Quote from: mcdouglasx on Today at 09:04:50 PM

Quote from: kTimesG on Today at 03:35:56 PM

I'm not preaching anything that's not in a 9th grade high school math book where I live.

Do you think this is the best way to win the debate? You're going off on a tangent and I don't think you're the one making shadow-posting (when a user overlays previous chats with long text to hide old chats, usually using alternate accounts).
I'd like you to give real mathematical proofs, like I did, this is a post to share ideas about the puzzle, not to impose our own, just because.

Lol, you are basically asking for a mathematical proof that all possible combinations stand an equal chance. But this is by definition, and everything else is simply a consequence.

Look, I'm not using ChatGPT or AI, because I have a working brain instead, and a finished college. But that should not matter here.

I don't really understand where you are starting to disagree with the below statements. Maybe you can enlighten whoever reads our debate, where exactly the movie starts to break:

1. RIPEMD-160 is a hashing algorithm, and as such, just like you said, implies an uniform distribution. If you don't agree, why not?

2. All hashes are independent of one another, e.g. any single time you do a hash, you have an equal chance of getting any one of the possible 2**160 hashes. Agree? If not, why?

3. For some random set of events (let's say, of length 2**66), this means any one of those 2**66 events can end up with any one of the possible 2**160 hashes. Agree? If not, why?

4. If you agree with 3, then there are 2**160 (possible hashes) raised to the power 2**66 (number of events) possible combinations that exist. Therefore, the random samples set (of length 2**66) is somewhere in these astronomically huge possibilities. Agree? If not, why?

5. If you agree so far, then you accept the fact that all of those 2**66 hashes in your set may all have the same value (because this is a valid combination). Agree? If not, why?

6. In the same way, all of the hashes may be different, or you may have an arbitrary number of repeated values. Agree? If not, why?

7. If you agree so far, then why would you think that your specially chosen set has some whatever hash appearing just once, or twice, or whatever amount of times? OK, you will say that it is less likely for that to happen, because 2**160 is much greater than 2**66, and you would be completely right about that.

In fact, this is exactly what a CDF (cumulative distribution function) and PMF (probability mass function) computes: These functions take three parameters:

- number of trials (e.g. 2**66 events)
- a number for which you'd like to know how many times (or up to how many times) some variable can occur (be a success) after those number of trials were performed in a randomly, uniform, independent manner
- the probabilities of success of the event(s) (in our case, all hashes have the same chance of occurring, every time)

If you notice, none of the above ever implies any kind of events order or events inter-dependency, because of #1 in effect. All that matters is how often you would expect something to occur, or to repeat itself.

Now, CDF and PMF directly take into account ALL POSSIBLE combinations that can occur. Because we have 2**160 raised to 2**66 possibilities (a number that is 160 x 2**66 bits in size), computing the CDF is impractical, if not impossible without a supercomputer).

However, you can play around with much less parameters, and observe the resulted probabilities dependning on what arguments you wish to get a result for.

You can use Python with scipy.stats.binom to play around with smaller values, and test your theories on rolling a dice. Or playing a virtual lottery. If you manage to somehow predict the dice rolls, or predicting the lottery numbers (just because, well, uniform distribution and etc.), you are my hero!

In reality, all you'll get is that independent events running off an uniform distribution simply produce noise, and there is no actual uniformity in the results. You will get different results every time. What I don't understand, is why would anyone think that things change at the slightest, when the size of the problem changes (like, for example, using the first however few first bits of a hash). They do not, and the only real answers to questions are only found if fully determining the values in the set (in english: full range scanning).