Bitcoin puzzle transaction ~32 BTC prize to who solves it

[bibilgin]

bla bla bla...

Okay master, we are going to the psychologist now.
Yes, I am crazy.

I am the one apologizing, right. (Did you really apologize?)
How much did it upset you when the thing you said was impossible came true?
Are you still aware that you are psychologically defeated with questions like vast.ai, $2500, impossible in your head? I wonder if the psychologist will answer this?

Here in front of everyone's eyes,
You made a mistake, You apologized, You said impossible (It turned out to be true.)

Now when you think that I found the 68th wallet, I guess your scenarios are clear.

Vast.ai used it.

Spent $250,000.
He is a very lucky person.
Swiss army knife.
It shows the truth twice a day, every hour.
I don't believe it, there is another power behind it....

I am sure he will say one of these. So you are a defeated person right now, accept this first.

[mcdouglasx]

It is a common characteristic of hashes that different numbers can produce the same initials. For example, as in the case you provided, 1BY8GQbnH7ny3ZbLpmVtBZLZ2Lw7528UfK. This is not surprising.


If you really think it has some  connection with the private key, then can you explain why the Bech32 and SegWit addresses of the same private key don’t have the same initials, while only the legacy address does? Does legacy address  have some magic in it?

 I never got answer to this question @ bibilgin & Mcdouglasx after reading your comments i believe you guys are expert in this so can please explain how come probability come into this if there's no connection in any shape or form?


It's called coincidence and not a probability

Each private key acts as an independent random event, and the probability of finding a specific prefix in an address does not change regardless of previous results. This is what confuses most people since the probability of an individual event (such as finding prefix "X") remains the same on each attempt.

However, when we look for the match of multiple prefixes in a dataset example 1:256, we enter the realm of compound probability.

This is the part most intuitively ignore and therefore fail:

-Probability of finding "a": 1/16

-Probability of finding "ab": 1/256

-Probability of finding two "ab" in 256 attempts: approximately 0.18

As you can see, if you find "ab" in an early shot, the probability of finding another "ab" in the next 256 attempts is very low. This would allow you to skip those subsequent attempts without losing significant precision, since although each attempt is still an independent event, the chances of it being there are very low.

For this reason, in probabilistic searches you can skip unlikely private keys. It's not that the private key has a relationship with the hash of the address, but that each key is essentially an independent shot, and we would omit the less likely shots.

[bibilgin]

I already told him he wasted his time, energy & electricity. I think he finally got a reality check now 😂

I never got answer to this question @ bibilgin & Mcdouglasx after reading your comments i believe you guys are expert in this so can please explain how come probability come into this if there's no connection in any shape or form?


It's called coincidence and not a probability

After writing the first article, I don't understand how you wrote the second article?

First of all, let me say this, KTimesG said there were 2 of us.

But for the last 2-3 days, I have been receiving questions and messages from many people in my mailbox. This proves that there are people who respect or believe in my work.

Now, I have actually proven your question many times before.

1- I have repeatedly stated the range of the 67th wallet = between 6D and 77
2- I gave 1 example of a decimal value difference. =
https://bitcointalk.org/index.php?topic=1306983.msg64873290#msg64873290
3- I proved what KtimesG said was impossible or didn't believe.

4- Let me even prove it. Now, do a range scan. Determine the range yourself.

Send me the wallets that come up in that range. I will tell you where the wallet with the prefix 1MVDYgVaSN is on AVERAGE PROBABILITY. After proving this, I hope you will be convinced.

I will not do more because.

Another topic;

KTimesG - As its name says, it has spent years reading, archiving and gaining knowledge on K, N and G. It has acquired serious information on creating EC, Kangaroo, mixed and BTC wallets.
In fact, the reason for its research was to search for a clue, a vulnerability in the system. It spent years on this subject, but could not find any vulnerability.

My work is on the calculation of the COMPOSITE PROBABILITY of encryption systems. The result tells me in which direction I am really looking for.

About 6 hours ago, I found the 19th 1MVDYgVaSN prefix. Has anyone been able to find 19 1MVDYgVaSN prefixes in this amount of time with the power of my hardware? I also have 1 prefix, 1MVDYgVaSN6. (2 pieces as given by Zahid888.)

[bibilgin]

This is commendable of course. In your opinion, where is 68 in what range? And you can publish several 1MVDYgVaSN.

Unfortunately, my friend, I do not want to share this information.

Or rather, I want to share what I have with people who give me another prefix in return.

I can only give a little information about the 68th wallet range. It is located above C0.

[kTimesG]

This is the part most intuitively ignore and therefore fail:

-Probability of finding "a": 1/16

-Probability of finding "ab": 1/256

-Probability of finding two "ab" in 256 attempts: approximately 0.18

As you can see, if you find "ab" in an early shot, the probability of finding another "ab" in the next 256 attempts is very low. This would allow you to skip those subsequent attempts without losing significant precision, since although each attempt is still an independent event, the chances of it being there are very low.

Here's your fallacy (which goes in repeat mode with you):

You didn't take into account that the probability to not find "ab" at all in 256 attempts is 36%.

You didn't take into account that the probability to find "ab" more than once is not "very low", but at a good 28% (100 - 36 (for 0) - 36 (for 1)). Because you never added the probabilities for it to appear 3 times, 4 times, 5 times, and so on (up to 256 times). You only base some claims on the fact that it's very unlikely to appear "a second time", but it can also appear more than 2 times, not necessarily just one more time.

So while in principle you may be on to something (though nothing more than simply observing the normal behavior of a uniform random variable), your calculations are off quite a bit, to the point that the combined probabilities of failure once you go skipping over and  over again, accumulate, and they accumulate in the fashion that you like really much: compounded.

[bibilgin]

Thank you. You wrote above C0 now everyone is looking above. But still thanks for the answer.

Here is another good information for you, it remains under E2. 68th wallet

[bibilgin]

Here's your fallacy (which goes in repeat mode with you):

You didn't take into account that the probability to not find "ab" at all in 256 attempts is 36%.

You didn't take into account that the probability to find "ab" more than once is not "very low", but at a good 28% (100 - 36 (for 0) - 36 (for 1)). Because you never added the probabilities for it to appear 3 times, 4 times, 5 times, and so on (up to 256 times). You only base some claims on the fact that it's very unlikely to appear "a second time", but it can also appear more than 2 times, not necessarily just one more time.

So while in principle you may be on to something (though nothing more than simply observing the normal behavior of a uniform random variable), your calculations are off quite a bit, to the point that the combined probabilities of failure once you go skipping over and  over again, accumulate, and they accumulate in the fashion that you like really much: compounded.

Now I ask you.

After finding the 1st ab, I calculate to find the 2nd ab.
I calculate to find the 3rd ab based on the 1st and 2nd ab.
After finding the 1st, 2nd and 3rd ab, I find the 4th ab.

Now, are you saying that these are not probabilities but predictions?
Are you saying that probability never changes?

Or are you saying that there is an independent probability but that there will be no conditional probability?

[kTimesG]

Now I ask you.

After finding the 1st ab, I calculate to find the 2nd ab.
I calculate to find the 3rd ab based on the 1st and 2nd ab.
After finding the 1st, 2nd and 3rd ab, I find the 4th ab.

Now, are you saying that these are not probabilities but predictions?
Are you saying that probability never changes?

Or are you saying that there is an independent probability but that there will be no conditional probability?

Once you find "ab", then the probability is no longer a probability, it's a certainty. It just proves, in long-term average, that the probability / prediction was correct. So, once you find an "ab", that is great and all, but it doesn't affect at all the chances of the interval that follows. You're still at a 36% that no "ab" will be found in the next 256 samples. And still at the same probability of 1/256 that it will be found in the next sample.

Results are only useful to prove predictions, not the other way around. Maybe you found some way to use the conditional probability, as you call it, into the mix, and maybe that helps to find better prefixes, I'm not saying it doesn't. But that doesn't change at all the chances that you can always skip the key you're actually looking for - it's the same.

[mcdouglasx]

This is the part most intuitively ignore and therefore fail:

-Probability of finding "a": 1/16

-Probability of finding "ab": 1/256

-Probability of finding two "ab" in 256 attempts: approximately 0.18

As you can see, if you find "ab" in an early shot, the probability of finding another "ab" in the next 256 attempts is very low. This would allow you to skip those subsequent attempts without losing significant precision, since although each attempt is still an independent event, the chances of it being there are very low.

Here's your fallacy (which goes in repeat mode with you):

You didn't take into account that the probability to not find "ab" at all in 256 attempts is 36%.

You didn't take into account that the probability to find "ab" more than once is not "very low", but at a good 28% (100 - 36 (for 0) - 36 (for 1)). Because you never added the probabilities for it to appear 3 times, 4 times, 5 times, and so on (up to 256 times). You only base some claims on the fact that it's very unlikely to appear "a second time", but it can also appear more than 2 times, not necessarily just one more time.

So while in principle you may be on to something (though nothing more than simply observing the normal behavior of a uniform random variable), your calculations are off quite a bit, to the point that the combined probabilities of failure once you go skipping over and  over again, accumulate, and they accumulate in the fashion that you like really much: compounded.

I already explained, the math backs me up, you can try to refute it but anyone who reads and knows the subject will understand what I mean. Neither you nor I can alter that. If the probabilities of finding 2 are low, what importance do the probabilities of finding 3 or more have if they decrease exponentially? And I care even less about the probabilities of finding nothing, as I am omitting that unlikely space.

Regarding how I correct the margin of error, I already explained. In a very unlikely case (very, very, very unlikely) that it misses the target, my script simply recalculates the database, omitting a smaller percentage until it finds the target. Consequently, in the worst-case scenario, which is extreme (omitting 0%, target next to another identical prefix), I will find the target 100% of the time. My script just prioritizes the most probable moves. So I don't understand your battle for ego.

Every time you try to be right, you fail more.

I am only defending my idea: I don't know what Bibilgin does, and it could be something entirely different. Since I don't know how he does his calculations, I can't comment on whether it's right or wrong, but I give him the benefit of the doubt because he hasn't disclosed his method.


edit:

You didn't take into account that the probability to not find "ab" at all in 256 attempts is 36%.

My script is based on the premise of 30% to minimize risks. I don't know why you think this hasn't been considered beforehand.

[dastic]

In the meantime i found out that machine learning can compute correct 3 hex digits of a privkey corresponding to the target pubkey... not much but interesting

[bibilgin]

Results are only useful to prove predictions, not the other way around. Maybe you found some way to use the conditional probability, as you call it, into the mix, and maybe that helps to find better prefixes, I'm not saying it doesn't. But that doesn't change at all the chances that you can always skip the key you're actually looking for - it's the same.

You say that this can happen with conditional probability.

I didn't say I won't skip the key. I didn't say it definitely won't happen. I didn't say it's impossible.

Do you accept this?
With this level of hardware power, if you had the knowledge to calculate the locations of the Prefixes.

1- Would you prefer the entire range or RANDOM scanning?
2- Would you chase the 1 in 1024 chance, even though you know the locations of close to 1024 with average probability?

I would be happy if you give a short answer. 1? 2?

If you choose 1, I can say a lot of words for you.
If you choose 2, you are thinking right. I wish you success in your studies.

[kTimesG]

With this level of hardware power, if you had the knowledge to calculate the locations of the Prefixes.

1- Would you prefer the entire range or RANDOM scanning?
2- Would you chase the 1 in 1024 chance, even though you know the locations of close to 1024 with average probability?

I would be happy if you give a short answer. 1? 2?

If you choose 1, I can say a lot of words for you.
If you choose 2, you are thinking right. I wish you success in your studies.

It depends on what I'm looking for, assuming your strategy works.

If I'm looking for an exact total match, I'd scan the full range, first to last key. No exceptions. It's simply the fastest, most efficient way to do it. It's also the most parallel-friendly way to do it at massive (and I mean 68-bit massive) scale, since there is no need for complex logic, like stopping 65536 threads in the middle of their work just because some prefix matches somewhere, and then setting things up again, recomputing ranges, etc. Maybe you do that manually, which is a time waste, which can be used to simply scan more stuff, IDK.

---

If I'm after a nice prefix but I don't care about a perfect match, just find it really fast - I'd go for 2. But in all honesty I would not even need to go for 2 - because a VanitySearch would run many times faster, for this purpose. For example, an RTX 4090 that searches for some target prefix, hard-coded, and using endomorphism, can do 9.6 billion hashes per second, and find a key having the desired prefix.

And here is why I wouldn't use 2 to find a complete match (I said this already on some of your first posts): even if you take that "1 in 1024" route, there is a chance you don't find what you wanted, but something else. But the overhead of repeating the process to exhaustion (starting over with the second "1 in 1024 chances", then the third ... then the millionth, etc), overall, results in a longer process than simply scanning the range start to finish. And the probability is identical in both cases, because you can't cheat your way out of searching linearly for something that is random. You may disagree here - it's OK.

Meanwhile, Bram already started hunting down 68 for several days now, since the crowd funding was completed, so I guess you should hurry up.

[bibilgin]

<<<cut>>

Look, this long answer really confuses you and shows that you are trying to show different results correctly.

How can you say that scanning the entire range with the work I did will take the same time?

I will ask you a question that you will answer very briefly.

You have 10m/key power.
ABC1234567890 - ABC9875643210
to find a wallet between these 2 hexes.

The result of the calculations;

ABC31-ABC32
ABC45-ABC47
ABC58-ABC60
ABC6B-ABC6D
ABC7F-ABC83
ABC90-ABC91

1- Scanning these ranges takes less time? (Even if you know the probability of skipping the wallet.)

2- Or the entire range or the random search result?

Give a very short answer, which one would you approve? 1 or 2?

Don't confuse people by writing long sentences.

Edit;

By the way, I don't think Bram will spend that much money again. Because his team has problems, I guess.

[brainless]

Thank you. You wrote above C0 now everyone is looking above. But still thanks for the answer.

Here is another good information for you, it remains under E2. 68th wallet

In my probability d and 8 , key not in these range

[kTimesG]

How can you say that scanning the entire range with the work I did will take the same time?

I said it will take more time, not the same time.

Yes, I'd scan the range (in some amounts of distributed different subranges, in some whatever order), with 100% confidence of finding the key along the way, at some point.

Your way means having less than 100% confidence after going through some ranges, and when things don't work as expected, then what? Rinse and repeat. More effort to build up to 100%, yes. And a gazillion intervals starting off from weird offsets. Keeping track of what was scanned and what not. Everyone understood already for a long time what you are doing, you don't really need to explain so many times.

But you do you.

[bibilgin]

I said it will take more time, not the same time.

Yes, I'd scan the range (in some amounts of distributed different subranges, in some whatever order), with 100% confidence of finding the key along the way, at some point.

Your way means having less than 100% confidence after going through some ranges, and when things don't work as expected, then what? Rinse and repeat. More effort to build up to 100%, yes. And a gazillion intervals starting off from weird offsets. Keeping track of what was scanned and what not. Everyone understood already for a long time what you are doing, you don't really need to explain so many times.

But you do you.

Ok master, everyone understands you now. lol

Everyone understood that it is more logical to start with the ranges you calculated and search for other ranges as you said. (except you)

It will start with the calculated ranges, how many people do you think there are here?

Remember;
1- You have many competitors, their hardware, software or money power is more than you.
2- Everyone has their own special thoughts and strategies.
3- After all these talks, I understood that the people who will agree with you are only those who make decisions because you give information about the kangaroo system and do not really understand what you wrote.

Do you have any work on this subject? Have you guessed any range? Or have others asked you for the prefix? Have you ever proven yourself on something? Have you won a bet?

The Swiss army knife showed the truth. But you lost. Now accept this.

If we were to vote here, you would not be able to get many votes for the idea of ​​scanning the entire range as you said.

[kTimesG]

I said it will take more time, not the same time.

Yes, I'd scan the range (in some amounts of distributed different subranges, in some whatever order), with 100% confidence of finding the key along the way, at some point.

Ok master, everyone understands you now. lol

Everyone understood that it is more logical to start with the ranges you calculated and search for other ranges as you said. (except you)

It will start with the calculated ranges, how many people do you think there are here?

Remember;
1- You have many competitors, their hardware, software or money power is more than you.
2- Everyone has their own special thoughts and strategies.
3- After all these talks, I understood that the people who will agree with you are only those who make decisions because you give information about the kangaroo system and do not really understand what you wrote.

Do you have any work on this subject? Have you guessed any range? Or have others asked you for the prefix? Have you ever proven yourself on something? Have you won a bet?

The Swiss army knife showed the truth. But you lost. Now accept this.

If we were to vote here, you would not be able to get many votes for the idea of ​​scanning the entire range as you said.

WTF...

I'm not looking at neither any address puzzles, nor the 135 puzzle. They will be solved long before I even have 1% chances of success, because:

1. RetiredCoder has a much faster ECDLP solver than anything that was ever made public. He's also a millionaire, you know? He'll solve 135 by end of this year for sure.

2. A retired cryptographer with a seed recovery business, that he owns, wrote this silly brute-force key cracker, that's also faster than anything that was ever made public. All of his investors made up a 125% profit in 67 days, you know?

These guys are your competition, not me. They are busy writing and improving their code, not fighting on a forum. If I would be in their position, trust me, you wouldn't even need to read my replies, as they would not exist - I'd be busy with actual work, you know.

Have you ever, ever, ever thought about WHY neither of those two guys have made public their software? Their actual software, not proofs of concept. I'll give you a hint: because those programs would have a value in hundreds of thousands of dollars, and millions of dollars in long-term.

Skills are valuable, but they must be met with actual results, not fantasies. I could very easily prove to you that your CUDA apps that you use are so slow, that they are laughable. And I can also show mcdouglasx that my kangaroo speeds are also very impressive, but neither of these things would help anyone with anything, except some dudes that would use them on vast.ai to produce some profit. This is, I can guarantee you, the main reasons you won't see RetirecCoder's or Bram's code anywhere any time soon.

[jedi12345]

This is pretty much the problem I'm having.

I can narrow the 68 puzzle down from.

295,147,905,179,352,825,856 Combinations to

402,120,015,509,709,217       Combinations

With a estimated 35-40% chance one will be right.

but even with 3 zeros removed its still requires a huge amount of compute.

I just assumed with only 0.15% of all possibilities needed it would take a regular 3090 like one week, but even with 8 3090s it would still take months.

I have no idea how people plan to pool mine this thing, will probably cost more in power than it is worth. unless you have a asic + low cost power and custom code.

if 8 3090s can check say 30billion keys a sec total (its not just a single 256 hash so prob cant even do that). it would take me about 140 days to check my guess.

so if you had 8000 3090 you could check every combo in 140 days estimate. From memory when I had my mining farm 60 Gpus were about $1000 a month in power but only using 100watts each, so your looking at about $200,000 a month in power or 1.2m to check every combo and also the loss in gpu value as tech devalues. Lets say 10% loss in value, or 800gpus, so like another 800k probably.

So for $2 million dollars you can crack this thing 

[WanderingPhilospher]

This is pretty much the problem I'm having.

I can narrow the 68 puzzle down from.

295,147,905,179,352,825,856 Combinations to

402,120,015,509,709,217       Combinations

With a estimated 35-40% chance one will be right.

but even with 3 zeros removed its still requires a huge amount of compute.

I just assumed with only 0.15% of all possibilities needed it would take a regular 3090 like one week, but even with 8 3090s it would still take months.

I have no idea how people plan to pool mine this thing, will probably cost more in power than it is worth. unless you have a asic + low cost power and custom code.

if 8 3090s can check say 30billion keys a sec total (its not just a single 256 hash so prob cant even do that). it would take me about 140 days to check my guess.

so if you had 8000 3090 you could check every combo in 140 days estimate. From memory when I had my mining farm 60 Gpus were about $1000 a month in power but only using 100watts each, so your looking at about $200,000 a month in power or 1.2m to check every combo and also the loss in gpu value as tech devalues. Lets say 10% loss in value, or 800gpus, so like another 800k probably.

So for $2 million dollars you can crack this thing 

Your numbers are way off. Even in your power costs for running 60 GPUs at 100w each, unless you live in a place where power costs are extremely high, which could be. At 15 cents per (kWh) costs, you'd be looking at $684 a month. For a $1000 a month, you'd be looking at 23.1 cents per kWh. That's pretty high IMO.

If you base the average solution time (the keys checked to actually find the solution) off of 67's, and you had to rent every GPU through a public supplier, you'd be looking at about $686,000 to solve 68. A lot less than the $2,000,000 you spoke of. If you owned your own GPUs, and had costs of 15 cents per kWh, you'd be looking at costs of around $206,000. These numbers are based off of using 256 x RTX 4090s, at 6.8 BK/s, which there are software versions out there that get over 7 BK/s, so I went a little lower.
2^67 * 57% / (6800000000 * 60 * 60 * 24 * 256) = 559 days

Unless, my numbers are way off, or I am not understanding what you are saying, both could be true lol.

[bibilgin]

<<<cut>>>

LOL.. LOL .. LOL...

What are you doing here? Other than being an opposition to everything? What are you good for? Are you hanging out to satisfy your EGO? Is this how you treat your psychology?

The 2 people you mentioned,
RetirecCoder = The only individual with only money power, someone who does not have any software that reaches different speeds. He broke it ONLY with his MONEY POWER.

Bram = Someone who is part of a team, I still have doubts about him being a team member. Because he may have requested these signing events from his team. Also, this team has made serious investments for this business.
He does not have any software that reaches more speed.

You = Are you trying to play this duo? What can you get? I will tell you. Nothing.

Now step aside, stay quiet. Don't reply to people to satisfy your EGO.