4 ms·

> you now have twice the odds of being the one that gets it This is not the right way to think about it. Someone else publishing a valid block doesn't deprive

by tromp 4y ago

> you now have twice the odds of being the one that gets it This is not the right way to think about it. Someone else publishing a valid block doesn't deprive you of a reward. You just continue mining on top of their now latest block. When a miner's hash meets the difficulty target, they have a valid block that they publish, and will collect its reward (except in the unlikely case of another miner finding a valid block at almost exactly the same time). The amount of rewards a miner collects thus depends solely on how often their hashes meet the difficulty target, irrespective of the total hashrate.

lazide 4y ago
Eh, we're saying similar things, but you're wrong in a fundamental detail - and you're making it really confusing for everyone who doesn't understand what is really going on. The rewards a miner collects do not depend solely on how often their hashes meet the difficulty target. Otherwise, every miner could keep mining prior blocks forever until they got a hash that matched, and collect the reward for that block. But that isn't allowed to happen. The first miner that gets a block accepted by the network with a valid hash wins. A block can only be mined once, by one miner. Otherwise, transactions would be double spent, or blocks double mined. Your individual odds of being that miner are directly proportional to the amount of hash power being used to mine the current, most recent block. The window of time you have (aka the 'time until the lottery drawing') is until someone finds a winning hash to the current block. When there is a conflict/split in the blockchain, the longest/highest chain of blocks wins, and everything on the 'split' chain gets ignored. The algorithm bitcoin uses to mine is a rather simple one. 1) Collect validated transactions 2) Add the block metadata, such as prior/highest current block (this is what makes it a block chain), this miners metadata (if any), and the block reward address if this block wins. 3) Add/Change the blocks random nonce 4) Construct a block by putting all these together into the binary format. 5) Hash it 6) If the produced output hash has the right number of zeros in it (really) to match the difficulty target, it is a winning block. Publish it, and if the network accepts and builds more blocks on top of it, you can spend your reward. Otherwise, go back to 3. The difficulty target is partitioning up the bitspace to determine the odds of 'winning' blocks, just like I'm describing using the lottery ticket analogy. There is no way any miner can change their odds here (or 'try to meet the difficulty target'), unless they collude with others to avoid already tried hashes/nonces for the same exact block values, which they have negative incentive to do. In fact, independent miners have every incentive to make their opponents try the things they already know won't work. Removing half the hash power would double the chances (all other things remaining the same) of the remaining hashers winning, as there is now half the number of competitors randomly trying to get a winning hash, as the winning hashes are randomly distributed. It also doubles the time to find a winning hash, all other things remaining the same, as there is now half the number of players trying random values. (or half the number of random attempts to find the right value). Unless someone has broken SHA-256, Every active miner has the same odds on any hash attempt to get the winning hash value. That means everyone has to try every value within that space. And they won't know what value they'll get from hashing until they do the actual hashing (aka do the work). That's pretty fundamental to the protocol. Increasing/changing difficulty targets isn't linear in impact, but that's not related to what is being discussed here, as the target itself isn't changing. Just available hashing power.
- Dylan16807 4y ago
  > Eh, we're saying similar things, but you're wrong in a fundamental detail - and you're making it really confusing for everyone who doesn't understand what is really going on. Which detail? > The rewards a miner collects do not depend solely on how often their hashes meet the difficulty target. Otherwise, every miner could keep mining prior blocks forever until they got a hash that matched, and collect the reward for that block. But that isn't allowed to happen. Obviously you have to mine based on the current block. Nobody suggested otherwise. It doesn't affect your odds of finding blocks because it only takes a fraction of a second to switch to the new one. > The first miner [...] and everything on the 'split' chain gets ignored. Nobody suggested otherwise. > The algorithm bitcoin uses Yes. > There is no way any miner can change their odds here Nobody suggested otherwise. > (or 'try to meet the difficulty target') That just means doing hashes and hoping you win. > Removing half the hash power would double the chances (all other things remaining the same) of the remaining hashers winning, as there is now half the number of competitors randomly trying to get a winning hash, as the winning hashes are randomly distributed. It changes the percent of blocks a miner wins. It does not change the number of blocks per hour they win. And that's the only number that matters. (Both expressed as expected averages, of course.) > Unless someone [...] Just available hashing power. Nobody suggested otherwise.
  - lazide 4y ago
    Good luck!

by tromp 4y ago

lazide 4y ago

Eh, we're saying similar things, but you're wrong in a fundamental detail - and you're making it really confusing for everyone who doesn't understand what is really going on. The rewards a miner collects do not depend solely on how often their hashes meet the difficulty target. Otherwise, every miner could keep mining prior blocks forever until they got a hash that matched, and collect the reward for that block. But that isn't allowed to happen. The first miner that gets a block accepted by the network with a valid hash wins. A block can only be mined once, by one miner. Otherwise, transactions would be double spent, or blocks double mined. Your individual odds of being that miner are directly proportional to the amount of hash power being used to mine the current, most recent block. The window of time you have (aka the 'time until the lottery drawing') is until someone finds a winning hash to the current block. When there is a conflict/split in the blockchain, the longest/highest chain of blocks wins, and everything on the 'split' chain gets ignored. The algorithm bitcoin uses to mine is a rather simple one. 1) Collect validated transactions 2) Add the block metadata, such as prior/highest current block (this is what makes it a block chain), this miners metadata (if any), and the block reward address if this block wins. 3) Add/Change the blocks random nonce 4) Construct a block by putting all these together into the binary format. 5) Hash it 6) If the produced output hash has the right number of zeros in it (really) to match the difficulty target, it is a winning block. Publish it, and if the network accepts and builds more blocks on top of it, you can spend your reward. Otherwise, go back to 3. The difficulty target is partitioning up the bitspace to determine the odds of 'winning' blocks, just like I'm describing using the lottery ticket analogy. There is no way any miner can change their odds here (or 'try to meet the difficulty target'), unless they collude with others to avoid already tried hashes/nonces for the same exact block values, which they have negative incentive to do. In fact, independent miners have every incentive to make their opponents try the things they already know won't work. Removing half the hash power would double the chances (all other things remaining the same) of the remaining hashers winning, as there is now half the number of competitors randomly trying to get a winning hash, as the winning hashes are randomly distributed. It also doubles the time to find a winning hash, all other things remaining the same, as there is now half the number of players trying random values. (or half the number of random attempts to find the right value). Unless someone has broken SHA-256, Every active miner has the same odds on any hash attempt to get the winning hash value. That means everyone has to try every value within that space. And they won't know what value they'll get from hashing until they do the actual hashing (aka do the work). That's pretty fundamental to the protocol. Increasing/changing difficulty targets isn't linear in impact, but that's not related to what is being discussed here, as the target itself isn't changing. Just available hashing power.

Dylan16807 4y ago
> Eh, we're saying similar things, but you're wrong in a fundamental detail - and you're making it really confusing for everyone who doesn't understand what is really going on. Which detail? > The rewards a miner collects do not depend solely on how often their hashes meet the difficulty target. Otherwise, every miner could keep mining prior blocks forever until they got a hash that matched, and collect the reward for that block. But that isn't allowed to happen. Obviously you have to mine based on the current block. Nobody suggested otherwise. It doesn't affect your odds of finding blocks because it only takes a fraction of a second to switch to the new one. > The first miner [...] and everything on the 'split' chain gets ignored. Nobody suggested otherwise. > The algorithm bitcoin uses Yes. > There is no way any miner can change their odds here Nobody suggested otherwise. > (or 'try to meet the difficulty target') That just means doing hashes and hoping you win. > Removing half the hash power would double the chances (all other things remaining the same) of the remaining hashers winning, as there is now half the number of competitors randomly trying to get a winning hash, as the winning hashes are randomly distributed. It changes the percent of blocks a miner wins. It does not change the number of blocks per hour they win. And that's the only number that matters. (Both expressed as expected averages, of course.) > Unless someone [...] Just available hashing power. Nobody suggested otherwise.
- lazide 4y ago
  Good luck!