Calculating the probability that the Universe is finite

[Buzz Bloom]

TL;DR Summary

The calculation is based on two numbers: (1) the value for the curvature density, and (2) the +/- error value. Both of these values are given in equation 47b on page 40 of the reference listed in the body. The calculation is also based on three assumptions: (1) the probability distribution is Gaussian, (2) all four Ωs can have a range of values (to find a best fit to database values), and (3) that the universe is not and cannot be flat.

Reference

https://www.cosmos.esa.int/documents/387566/387653/Planck_2018_results_L06.pdf​

I note that the use of Gaussian probabilities is mentioned many times in the reference. However in many discussions via posts in many threads, there seems to be a consensus that the distribution is actually only approximately Gaussian, so the results of the calculation presented will likely not be as accurate as it is shown to be.

The equation 47b is

Ω_k = 0.0007 +/- 0.0019 .​

This means that the integral representing the probability for

0 < Ω_k < 0.0007​

is

(1/√π)∫₀^7/19 e^-x2 dx = 0.19882 .​

The integral for

0.0007 < Ω_k < ∞​

is

(1/√π)∫_7/19^∞ e^-x2 dx = 0.5 .​

The probability P_fu that the universe is finite is

P_fu = 0.19882 + 0.5 = 0.69882 .​

The probability P_iu that the universe is infinite is

P_iu = 0.30118.​

Note that the probability that the universe is flat is 0 because the probability for a single value 0 would be calculated by an integral from 0 to 0.

This the first of a series I plan to calculate. Next is calculating the expected value of Ω_k and the corresponding expected value of the radius of curvature. I am guessing (without yet having started to calculate) that the Ω_k answer will not be very far from 0.0007, but also larger and not by a trivial amount.

EDITED 2/8/2022

 

[anuttarasammyak]

Finite with almost 70% provability interests me. I thought wrongly that the Universe had been revealed to be infinite or open. Thanks.

 

[sysprog]

I think that it's an untestable hypothesis that is not validly subjectable to probabalistic analysis $-$ there is no legitimately comparable phenomenon set available to provide a foundation for such analysis.

 

[PeterDonis]

I note that the use of Gaussian probabilities is mentioned many times in the reference.

That is for the assumed distribution of errors in the measurements underlying the calculations. It is not in any way a claim that there is a meaningful "probability distribution" for the spatial curvature of the universe, or that the results given in the paper express the parameters of such a distribution. You can of course compute the portion of the total area under a Gaussian curve that corresponds to values in or outside some range. But that is very different from the claim that the result of such a computation is meaningful as a probability that the universe is positively curved/spatially finite.

 

[PeterDonis]

Finite with almost 70% provability interests me.

"70% probability" is not the same as "70% provability", which is an even stronger claim. (But even the weaker claim is not meaningful in this context.)

I thought wrongly that the Universe had been revealed to be infinite or open.

I don't think any cosmologist has claimed that this question is settled. Cosmologists often use exact spatial flatness in their models because it is the simplest case to model mathematically, and as long as "exactly flat" is within the error bars of the measurement, this modeling convention is justifiable. But it is not at all the same as claiming that we know our universe actually is spatially flat. Since error bars are always finite, if the error bars include exact spatial flatness, they will also include some region of very small positive and very small negative curvature. And as long as that continues to be the case, the question will remain open. The only way to resolve it for sure would be for the error bars to move away from spatial flatness, in either the positive or negative direction, far enough to no longer include exact spatial flatness. But that has not happened.

 

[PeterDonis]

Next is calculating the expected value of Ω_k and the corresponding expected value of the radius of curvature. I am guessing (without yet having started to calculate) that the Ω_k answer will not be very far from 0.0007, but also larger and not by a trivial amount.

Your guess is obviously false. The value you refer to in the paper is the expected value of $\Omega_k$ given the data (i.e., the mean of the Gaussian distribution obtained from all the measurements with their assumed Gaussian distribution of errors). The plus or minus value given for that $\Omega_k$ value is just the standard deviation of the Gaussian distribution.

 

[PeterDonis]

Note that the probability that the universe is flat is 0 because the probability for a single value 0 would be calculated by an integral from 0 to 0.

Yes, and since the universe could in fact be exactly flat, this correct statement is one way of seeing why any intepretation of calculations such as you have been making in terms of "the probability of the universe having such and such curvature" is meaningless.

 

[Orodruin]

That is for the assumed distribution of errors in the measurements underlying the calculations. It is not in any way a claim that there is a meaningful "probability distribution" for the spatial curvature of the universe, or that the results given in the paper express the parameters of such a distribution.

Well, cosmologistst typically do Bayesian statistics (and definitely in this case as they are using CosmoMC). This means that the posterior inherently is considered a probability distribution. It is of course dependent on whatever prior went into the analysis and this needs to be considered in the interpretation. In a model where a continuous prior for curvature is natural, it should not come as a surprise that you get zero probability for exactly flat out. On the other hand, a model where you predict exactly flat would also have that as an output. The more interesting computation would be to compare the Bayesian evidence for these two models, which would give you an idea of ”how much should this data make us lean towards one or the other model”. However, the outputis cerainly a probability distribution. You just have to be very careful with what conclusions you draw from it.

 

[PeterDonis]

The more interesting computation would be to compare the Bayesian evidence for these two models

I think that would depend on which two models you try to compare. If you try to compare a model that says "the universe is exactly flat" with a model that says "the universe is positively curved", unless the latter model makes some particular prediction about what the curvature parameter should be, you're back to a continuous prior for the curvature parameter, which gives the "exactly flat" model zero probability and so isn't a meaningful comparison.

But if you could compare two models, one which says "the universe is exactly flat" and one which says "the universe is spatially curved, but the curvature parameter is very small so it's hard to distinguish from flat", then perhaps you could do a useful Bayesian analysis on those grounds. The problem here is that we don't have a model (at least AFAIK) that makes the first prediction. Our best current model, which is Lambda CDM plus inflation, makes the second sort of prediction (and it doesn't specify whether the very small spatial curvature is positive or negative). I'm not aware of any other model with which we could compare it in this way.

the output is cerainly a probability distribution. You just have to be very careful with what conclusions you draw from it.

Yes, agreed. I was talking about the "very careful" part.

 

[Orodruin]

I think that would depend on which two models you try to compare. If you try to compare a model that says "the universe is exactly flat" with a model that says "the universe is positively curved", unless the latter model makes some particular prediction about what the curvature parameter should be, you're back to a continuous prior for the curvature parameter, which gives the "exactly flat" model zero probability and so isn't a meaningful comparison.

No, this is not how Bayesian model comparison works. What you would do is to compare the Bayesian evidence of the respective models:
- A model where the distribution of the curvature parameter is a delta function at zero.
- A model where you have a continuous curvature prior.
The ratio of the Bayesian evidence for the models (computed separately) in essence tells you how to update your relative belief in these models. Unless your second model restricts the curvature parameter to be very close to flat, I suspect this would come out strongly in favor of the flat model.

 

[PeterDonis]

What you would do is to compare the Bayesian evidence of the respective models:
- A model where the distribution of the curvature parameter is a delta function at zero.
- A model where you have a continuous curvature prior.

I'm not sure you can compute meaningful Bayesian evidence with the first model. Wouldn't a delta function prior automatically give a zero posterior for any observation except a curvature of exactly zero?

Or are you saying the actual prediction of the first model for what data would be observed, based on finite measurement accuracy, would be some continuous Gaussian distribution of observations with a mean of zero and a narrow standard deviation?

 

[Orodruin]

I'm not sure you can compute meaningful Bayesian evidence with the first model. Wouldn't a delta function prior automatically give a zero posterior for any observation except a curvature of exactly zero?

Yes, but that is not what evidence is. The Bayesian evidence is in essence the likelihood of the data given the model. In the case of a model with fixed parameters, this is trivially the likelihood of the data given those parameters. In a model where the parameters are not fixed, it is the integral of the likelihood over the parameter space weighted by the prior. The evidence ratio is in essence the ratio of how likely each model was to produce the observed data (of course experimental errors and uncertainties need to be included here).

 

[PeterDonis]

The Bayesian evidence is in essence the likelihood of the data given the model.

Yes, I understand that. I'm just not sure how you get a nonzero likelihood of any data other than "curvature exactly zero" from the delta function model without some additional assumptions, to do with finite accuracy of measurement if nothing else, so that there is some continuous distribution of predicted data. But then I'm not sure what the difference between this model and a model whose prior for curvature is a narrow Gaussian instead of a delta function.

Perhaps it would help if you described in more detail how you envision computing the Bayesian evidence for each model from the data given in the OP.

 

[Orodruin]

Yes, I understand that. I'm just not sure how you get a nonzero likelihood of any data other than "curvature exactly zero" from the delta function model without some additional assumptions, to do with finite accuracy of measurement if nothing else, so that there is some continuous distribution of predicted data. But then I'm not sure what the difference between this model and a model whose prior for curvature is a narrow Gaussian instead of a delta function.

That is not how the measurements work though. The data is significantly more complicated in this case, but let us assume that you measure the parameter with some experimental uncertainty s and central value 0. The likelihood of obtaining a measurement x is then the Gaussian likelihood at x with standard deviation s. Again, this is not about the posterior distribution of parameters in the model - the posterior within the model will obviously still be a delta function at zero. However, this is not what you use for model comparison. You use the Bayesian evidence.

For the model with continuous parameter $y$, the Bayesian evidence would be given by
$$
\int L(x,y,s) \pi(y) dy.
$$
Here $L$ would be the likelihood of measuring $x$ when the expectation is $y$ and the standard deviation $s$ and $\pi(y)$ is the prior within the continuous model.

 

[PeterDonis]

let us assume that you measure the parameter with some experimental uncertainty s and central value 0

In other words, yes, you do have to add an additional assumption about experimental uncertainty. As I said in my previous post. And then, as I said in my previous post, I'm not sure how this model is any different than a model with a continuous prior for the curvature parameter, in terms of the likelihood of observing some particular data.

this is not about the posterior distribution of parameters in the model

Yes, I know that. I'm not talking about the posterior distribution in the model, I'm talking about Bayesian evidence, i.e., the likelihood of the data given the model.

 

[Buzz Bloom]

Hi @PeterDonis and @Orodruin:

I do very much appreciate your posts. Unfortunately I am not educated well enough to grasp implications based on Bayesian methods. However, I can visualize a process of producing a numerical error result corresponding to a particular choice of values for each of the four Ωs (which sum to unity) based on a reference database consisting of pairs of numbers: (1) for the redshift of light from a distant galaxy, and (2) for the distance to the galaxy measured by the brightness of a particular kind of supernova.

A process might work as follows. (I do not claim that such a process is actually used.) One might randomly choose a value for each of the four Ωs (which sum to unity), and then using the Friedmann equation with each of the database pairs to calculate a corresponding value of lookback time based on the z value. (I have just tried to find on the Internet the formula for the lookback time involving an integral, but I failed to find it.) The lookback time times c gives a distance value. Let E be the difference between this value and the distance value in the database pair. The process is repeated multiple times to find the combination of Ωs which gives the minimum of the sum of the E values for a specific four Ωs. These minimum values are the means of a distribution, and the standard deviation of the values can also be calculated. I do not know if this distribution would be Gaussian, but I would expect there would be a sufficient closeness to Gaussian for the purpose of calculating reasonable expected values while assuming a Gaussian distribution.

I understand that it may be a reasonable interpretation that such an approach would not be applicable to cosmology. However, I am unable to visualize any reason for this. I would much appreciate what either of both of you might say to explain such a reason.

ADDED
I did the math for lookback time T. It took me quite a while, but I think I have it correctly.

T = (1/H₀) ∫_a(T)¹ F(a) da​

where

a(T) = 1/(z+1),​

and

F(a) = (Ωr/a² + Ω_m/a + Ω_k + a²Ω_Λ)^-1/2 .​

Regards,
Buzz

 

[Orodruin]

In other words, yes, you do have to add an additional assumption about experimental uncertainty.

All measurements cone with uncertainty. If not there would be no point in discussing statistics.

And then, as I said in my previous post, I'm not sure how this model is any different than a model with a continuous prior for the curvature parameter, in terms of the likelihood of observing some particular data.

It is not, why would it be? If you find a model where it is natural to get values very close to flatness (such as inflation), it is going to be very very difficult to separate from a model with exact flatness by construction. However, if you have a model where a prior that is reasonably flat over a large region of non-flatness is natural (such as not having inflation), then such a model will be disfavored by data relative to a model with exact flatness.

 

[PeterDonis]

If you find a model where it is natural to get values very close to flatness (such as inflation), it is going to be very very difficult to separate from a model with exact flatness by construction.

Yes, and this is the sort of model I was talking about in the earlier post of mine that I referred to: a model whose prior for curvature is a very narrow one centered on flatness. Of course a model whose prior for curvature is much wider is strongly disfavored by the data.

 

[PeterDonis]

All measurements cone with uncertainty.

Yes, but the effect of measurement uncertainty on the likelihood of particular data given the model can vary widely by model. For example, given a model whose prior for curvature is a very wide distribution, the effect of measurement uncertainty on the likelihood of particular data for this model will be very small. Whereas, given a model whose prior for curvature is a delta function at zero, the effect of measurement uncertainty on the likelihood of data is highly significant: without measurement uncertainty, the likelihood of any data other than exactly zero given this model is zero.

 

[Buzz Bloom]

Yes, and since the universe could in fact be exactly flat, this correct statement is one way of seeing why any intepretation of calculations such as you have been making in terms of "the probability of the universe having such and such curvature" is meaningless.

Hi @PeterDonis:

I do not understand why, nor can I imagine why,

(1) "the universe could in fact be exactly flat", and​

(2) the calculation of probabilities for positive and/or negative values of Ω_k​

can not relate to the possibility "1" lead to your conclusion:
"the probability of the universe having such and such curvature" is meaningless.

I would very much appreciate reading your reasoning about this. It seems to me that the method does calculate any small continuous range of values, but no single value can be calculated. This does not seem to be a reason why what it can calculate is not a useful understanding of an approximate cosmological fact, particularly since there is no observable evidence, and currently also no current anticipated method to gain future evidence that the universe is definitely flat.

Regards,
Buzz

 

[PeterDonis]

I do not understand why, nor can I imagine why,

(1) "the universe could in fact be exactly flat", and​

(2) the calculation of probabilities for positive and/or negative values of Ω_k​

can not relate to the possibility "1" lead to your conclusion:
"the probability of the universe having such and such curvature" is meaningless.

Because the relevant calculation is not the calculation of a probability distribution for $\Omega_k$ but a calculation of a probability distribution for models. The statement "the universe could be exactly flat" is not a statement about one possible place where our actual universe could fall within the distribution of possible values for $\Omega_k$ that you calculated. It is a statement about possible models: it says that one possible model of the universe is that there is some constraint that forces it to be exactly spatially flat, always.

Your calculation says nothing about the relative likelihood of such a model as compared to, say, an inflation model in which the spatial curvature can start out anywhere and gets driven to be very close to flat by inflation. (Note that in such a model, the universe is not exactly spatially flat.) But that is the sort of relative likelihood we have to assess if we want to answer the title question of this thread.

no current anticipated method to gain future evidence that the universe is definitely flat

We could never gain direct evidence that the universe is exactly flat, because our measurements always have some finite uncertainty. Any belief we might have about the universe being exactly flat (as opposed to just so close to flat that we can't measure any difference) would have to be based on our belief in some model in which that was a constraint, and such a model would have to be established by other evidence besides measurements of spatial curvature.

 

[Ibix]

I would very much appreciate reading your reasoning about this.

Fundamentally, the point is that "the probability that the universe is closed" isn't a complete concept. The universe is what it is, it can't be anything else, so probability theory doesn't apply. It's like tossing a coin and seeing it land heads and asking what's the probability that it landed heads. It's not really a meaningful question because that toss is a done deal - probability is for things where several outcomes are possible. What you are really trying to calculate is "the probability that the universe is closed given our current understanding of theory and our data" - which is in the domain of probability theory. That's like me tossing a coin and hiding the result from you. The toss is a done deal, but you don't know the outcome.

Bayes' Theorem is your tool for this. Here's an example. I get a coin from your wallet and toss it ten times, getting nine heads and one tail. Do you think that the coin is fair?

The data implies a mean of 0.9 with a standard error of 0.3. That's quite a long way from 0.5, so a naive view might be that your best estimate is that the coin is biased. But you have to admit that nine heads could just be luck from a fair coin.

But, come on! I specifically said I took the coin from your wallet, so it's probably legit legal tender, and national banks (and vending machine companies!) are really finnicky about precision manufacture. If one particular type of coin were biased due to manufacture that'd be on Wiki or Mythbusters by now. So if it's biased it must have been damaged somehow, and (a) coins are pretty tough, and (b) damaged coins go out of circulation fairly quickly these days. How did it get damaged? And how unlucky would we have to be to draw a biased coin to do our test?

This kind of stuff is a priori information about the coin: it almost certainly is fair. We have two models in mind - a mint coin and a damaged coin. And initially we believe that we are hugely more likely to have a fair coin than an unfair one. But then we do our experiment and get a fairly extreme result. That should weaken our belief that this is a fair coin, but probably not much. Bayes' theorem will tell you what your a posteriori beliefs should be, given your a priori beliefs and the data from the experiment.

You can apply similar reasoning to your actual problem here. You have three models in mind - open, flat, and closed universes, and you have data in the form of a measure of curvature with error bars. You need to write down your prior beliefs, then see how the data affects them. Since what the data says is pretty equivocal, it probably won't change anything much.

 

[PeroK]

The universe is what it is, it can't be anything else, so probability theory doesn't apply. It's like tossing a coin and seeing it land heads and asking what's the probability that it landed heads. It's not really a meaningful question because that toss is a done deal - probability is for things where several outcomes are possible.

This is not necessarily true. Probability theory can be used to estimate the likelihood of something that is definite but unknown.

In terms of the "probability that the universe if finite", then I would say (as a frequentist) that may not be a valid use of probability theory. This is where Bayesian probability theory is able to conjure a probability from circumstances where a frequentist would not! See, for example:

https://ocw.mit.edu/courses/mathema...pring-2014/readings/MIT18_05S14_Reading20.pdf

The real issue is (for the frequentist) is that you must assume a distribution. Then it's a case of applying the standard theory of conditional probabilities.

Let's look at the example of a coin that comes up heads nine times out of ten.

We can test the hypothesis that it is a fair coin. And, we can test any other hypotheses. But, it's difficult to give a probability that it is a fair coin. The reasoning is as follows:

1) If we know that all coins are fair, then it must be a fair coin.

2) If we know (or assume) that coins come in three equally-likely flavours: fair coins; coins that are 90% biased towards heads; and, coins that are 90% biased towards tails - then we can crank out the probability that this is a fair coin. (Assuming the original choice of coin was equally likely to be anyone of these three.)

3) If we know or assume that biased coins are rare, then we can calculate a different number. And, the number depends on how rare we think they are.

What we can't do, therefore, is give a number to the probability that it is a fair coin without assuming the distribution of fair coins against biased coins.

Bayesian methods can, however, go further than this. Personally, however, I'm not convinced that the numbers that come out of such calculations are meaningful.

 

[Ibix]

This is not necessarily true. Probability theory can be used to estimate the likelihood of something that is definite but unknown.

Yes - that's why I gave the example of a hidden coin toss. Perhaps I should have said that probability is for when several outcomes are possible based on the information you have at the time.

 

[PeroK]

Yes - that's why I gave the example of a hidden coin toss. Perhaps I should have said that probability is for when several outcomes are possible based on the information you have at the time.

I think a probability of $0$ or $1$ is just as valid as a probability of $0.5$!

 

[Buzz Bloom]

Because the relevant calculation is not the calculation of a probability distribution for but a calculation of a probability distribution for models.

Hi @PeterDonis:

I understand that the calculation of a Friedmann model has as a result values for five variables: H₀ and the four Ωs. Also, for each variable, a +/- error range is given together with a value for the degree of confidence. The confidence can be defined in terms of standard deviations. I may have misunderstood what I read in the Planck 2018 paper, or confused what I remember, but I think the confidence values were defined as one standard deviation. It may be that the H₀ variable was not calculated specifically for the Planck paper, but was taken form another source. The issue we are disagreeing about is whether (1) each of the five variables have values for mean and standard deviation, or (2) the numbers presented in the form x +/- y have a different interpretation. I think we agree that the values depend on data used specifically for this paper, and that it not intended to be an all time final result. What I cannot grasp is what specifically the values could mean as a "(2) different interpretation".

Perhaps a simpler topic would help me to understand the meaning. The calculated value for H₀ has been (and continues to be) calculated based on a database collection of pair of values: z and distance (z values corrected if necessary for our own velocity with respect to CMB or other standard). The calculation gives a value for H₀ and a +/- error value (or two values, one for each of + and -). Let us assume for the purpose of this discussion that it is a single +/- error value. The H₀ result seems to me to be a mean of a probability distribution, and the +/- error is a standard deviation, or a specified multiple of a standard deviation. With these values one could calculate the probability that H₀ is within a specified range of values, with the understanding that such a calculation value is likely to change when new data produces a different mean value and standard deviation. We possibly also agree that a change implies that the values are not a reliable mean and standard deviation, but we might disagree whether they are useful for this purpose.

Regards,
Buzz

 

[Buzz Bloom]

Fundamentally, the point is that "the probability that the universe is closed" isn't a complete concept. The universe is what it is, it can't be anything else, so probability theory doesn't apply.

Hi @ibis:

I do not understand the logic in the above quote. It is certainly unknown, and possibly unknowable whether or not the universe is hyperbolic, flat, or hyper-spherical. However, even if it is what it is, and we do not know what it is, we do have a clue. The clue is what I interpret to be the mean and standard deviation values of Ω_k. The clue is incomplete in that "flat" implies a specific value (zero), and the probability distribution clue cannot deal with that, or at least I am unable to understand how it might.

However, there is a related question that does have a calculable answer:

"Is the universe finite or infinite?"​

I just cannot understand why something that is what is denies the possibility that among the possible things it might be, each possibility has a probability. If you agree with this, how can it be that "probability theory doesn't apply"?

Example: I have a standard shuffled deck of 52 cards. You pick a card without looking at it. It is what it is. There is a certain probability (25%) that the suit of this card is Spades. There is a certain probability (7.692...%) that the denomination of this card is a Jack. There is a certain probability (1.923...%) that the card is the Jack of Spades. Why do these probabilities "not apply"?

Regards,
Buzz

 

[PeroK]

Example: I have a standard shuffled deck of 52 cards. You pick a card without looking at it. It is what it is. There is a certain probability (25%) that the suit of this card is Spades. There is a certain probability (7.692...%) that the denomination of this card is a Jack. There is a certain probability (1.923...%) that the card is the Jack of Spades. Why do these probabilities "not apply"?

Regards,
Buzz

What if you have a pack of cards and you don't know how many suits there are or how many cards there are in each suit? You just get to look at a few cards.

Of course, you can come up with a number, such as 25% or 1%, but unless you have sufficient knowledge of the data you are dealing with such numbers may be meaningless.

 

[Buzz Bloom]

What if you have a pack of cards and you don't know how many suits there are or how many cards there are in each suit? You just get to look at a few cards.

Of course, you can come up with a number, such as 25% or 1%, but unless you have sufficient knowledge of the data you are dealing with such numbers may be meaningless.

Hi @PeroK:

Are you implying that the value of Ω_k and its +/- standard deviation value are meaningless numbers? If so, what makes them meaningless? Also are the correspond numbers for H₀, Ω_r, Ω_m, and Ω_Λ also meaningless?

Regards,
Buzz

 

[PeroK]

Hi @PeroK:

Are you implying that the value of Ω_k and its +/- standard deviation value are meaningless numbers? If so, what makes them meaningless? Also are the correspond numbers for H₀, Ω_r, Ω_m, and Ω_Λ also meaningless?

Regards,
Buzz

A measured or estimated energy density is not meaningless. But, say, a statement that says that there is a 30% probability that the universe is finite may be a meaningless statement.

In general, probabilities only make sense given some specific assumption(s).

 

[PeterDonis]

the calculation of a Friedmann model

Is not the relevant calculation, because what you mean by "a Friedmann model" here is one particular Friedmann model with particular unknown values for all the parameters, and you're trying to estimate what the parameters are from the data. In other words, you're not comparing models, you're fixing the model and estimating its parameters, and you're assuming that all of the parameters you are estimating are free, to be estimated from the data.

But if I propose a different model, which is still described by a Friedmann equation but in which there is a constraint that forces one particular parameter, $\Omega_k$, to have the exact value zero always and everywhere, then that changes how you estimate the other parameters. And the calculations you are doing do not help you at all in comparing the likelihood of those two different models, the one I propose and the one you refer to in what I quoted from you above. If all you have is the parameter estimation data, the best you can do is say that neither model is ruled out by the data (since $\Omega_k = 0$ is within the error bars of our current measurements). You can't give any relative likelihood.

 

[PeterDonis]

However, there is a related question that does have a calculable answer:

"Is the universe finite or infinite?"​

I just cannot understand why something that is what is denies the possibility that among the possible things it might be, each possibility has a probability.

The question is how you estimate that probability. The point is that you can't estimate it just from the $\Omega_k$ calculation.

I have a standard shuffled deck of 52 cards.

Yes, which means you have fixed the model, and that model already has known probabilities for all possible cards.

But the position we are in in trying to answer the title question of this post is not like that. We do not know what the correct model for the universe is. We can pick a model that looks like it might be right, and estimate parameters from that, but that does not answer the title question of this thread, because it does not answer the question of how likely that particular model is to be right among all the possible models that are consistent with the data.

It's as if, as @PeroK said, there are many different possible decks of cards and we don't know which one the card whose properties we are trying to calculate the probability of was drawn from. Calculating the probability of drawing a Spade from a standard deck does not help answer that question, because it doesn't tell you the likelihood that the deck is a standard deck to begin with.

 

[PeterDonis]

Are you implying that the value of Ω_k and its +/- standard deviation value are meaningless numbers? If so, what makes them meaningless? Also are the correspond numbers for H₀, Ω_r, Ω_m, and Ω_Λ also meaningless?

Of course they're not meaningless; they are estimates of those parameters given a particular assumption about the underlying model. But that in itself, as has been pointed out, is not enough to answer the title question of this thread.

 

[Buzz Bloom]

In general, probabilities only make sense given some specific assumption(s).

Hi @PeroK:

I would veru much appreciate your listing a few examples of probabilities making sense based on given some specific assumptions.

Regards,
Buzz

 

[PeroK]

Hi @PeroK:

I would veru much appreciate your listing a few examples of probabilities making sense based on given some specific assumptions.

Regards,
Buzz

If we stick to cosmology:

1) The probability that Betegeuse will go supernova in the next 100 years, as observed from Earth.

2) The probability that Dark Matter particles will be discovered in the next 20 years.

3) The probability that extraterrestrial life is discovered in the next 50 years.

4) The probability that universe is proved to be finite (or that a finite model is adopted) in the next 500 years.

Notice that if we remove the timescale, it makes things a lot less clear. If you bet that extraterrestrial life will never be discovered, then you can only ever lose that bet and never successfully win.

Likewise, if I bet that the universe is infinite, it's a bet that I (probably!) can't ever win. What would constitute proof of an infinite universe? I could only ever lose that bet.