Cover's paradox: Is 29 the larger of my two secret numbers?
45
closes Sep 30
40%
chance

I have picked out two secret numbers, one of which is larger than the other. I then randomly and uniformly selected one of these numbers using the random.choice() function in Python. This means that there is no bias in me trying to choose the "harder" number for this market, it was all done randomly. The number that was picked was 29. Your goal is guess whether 29 is the larger number or the smaller one, without any further details on what the other number cold be.

It seems like you have a 50/50 chance of getting it right, but Cover's paradox gives a strategy that has a >50% chance of successfully guessing whether the number you're shown is larger or not:

At close, I will reveal the other secret number and resolve based on whether 29 was larger or not. I will not bet on this market, since I already know the answer.

The following Sha256 hash encodes text that reveals what the other secret number is, so that you can verify it after close (I'll reveal what the input to the hash function was):

0c1d75ced9facbe2a1d2d05697dd12cb6cd5532e217c8d16594f69f251e17247

Get Ṁ500 play money

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There is no paradox, this is just regular Bayesian prediction

6 replies

@Loop "Paradox" is a subjective term. In this case, it's a veridical paradox, so if you already found the result intuitive, there's no paradox for you. But for many people, the result runs contrary to their intuitions, so it feels paradoxical to them.

Loopbought Ṁ5 of NO

@JosephNoonan Every definition can be called 'subjective', but we have to communicate.

I think the only way to find this unintuitive is to misread the question or not think it through, but then maybe I'm overestimating people?

@Loop I didn't mean that the definition of "paradox" was subjective. I meant that what's counterintuitive to one person isn't to another. You are right that for someone well-versed in Bayesian reasoning, it's not unintuitive that you can guess with >50% probability, although I still think it's a surprising result that there's a strategy that works no matter what the secret numbers are (from Bayesian reasoning alone, I would expect that you could guess with >50% accuracy only by assuming a distribution of the secret numbers as a prior, calculating whether the first number revealed is larger than the median of that distribution, and then hoping your prior was calibrated well enough).

Looppredicts NO

>I didn't mean that the definition of "paradox" was subjective. I meant that what's counterintuitive to one person isn't to another.

(nitpick: it doesn't always give you >50%, with some secret number distributions it will be exactly 50%) Edit: this is wrong

(nitpick: it doesn't always give you >50%, with some secret number distributions it will be exactly 50%)

Not if you choose a random number based on a distribution with support over the entire real line. No matter what distribution is used to pick the two secret numbers, there will always be a nonzero probability that the random number you choose is in between them (with the exception of a distribution that guarantees that the two secret numbers will be equal, which isn't allowed because the paradox states that they are distinct). And because it is randomly and uniformly chosen which of the two secret numbers will be shown to you, there is always an exactly 50% chance that you will guess right, given that the random number you choose isn't between them. So overall, the probability of guessing right is 50%*(1-P(between)) + P(between) = 50%*(1+P(between)), which is always strictly greater than 50%.

Looppredicts NO

@JosephNoonan ah shit I had not watch the whole video and didn't consider stochastic strategies

~

Here is a weirder paradox: there are 2 slips of paper A and B; each one has a natural number written on it. You have to guess which one is bigger. You're allowed to look at one of the papers. You read the number on paper A. No matter what the number is, there are an infinite number of numbers greater than it and only finite smaller than it. So you'd think that B is bigger. You should switch, right? But it doesn't even depend on what the number is, so you might as well have made the switch before looking at A. It's a symmetrical situation, but somehow it seems you should pick B!

jcbpredicts YES

@ItsMe the problem is that "there are an infinite number of numbers greater than [A] and only finite smaller than it" does not imply anything about the likelihood of B being bigger. If you knew that the two numbers were uniformly selected from the naturals, that would follow, but there is no uniform distribution over the naturals!

jcbpredicts YES

I found the discussion on https://johncarlosbaez.wordpress.com/2015/07/20/the-game-of-googol/ illuminating, especially this comment from Greg Egan:

You can sum up the entire strategy as: Ensure that, the greater the number you see, the greater your chance of sticking with it. It’s very easy to see why that should work, and not much harder to see how to make it happen.

(His implementation is slightly different from the one in the video, but I think equivalent.)

(I do find it hard to see how this works because I suck at thinking about the reals; like, you can't actually select a real uniformly at random on (0,1)--or even a rational! but ignoring such impracticalities as we like to do in math, it makes sense.)

@jcb Yeah, in real life practice, you just have to use a computer to generate the number, which technically is just a uniform distribution on {n/2^52} for n from 0 to 2^52-1, or something along those lines. So in real life, you can end up with an exactly 50% probability if the person choosing the numbers decides to make them so close together that the number your computer chooses is guaranteed to not be in between them. Maybe you could use some quantum process to choose a number from a truly continuous distribution instead of just an approximation of one.

If the process with which the two numbers are created is truly unknown, then the probability to guess a number that is between them is 0%, because any finite interval is infinitely smaller than the number line. So that gives 50% probability of winning in purely mathematical game.

Here is how >50% achieved. We KNOW that people are likely to create number closer to zero and closer to normal distribution and that is the additional info. The statement in the video that "we have Absolutely no information" is hilariously incorrect. Instead of working with infinite line (where truly random generation is undefined) we work with some finite interval. Humans are more likely to pick 3.7873 than 8466363677585857636526647485. So if we make an assumption that opponent's number are limited to -1M<x<1M, the the probability to hit between the numbers is non-zero. Knowing the person better we could narrow it down more and get better chance to hit between numbers.

Also it is important to pick our number before we see one of the opponents numbers. If you do it later, then you do not make your independent interval assumption, and if you do it later then you are stuck back to 50/50.

So, there is no paradox. It is just unaccounted knowledge. You have 50% in mathematical game but >50% in psychological game (where you account for the knowledge).

21 replies

Another bit of knowledge: people are likely to pick from the same class (of course that breaks forever as soon as they ever see this or a similar statement, but works with kids). If one has 3 digits, the other might also have 3 digits. That limits to intervals like 10-99, and you just answer the question whether the one named number is in the lower part of the interval or in the higher.

Fionpredicts NO

@KongoLandwalker thanks, I appreciate your explanation of the "paradox". It's pleasing that it has such a simple resolution!

One complication with playing the psychological game is that, while humans are indeed more likely to pick simple numbers, this doesn't necessarily mean small. If you ask people "think of a number, any number" I think more of them will say "one billion" than "3.7873"

I suspect Joseph will have come up with some interesting way of "picking out" the two numbers. But I don't have any good guesses for what that way might be.

@KongoLandwalker This is incorrect. If you pick out a random number from a normal distribution, the probability of the number coming from any given interval is positive. So there is always a nonzero probability of picking your random number between the two secret numbers. That probability is larger if the numbers are close to zero, but it is always nonzero, no matter what they are.

The actual resolution to the paradox is that it's veridical. It simply is the case that there's an algorithm that always has P(correct guess | secret numbers are x and y) is greater than .5 for all x and y.

@JosephNoonan you did not read carefully. I said that the First person is likely to use normal distribution and Then the probability is >50%.

The fact that distribution is normal (any other specific) is already the bit of knowledge which diverts us from 0.5.

Do you agree that 0.(9) is equal to 1? Do you agree that 0.5 + 0.(0) is equal to 0.5? If there was uniform distribution possible, then we would operate with 0.(0) for any given finite interval, do you agree?

@KongoLandwalker No, it doesn't matter what distribution the first person uses. They could use a funky distribution with a mean of googleplex (or better yet, no mean at all), and the second person would still have a >50% chance of guessing right. It works for absolutely any method used to choose the two secret numbers. Also, the second person doesn't have to use a normal distribution either, they can use absolutely any distribution, as long as its pdf is supported everywhere.

The reasoning for this has nothing to do with what distribution the secret numbers are chosen from. As I mentioned before, regardless of what x and y are, the probability of a random variable Z (distributed according to any nonvanishing pdf on the reals) being between x and y is always positive, unless x=y (and for the paradox, we're not allowed to have x=y). So if the first person already has two numbers in mind, then the second person has a guaranteed >0% chance of choosing a number between them, and if they're going to choose the number according to some distribution, the second person still has a >0% chance of guessing between them (just average the probability of choosing a number between x and y over all possible combinations according to the first person's distribution).

If there was uniform distribution possible, then we would operate with 0.(0) for any given finite interval, do you agree?

A uniform distribution over the reals isn't possible. And even if it was, that still wouldn't really matter. No matter what two numbers were picked from that distribution, the second person would still have a >50% chance of getting it right. I suppose you could argue that, even though the probabilities conditional on given values of x and y are >50%, they would average to 50%, but that would require taking the average of a function over a nonexistent distribution, which is impossible.

Fionpredicts NO

@JosephNoonan great, now I'm back to being confused! :D

So regardless of how the two numbers were generated, there are some choices the guesser can make (such as a normal distribution) such that the chance of being between the two numbers is >0%? I guess this is because the normal distribution falls off rapidly enough that no matter which bit of it the two numbers fall in, there is an infinite stretch of number line that is contributing a finite amount to the probability density of the distribution. So the bit between the two numbers would be an infinitesimal part of some uniform-infinite distribution (if such a thing is definable), but a finite amount of a normal distribution.

Is that right?

@Fion The reason is that there are distributions (the normal distribution being the most obvious one) that are supported on the entire real line. Meaning for any two distinct numbers, the probability of being between them is always >0. The exact probability depends on the distribution and the numbers, but it's always >0.

Also, an important part of the paradox is that I have to randomly choose which of the two secret numbers to show you. If I got to decide which one to show, then I could easily subvert the strategy by choosing the number that you're more likely to be on the wrong side of. For example, if I expect you to choose your random number from the standard normal distribution, I could choose my numbers to be googleplex and googleplex+1, and then show you googleplex, and you would have basically a 0% chance of getting it right. However, since I have to randomly pick which number to show you, I'm just as likely to show you googleplex+1, in which case you have a nearly 100% chance of getting it right. But the difference from 0% in the first case is slightly larger than the difference from 100% in the latter case, making your chance of guessing right overall slightly greater than 50% (though the difference in this case would be astronomically small).

@Fion Basically, it goes like this:

Let X be your randomly chosen number, and A and B be the secret numbers with A < B. No matter what A and B are, there is some nonzero probability that A < X < B, since you have chosen X from a distribution with support everywhere. If this happens, then you have a 100% change of getting it right. On the other hand, if X ∉ (A,B), then there is a 50% change of getting it right: If X < A, you will guess right iff B is the chosen number, and if X > B, you will guess right iff A is. So the probability of getting it right is P(A<X<B) + 50%*[1-P(A<X<B)], or 50%*[1+P(A<X<B)], which is of course greater than 50%. (I left out the case where X = A or X = B, but this has a 0% chance of happening for most distributions, so it won't affect the probability unless you choose a distribution with nonzero mass at A or B. If it does happen, you can just choose randomly whenever X is equal to the revealed number).

Of course, this is the probability of getting it right, given particular values of A and B. But since it is greater than 50% no matter what A and B are, there is no way for A and B to be chosen that doesn't give you a slight edge. If there is some distribution of possible values for A and B, then your probability of getting it right is just the average of 50%*[1+P(A<X<B)] over that distribution, which will still be >50%.

@Fion the person straight up ignores the existance of "infinitely small values". Multiplication of finite (100% for example) and infsmall is infsmall. 50%+100%*0.(0)=50%.

But he says it will be ...=x>50% in the equation above.

Funny, but that is the same mistake as in Snake eyes with "case of infinite dice throws with no deaths". People give non-zero weight to infinitely small cases. (But in snake eyes people tried to give inf big weight to the case).

According to the video, you don't even have to generate any number. Just always compare to minus five. That is because the boundaries of generator are arbitrary, why not set them close to each other to skip the step of generating. And -5 is not any worse than any other number.

Then let another guy to play guesser simultanously with you. But he with the same level of "arbitrary" picks a number of google^google and compares to it. If there was any paradox existing, then you both would have >50% chance.

If you see a paradox, it is more likely somebody made an unnoticable for others wrong assumption.

Method of proving by contradiction would not exist if unsolvable paradoxes existed.

the person straight up ignores the existance of "infinitely small values".

"Infinitely small values" don't exist. Not in standard probability theory, anyway.

50%+100%*0.(0)=50%.

But he says it will be ...=x>50% in the equation above.

Nothing in the equation above says that 50% + 100% * 0.(0) will be >50%. P(A<X<B) is always greater than 0, that's the whole point.

According to the video, you don't even have to generate any number. Just always compare to minus five.

No, this doesn't work. The point of choosing a random number is so that, no matter what the secret numbers are, or what distribution they are selected from, you will always have a >50% chance of guessing right. For -5, you will have a 100% chance of guessing right if -5 is between the two numbers, but only a 50% chance otherwise, so you're not guaranteed to have a >50% chance.

That is because the boundaries of generator are arbitrary, why not set them close to each other to skip the step of generating.

What boundaries are you talking about? For the strategy to work, you have to pick a random number from an unbounded distribution.

Then let another guy to play guesser simultanously with you. But he with the same level of "arbitrary" picks a number of google^google and compares to it. If there was any paradox existing, then you both would have >50% chance.

If both guessers actually used the right strategy, then yes, you would both have a probability >50% of guessing right. If you used a normal distribution centered at -5, and he used one centered at googol^googol (you have to use a distribution for the strategy to work), then you would both have a >50% chance of guessing it right. What is supposed to be the problem with this?

Method of proving by contradiction would not exist if unsolvable paradoxes existed.

This isn't an unsolvable paradox, it's a veridical one. It is a formal mathematical theorem that a strategy exists to guess whether the revealed number is the larger one that will be correct with probability strictly greater than 1/2. The solution to the paradox is that the intuition telling you that this shouldn't be possible is simply wrong.

@KongoLandwalker To clarify the point about just choosing -5, if you assume that the secret numbers are picked out by some process that gives a nonzero chance of -5 being between them, then yes, just picking -5 works as a strategy. But Cover's result is stronger than that - it shows that you can use the same strategy, and it will work for any distribution of values for the two secret numbers, including ones where there is a 0% chance of -5 being in between them. This will even work if the person choosing the numbers knows the strategy you are using and is adversarial. The best they can do is pick two numbers that make your probability of being right extremely close to 50% (like googol and googol+1), but they can never make it exactly 50%.

@Fion uniform infinite random generátor is not invented, but we can talk about its properties if it was real. For any number X we would never bé able to tell, whether the generated Z will be bigger or smaller. 0.5 that z>x and 0.5 that z<x for any x from real numbers.

Let's take x= finite A. We know, that z<A has 50%. Let's take any arbitrary finite B > A. Than z>B has 50%. We already reached 1, but have the interval between A and B. Then the interval has 0% of recieving the number from the generator. Because of the fact that A and B are finite and arbitrary, the probability of magical uniform value to fall into any finite interval is 0%. Interval for a line is like dot for a segment.

Like in darts, where any point has "higher than zero probability to be hit" because the integral has to be 1, but each point has 0 probability to be hit. Infinitely small values exist in probability theory.

@JosephNoonan in you model

1. you say you have >50% to guess

2. But you base your method on seeing one of the numbers. Before seeing the number your chance is 50%. (Imagine one card of the two is chosen but not revealed, and you try to predict whether it is the smaller one).

3. That means that seeing the number gives you some information to update tour probability. But the number does not give you information about the second number. Operation of comparison needs info about both, but you still have none about the second number. (In a mathematical game. In psychological, of course, you get a lot).

In my model 0.(0) = 0 and the probability is always 0.5. I did not recieve new info about the numbers relation, so I have no right to update Bayesian probability after seeing just a card.

uniform infinite random generátor is not invented, but we can talk about its properties if it was real.

It's not just that one hasn't been invented. It's logically impossible to have a uniform distribution over the reals. And the strategy is to choose the random value X according to a non-uniform distribution, so what is the point of talking about this nonexistent distribution? I guess you're trying to say that A and B are taken from this non-distribution, but there are two problems with that:

1. You are trying to say that Cover's result is invalid because it wouldn't be true if the secret numbers were taken from a distribution that doesn't exist. So what if it wouldn't be true in an impossible situation?

2. Even if we imagine this nonexistent distribution and try to reason about it, we would say something along the lines of "The probability of being in any finite interval is 0, and of being in a half-infinite interval (i.e., one that is bounded from only one side) is 1/2". Basically, we would remove the criterion of countable additivity for probability and replace it with finite additivity and countable subadditivity, while keeping the intuitive idea that the "distribution" should have an equal chance of picking a number greater than n or less than n for any given n. You should agree with all this because that is what you already claimed would be true of a uniform distribution over the reals. If we use this pseudo-distribution, then we find that, for any given X chosen from a normal distribution (or any other probability distribution), there is a 25% chance that A and B are both greater than X, a 25% chance that they are both less than X, and a 50% chance that one is less than X and the other greater than X. In the first two cases, there is a 50% chance of getting it right, depending on which one is chosen and revealed to the player. In the final case, there is a 100% chance of getting it right. So even if the secret numbers are somehow chosen not from an actual probability distribution, but from this pseudo-distribution, there is a 75% chance of getting it right if you use the strategy presented. That is actually way better than the probability usually is in real-life scenarios, and it comes from the fact that, in this pseudo-distribution, P(A<X<B) is quite large (50%).

Like in darts, where any point has "higher than zero probability to be hit" because the integral has to be 1, but each point has 0 probability to be hit. Infinitely small values exist in probability theory.

Assuming that space is continuous, no point has a higher than 0 probability to be hit in darts. They all have a probability of exactly 0 of being hit. You are confusing probability density with probability mass.

@JosephNoonan in you model

1. you say you have >50% to guess

Yes, except that there's no "model" here. As I have repeatedly said, this is true no matter how A and B are chosen.

1. But you base your method on seeing one of the numbers. Before seeing the number your chance is 50%. (Imagine one card of the two is chosen but not revealed, and you try to predict whether it is the smaller one).

The probability of guessing right is >50%, but yes, you have to see the number in order to make that guess. If you didn't get to see the number, the probability would be exactly 50%.

1. That means that seeing the number gives you some information to update tour probability. But the number does not give you information about the second number. Operation of comparison needs info about both, but you still have none about the second number. (In a mathematical game. In psychological, of course, you get a lot).

It doesn't need to give you any information about the second number. You aren't just trying to guess some property of the second number, you're trying to guess about a relation between the first and second numbers. Seeing just one of the numbers gives you some information about that.

You say that, and then don't give any contradiction. Unless you meant that point 3 above was a contradiction, but the problem there is that point three is just plain wrong. Your assumption that you need to get information about the second number in order to have information about the comparison is simply incorrect.

In my model 0.(0) = 0

This is completely irrelevant. We both agree that 0.0 repeating = 0, and standard probability theory, the "model" I am using, is based on this fact. You are the one who is trying to build some non-standard theory that violates this (e.g., pretending that there is a uniform distribution over the reals that has probability density 0 everywhere, and yet the total probability is still one - this is kind of like saying that 0.(0)=1)

probability density 0 everywhere, and yet the total probability is still one - this is kind of like saying that 0.(0)=1)

No, it is like saying that the integral from -inf to inf of 0.(0) can be 1. 1) Do you agree that in the pseudodistribution, that i described, every point has 0.(0) probability density? I used the distribution because it is the closest to represent absolute no information about opponent's distribution function.

It can be described as the limit of the process of moving bounds of uniform distribution to infinity. The integral at each moment of the process is 1, but pdf(x) goes to 0.(0) for any x.

2) There is nothing wrong with the word "model". So, in your model applied to my pseudodistribution, does the order in which {x, A, B} are generated, matter? Will you recieve the same 75% in each ordering?

(If we stop discussing the pseudodistribution, then we fall into the case of "known to have not infinitely small pdf" distributions, and as i said, that is the knowledge that increases the chance to guess. But I think there are infinitely more functions with infinitely small or 0 pdf's around human-comprehensible interval for which we can generate x. And there is limited (but still not countable set of distributions whete pdf in the interval is not zero, thus >50% winrate). So if we truly do not have any info, then those with 0.(0) are more likely, because their set is order of infinity bigger.).

No, it is like saying that the integral from -inf to inf of 0.(0) can be 1.

The integral from -∞ to ∞ of 0 is 0. So saying that the integral from -∞ to ∞ of 0.(0) is not 0 means that you are saying 0.(0)≠0. The specific reason I said that claiming that the integral is one is like claiming that 0.(0)=1 is because, by definition, 0.(0) is the sum of countably many zeros, while the integral from -∞ to ∞ of 0 can be split up into a sum of countably many zeros.

1) Do you agree that in the pseudodistribution, that i described, every point has 0.(0) probability density?

Every point in this pseudo-distribution has probability density 0 because there is no difference between 0 and 0.(0).

I used the distribution because it is the closest to represent absolute no information about opponent's distribution function.

It would be better to just use a Gaussian with a very high mean for this. At least then you'd have a real probability distribution.

It can be described as the limit of the process of moving bounds of uniform distribution to infinity.

The problem is that the limit of this process isn't a distribution.

2) There is nothing wrong with the word "model". So, in your model applied to my pseudodistribution, does the order in which {x, A, B} are generated, matter?

The only thing that matters is that x is chosen independently of the numbers revealed to you.

If we stop discussing the pseudodistribution, then we fall into the case of "known to have not infinitely small pdf" distributions

By which you mean actual distributions. There is no such thing as an infinitely small pdf. The pdf is either zero, or positive and non-infinitesimal.

But I think there are infinitely more functions with infinitely small or 0 pdf's around human-comprehensible interval for which we can generate x.

There are exactly zero functions with an infinitely small pdf anywhere. As for 0 pdf, it makes no sense to me to use a function with a pdf of zero at any location for where A and B could be chosen. But even that wouldn't matter. The strategy for guessing with a >50% win rate still works even if A and B are guaranteed to have magnitudes larger than Rayo's number. It works for any real distribution on A and B.

So if we truly do not have any info, then those with 0.(0) are more likely, because their set is order of infinity bigger.

I don't know what you mean by this. You seem to still be treating 0.(0) as distinct from 0 when they are the exact same number. "Order of infinity bigger" isn't even a meaningful description. Are you claiming that one set has a larger cardinality than another, or is larger by some other measure? If so, what sets and what measure? The set of distributions that have probability density zero everywhere is the empty set, which is smaller than any other set by pretty much any measure you can come up with.

@JosephNoonan I used "0.(0)" as a non-mathematical somewhat lexical substitution for "infinitely small value" (because in attempt of writing it down one would recieve unlimited zeros, but would not be allowed to stop). If that was not clear contextually, then i don't understand why you followed along with the notation without asking. I am sorry to confuse you.

The problem is that the limit of this process isn't a distribution.

Does Dirac delta function exist? Can it be distribution for some number-generator? Is it's integral equal to 1? What is its pdf in point 1?

What i suggested is the process, which is just the reverse of Ddf. You would recieve the same entity by smearing both uniform and normal distributions.

"Order of infinity bigger" isn't even a meaningful description.

How many points in a [0,1] segment? Infinitely many. How many points in [(0,0); (1,1)] square compared to the previous answer?

I see that you do not have experience with abstract thinking about the concepts I have used (infinitely small values in probability, limit of process, infinite sets comparison), so I will leave this dialog. Have a nice day.

(You don't have to answer any of questions here, but people usually like to have "the last word", so I will leave the opportunity up to you)

I used "0.(0)" as a non-mathematical somewhat lexical substitution for "infinitely small value"

You were referring to 0.0 repeating, right? That's not an infinitely small values, that's just zero. I went along because initially you seemed to agree that 0.(0)=0, but now you're changing your tune.

Does Dirac delta function exist? Can it be distribution for some number-generator? Is it's integral equal to 1?

Of course, the Dirac delta function is a well-defined probability distribution. There's a perfectly precise and formal mathematical theory that describes exactly what it is and how to integrate it (which of course gives 1). And that same mathematical theory is the one that says there's no uniform distribution over the reals.

What is its pdf in point 1?

The Dirac delta function doesn't have a pdf. Despite its name, it's not actually a function. If we were talking about real-valued functions, rather than probability distributions, and someone tried to bring up the Dirac delta function as a counterexample to some statement about such functions,it would be a complete non-sequitur. Kind of like how bringing up a nonexistent distribution is a non-sequitur.

What i suggested is the process, which is just the reverse of Ddf. You would recieve the same entity by smearing both uniform and normal distributions.

You suggested defining a uniform distribution over the reals as a limit of a sequence of distributions that doesn't have a limit. The difference between this and the Dirac delta function is that the Dirac delta function is the limit of a sequence of distributions that actually do have a limit. You can't just take an arbitrary sequence and expect it to have a limit. I think what you are confused is the difference between a distribution and a function. A distribution is not the same thing as its pdf, and the limit of distributions is not the same thing as the limit of the pdfs. The pdfs of uniform distributions don't have a limit as you shrink the range - they converge to 0 pointwise, except at x=0, where they blow up to infinity. But the distributions themselves do have a limit, the Dirac delta. Meanwhile, the pdfs of uniform distributions have a limit as the range increases (the zero function), but the the distributions themselves do not.

How many points in a [0,1] segment?

2^ℵ₀

How many points in [(0,0); (1,1)] square compared to the previous answer?

Also 2^ℵ₀. It's actually a well-known theorem that the square does not have more points than the line, they have the exact same number. And no, the square isn't a larger infinity - they're both the exact same infinity. Now of course, the square is larger in some ways. It has an area of 1, while the line has an area of 0. But that has nothing to do with the number of points. That's why I asked what you meant by "order of infinity bigger". If you said, "the square is an order of infinity bigger than the line", that doesn't tell me if you're claiming that the square has a larger number of points than the line (which is false), or has a larger area than the line (which is true), or something else, because there is no mathematical concept called "order of infinity bigger". You can't make up terms and then expect me to know what you mean. But I think the reason you did this is because you don't even know what you mean. You have an intuition that one thing is infinitely bigger than another but have no idea how to formalize this intuition, so you threw out a hopelessly vague phrase. If I'm wrong about this, then please clarify by giving an explanation of you meant using accepted mathematical terminology.

I see that you do not have experience with abstract thinking about the concepts I have used (infinitely small values in probability, limit of process, infinite sets comparison), so I will leave this dialog.

If you want to leave the dialog, that is fine, but this is a cowardly way to do it. You're just going to throw out a baseless insult and then try to end the conversation on that note so you don't have to defend it? I think the previous comments speak for themselves that I have a lot more experience than you with abstract reasoning about limits and infinite sets. As for infinitely small values in probability, I have enough experience to know that those aren't actually a thing in standard probability theory, and you don't. And I can guarantee based on your comments that you're not using a formal non-standard probability theory that allows for infinitesimals - you're just running on intuitions that you don't know enough about the actual math to be able to formalize, and you get incorrect conclusions as a result.

dionisosbought Ṁ10 of NO

@KongoLandwalker You don’t need infinitely small value, the probability could just be 0.

The probability of choosing any give number randomly on R is just 0 (if the distribution is a homogeneous distribution).

dionisospredicts NO

Now, we should take into account it is impossible, there is only a countable number of possible programs.
And it is impossible to have an homogeneous distribution on an infinite countable set.
(and if the set is finite, then you get a probability greater than 0)

dionisospredicts NO

deleted

The probability of choosing any give number randomly on R is just 0 (if the distribution is a homogeneous distribution).

The probability of choosing any given number on R is 0 using any continuous distribution. I think what you meant to say is that the probability of picking a number from any finite interval would be zero, but this violates countable additivity, and it would produce contradictory results.

For example, imagine there was a uniform distribution on R, where any finite interval was given a probability of 0, any half-infinite interval (e.g., (0,∞), (-∞,5]) had a probability of 1/2, and of course, the entire real line has a probability of 1. Let's choose two numbers from this distribution, A and B. What's the probability P(A>B|A>0)? We can calculate this in two ways. On the one hand, B has a 50% chance of being positive, and 50% of being negative. So P(A>B|A>0) = 50%*[P(A>B|A>0,B>0) + P(A>B|A>0,B<0)]. That latter probability is clearly 1, since A must be greater than B if A>0 and B<0, so we find that P(A>B|A>0) = 50%*[P(A>B|A>0,B>0)+1]. But P(A>B|A>0,B>0) has to be 50%, since the chance that A=B is 0%, and they're both selected independently from identical distributions (in this case, the "uniform distribution on the positive numbers"). So, we find that P(A>B|A>0) = 75%.

But now we calculate the probability in a different way. For any particular value of x, P(A>B|A=x) = 50% - after all, this is one of the stipulations of our "uniform distribution on the reals". In other words, A>B is completely independent of the value of A, and always has a probability of 50%. So, we can conclude that P(A>B|A>0) = 50%.

In order to avoid this contradiction, we would have to ditch a lot more than just countable additivity. We would have to ditch basic methods of probabilistic reasoning. The result would be something so different from actual probability that it hardly makes sense to even call it a pseudo-probability.

If you really want to extend the notion of probability to something that has a uniform distribution over the reals, the best way to do it is probably with infinitesimals, i.e., use a "probability" that doesn't have to be real-valued. This would get you out of the contradictions above, although it would be hard to define a good rule for assigning infinitesimal-based probabilities that doesn't leave a lot of sets with indeterminate probabilities. However, Cover's paradox would still apply to probability defined this way, even if you wanted to use some tricky prior distribution of the secret numbers to try to avoid it.

(Sorry for such a long response, most of this isn't really a direct reply to your comment and more of a "here's why we can't have a uniform distribution over the reals" for anyone still confused about it.)