such a probability. And the cosmologists are concerned.
I'm unsure but perhaps:
For each real number there is a universe in which that number is selected by Michael at random, and the real numbers are uncountable.
And even when a bijection between S and the set of natural numbers is specified, this only implies that S must be understood in terms of a monotonically increasing process rather than in terms of a completed basket of goods.
In both cases, the use-meaning of "infinity" should be understood to mean "finitization is decided by circumstances that are external to the specification logic".
So philosophical or practical questions about "infinite facts", as opposed to mathematical questions concerning the definition of mathematical infinity, should always be decided by elaborating assumptions until the facts concerned are "finitized". The presence of infinity in a non-mathematical question is only an indication that the question concerned isn't well-posed.
Bob's chances of getting it right are 1 out of infinity.
Quoting RogueAI
Its your thought experiment, that's a limitation you decide.
Quoting RogueAI
1 out of infinity still.
Actually, indirectly, it does.
It is actually a requirement that the (unknown) theory of the physical universe cannot predict its end.
If it could predict its end, then the physical universe is effectively a finite structure, which cannot participate in a multiverse. Furthermore, in that case, the physical universe would be entirely predictable from its theory. This means, for example, that free will cannot exist either.
The physical universe is finite but must be potentially infinite.
If this requirement is satisfied, then even the perfect theory of the physical universe will not be able to compute when the universe will disappear or for what reason or how exactly that will take place.
That is a question for you to answer: you are the one setting the parameters of your thought experiment. You seem to be unsure as to what you are asking: is this an abstract probability question or a physics/technology question? If it is the former, then you need not be concerned about practical limitations. If it is the latter, then you may as well scrap the whole thing, as it is obviously impractical.
Quoting RogueAI
The common-sense answer is zero, but as fishfry pointed out, it does not actually work in the commonly used mathematical probability theory (Kolmogorov probability). However, the question was not about mathematics. A mathematical theory of probability formalizes our pre-theoretical notions of what probability is, such as relative frequency or credence. The failure of one particular mathematical model to express something that can be expressed in an informal theory should give you a pause, but it is not a proof positive that the idea is altogether nonsensical.
1. With an infinite number of options, not all options can have equal probability.
2. As far as I can tell, most non-equal-probability algorithms for selecting a number preferentially choose numbers on the "smaller" side of things.
3. Thus, if you want to decrease the chance of someone randomly selecting your number (and we can assume, I guess, that these universes are numbered), choose a bigger number! As big as you can!
Universe 723645283928374827364817354871263498172736498712736481239 will be less likely to be randomly guessed than 100, given an infinite number of choices, I think.
Obviously there are some algorithms for selecting 723645283928374827364817354871263498172736498712736481239 more frequently than 100, but those algorithms are less likely to be the algorithm chosen, compared to an algorithm which makes 100 more likely.
(the above isn't a proof or a known fact, just an idea I intuit)
The first part of the address of alternative universe is the infinite cardinal ? of its isomorphism class. Next, you may (or may not) need an infinite ordinal ? to locate the universe within its class. So, if you know what two-tuple (?,?) Alice has used, you can follow her where she happens to be.
You will need the axiom of choice for this.
If the representation of the two-tuple (?,?) is not finite in size, it may actually be impossible to transmit it in finite time to the machine. But then again, in that case, Alice would not have been able to transmit it either.
I get that you're trying to explore statistics, but if one drills down into the OP, as written, it devolves into an exploration of the limits of machines to choose "randomly" among an infinite number of choices. Which is to say: wholly inadequate. Thus the answer is: not zero.
It is akin to the question: if an account holder can choose between 68 to the 8th power, number of possible passwords, what is the chance of guessing their password? This isn't a statistics question, it's a human behavior question.
What about the second party of the question, where I let the universe decide which world Alice teleports to?
Ah, that's an actual statistics question. The answer is it's a chance that is infinitely close to zero.
Is infinitely close to zero = zero? Like .999...=1?
If alternative universes in the physical multiverse are structurally similar to nonstandard models/universes in the arithmetical multiverse, then alternative universes are not similar to galaxies.
While the distance between galaxies is finite, there is no legitimate notion of distance between universes. The distance in between universes is "infinite" or "inapplicable" (whatever that may mean). It is not possible to reach them by physically traveling to them. With galaxies, you conceivably can.
You can see light arriving from some galaxies (but possibly not from all galaxies). You cannot see light arriving from another universe. It would never be able to bridge the gap. Intergalactic space is still filled with background radiation. Inter-universal "space" is not, because it is not even "space" as in exists within a universe.
These alternative physical universes actually influence each other, just like nonstandard models of arithmetic do, but the influence between universes is of a different nature than influences within a particular universe. For example, if a fact occurs in one universe but not in another, then this fact is fundamentally unpredictable. If a fact is predictable, however, then it occurs in all universes in exactly the same way.
Quoting tim wood
If it were possible to make such prediction, then the structure of the physical universe would be overly simple. It would be much simpler than the arithmetical universe. Its theory would have a single, finite model. Its theory would be able to tell you what exactly you are going to do next week, minute by minute. In that case, we are all just automatons.
God bless anyone who takes this drivel seriously.
Practically speaking? Yes. Conceptually? No.
I think we can make an even stronger statement: the only way that the usual probability rules (normalizability, additivity) can be satisfied on an infinite sample space is if all but a finite number of simple outcomes have probability zero.
I can try to give you an example of "an algorithm that gives all integers (without limit) SOME probability of happening" if you like.
edit. I'll just give it now, I just thought of one.
The algorithm is to produce a number in binary. Every step involves 2 flips.
flip 1: determine if the digit you're on should be a 0 or a 1, heads is 1, tails is 0.
flip 2: determine if you should STOP or generate another digit - heads means generate another digit, tails means stop.
so your first flip, you do Heads, so
1.
your next flip, you do Heads, so flip another.
your next flip, you do Tails, so
01.
your next flip, you do Heads, so flip another.
next flip is heads, so
101.
next flip is Tails, so stop.
final number in binary is 101, converted from binary to decimal is
5
This algorithm is theoretically capable of randomly producing ANY of the infinite sequence of integers, but it preferentially chooses lower numbers.
Not so (duh!)
Quoting flannel jesus
I am not sure I understand your algorithm and what probability distribution it gives to the integers, but clearly, there is any number of distributions that one can come up with for a countable set of disjoint events that gives all events a non-zero probability. For example, {1/2, 1/4, 1/8, ...}
Quoting flannel jesus
As for preferentially choosing lower numbers, I take it that you mean something like: "for any numbered event, there is a higher-numbered event with a lower probability," right? That has to be true, because otherwise there would be a lower bound on the probability of an unlimited number of events.
There's a 50% chance that my algorithm ends with one digit. There's a 25% chance my algorithm ends with two digits. There's a 1/8 chance my algorithm ends with 3 digits. Etc. Each additional digit is another 50% chance of stopping there.
And each individual digit has 50% chance of being a 1 or a 0.
But it looks like you agree that there's a way to generate a random number without it being true that "all but a finite number of simple outcomes have probability zero.", which is what my algorithm was trying to illustrate