Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
You should straight up go to Princeton and give some lectures about this before they call the police.
The advantage of prefixes is that it's not common to find multiple prefixes of a hash in a space x where their probability is 1/x.
Wrong. Finding more than one prefix having probability 1/N in N tries is exactly converging to a total of (1 - 2/e) = 26.42%, the more N grows. That is, Out of 100 tries, 28 will have at least two findings of he prefix.
If we have a lot of N-size tries (let's say, a billion), around 264 million of those will have more than one findings (e.g. they will contain either two, three, or more findings).
We can do the same calculul if you want more than two findings, up to the chances of fidning the entire N findings (hint: it's not zero chances, because the option exists, hence the probability for it also exists, which makes it a certain event if you do a shitload of tries).
Why don't you write that letter to Princeton, Stanford, etc?
1st grade fact: A uniform distribution only means that the target is equally distributed, not that all stopping criteria perform equally.
[/quote
Totally wrong. Unless, of course, you're writing your own definitions.
https://ecampusontario.pressbooks.pub/sccstatistics/chapter/the-uniform-distribution/
"The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur"
https://web.stanford.edu/class/archive/cs/cs109/cs109.1218/files/student_drive/3.5.pdf
"In the this lecture we will continue to expand our zoo of discrete random variables. The next one we will
discuss is the uniform random variable. This models situations where the probability of each value in the
range is equally likely, like the roll of a fair die."
[/quote
Totally wrong. Unless, of course, you're writing your own definitions.
https://ecampusontario.pressbooks.pub/sccstatistics/chapter/the-uniform-distribution/
"The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur"
https://web.stanford.edu/class/archive/cs/cs109/cs109.1218/files/student_drive/3.5.pdf
"In the this lecture we will continue to expand our zoo of discrete random variables. The next one we will
discuss is the uniform random variable. This models situations where the probability of each value in the
range is equally likely, like the roll of a fair die."
LOL, each stopping criterion changes the probability of success within the block, even with a uniform distribution.
p/quote
No, it does not at all. You are in a total error and just confuse people with your made-up statements of badly interpreted basic math.
p/quote
No, it does not at all. You are in a total error and just confuse people with your made-up statements of badly interpreted basic math.
Pruning by 3 hex prefix: success ≈ 63.2%
Stopping at a random point: success ≈ 50%
p/quote]
What you're pruning, the 63%, includes AT LEAST ONE orccurences, not "JUST ONE" occurrence.
So, what you're skipping, is basically either to find another hit, or maybe two, or maybe three, or maybe a shitload of them. You totally ignore this. It simply compensates back over time, because the hits you've missed, they will show up as non-hits later on.
Please stop lying. It's exactly 28.42% likely to find >= 2 hits of a 12-bit prefix in a 4086-bits set.
This case is a subset of the 63% of finding >=1 hits. If you're good at arithmetic, you'll notice that the probability of finding just A SINGLE HIT is thus 36%.
Which is exactly the same as the chance of not finding a hit. Identical chances. I guess why that happens? Maybe because the average will be 1, but it only does that because there are those 26% other cases, that help averaging out everything? Geeeeeez, math in action!
What you don't understand is basic maths. No one's crying here, I'm just keeping my promise: that every time you bring up this prefix non-sense here, I'll bring up the Scooby Doo method.
For hose new: the Scooby Doo method is an exceptional breakthrough, that extends and improves on the prefix theory, by simply removing the need to check for a prefix.
Simply skipping ahead by a chance of 1/4096, instead of doing a prefix check, yields identical results. Because this guy mcdouglasx is a fraud, and I can't understand why he's pushing total bullshit ahead over and over again, in total disregard of mathematical evidence. It seems all he wants is to get some credits, judging by his actions.
I'm gonna respond to you by saying that you have it wrong, and you can't simply extend these numbers the way you're trying to do, because this is not how a uniform distribution works.
For example, it's not the same thing to compute probability of success in a single 8192-bit range, versus 2 4096-bit ranges. So your math is wrong by definition, since the extrapolation you made has no corelation to a larger search space.
Dude, seriously, I'm cracking up. You're basically saying that, if 6 is rolled after 2 rolls, the game should stop, since we're at 63% chances in 4 more rolls.
When should those 4 rolls be skipped? You're saying they should never happen at all.
Stopping at a random point: success ≈ 50%
p/quote]
What you're pruning, the 63%, includes AT LEAST ONE orccurences, not "JUST ONE" occurrence.
So, what you're skipping, is basically either to find another hit, or maybe two, or maybe three, or maybe a shitload of them. You totally ignore this. It simply compensates back over time, because the hits you've missed, they will show up as non-hits later on.
Finding ≥ 2 prefix collisions in 4096 keys is very rare; if it were frequent, the hash function would be compromised (this doesn't mean there can't be cases where it happens, but the point is that for what the prefix lookup requires, it works).
Please stop lying. It's exactly 28.42% likely to find >= 2 hits of a 12-bit prefix in a 4086-bits set.
This case is a subset of the 63% of finding >=1 hits. If you're good at arithmetic, you'll notice that the probability of finding just A SINGLE HIT is thus 36%.
Which is exactly the same as the chance of not finding a hit. Identical chances. I guess why that happens? Maybe because the average will be 1, but it only does that because there are those 26% other cases, that help averaging out everything? Geeeeeez, math in action!
Missing the target due to premature collision occurs in the remaining ~36.8%, just as the script predicts. I don't understand why you flatly deny this fact.
Instead of continuing to cry or hesitate, accept it.
Instead of continuing to cry or hesitate, accept it.
What you don't understand is basic maths. No one's crying here, I'm just keeping my promise: that every time you bring up this prefix non-sense here, I'll bring up the Scooby Doo method.
For hose new: the Scooby Doo method is an exceptional breakthrough, that extends and improves on the prefix theory, by simply removing the need to check for a prefix.
Simply skipping ahead by a chance of 1/4096, instead of doing a prefix check, yields identical results. Because this guy mcdouglasx is a fraud, and I can't understand why he's pushing total bullshit ahead over and over again, in total disregard of mathematical evidence. It seems all he wants is to get some credits, judging by his actions.
Come on, man, you're taking all the fun out of mixing so much nonsense into a single post.
The 3-hex filter is 1/4096 in any space. It doesn't matter if the hash has 40 digits, the probability of matching the first 3 is always the same.
Limiting the range doesn't "fix" the math: the uniformity remains within that subspace, and the formula applies the same.
Going from 4096 to 2**256 only changes the scale. The exponential form is the same; the relative result remains the same.
kgtimes, you won't respond to this guy's statements, or you'll apply the "the enemies of my enemies are my friends" rule
.
The 3-hex filter is 1/4096 in any space. It doesn't matter if the hash has 40 digits, the probability of matching the first 3 is always the same.
Limiting the range doesn't "fix" the math: the uniformity remains within that subspace, and the formula applies the same.
Going from 4096 to 2**256 only changes the scale. The exponential form is the same; the relative result remains the same.
kgtimes, you won't respond to this guy's statements, or you'll apply the "the enemies of my enemies are my friends" rule
.I'm gonna respond to you by saying that you have it wrong, and you can't simply extend these numbers the way you're trying to do, because this is not how a uniform distribution works.
For example, it's not the same thing to compute probability of success in a single 8192-bit range, versus 2 4096-bit ranges. So your math is wrong by definition, since the extrapolation you made has no corelation to a larger search space.
Your analogy with dice tells me you still don't understand prefix searching. No, that's not what I'm saying. If you focused on understanding more and hating less, you would easily understand what the probabilistic prefix search method is based on. The fact that a six appears on the fifth roll doesn't mean that another 6 is less likely on the sixth; they are independent events. But it is true that if they tell you to roll a die 6 times, and the second time you got a 6, it's normal for you to bet that there won't be another 6 in the next 4 events. Although it could happen, statistically it doesn't change the probability of 1/6.
Dude, seriously, I'm cracking up. You're basically saying that, if 6 is rolled after 2 rolls, the game should stop, since we're at 63% chances in 4 more rolls.
When should those 4 rolls be skipped? You're saying they should never happen at all.