I keep seeing a lot of people that I think are confused how pooled mining works.
We aren't all doing part of a big problem, and a round is not just one block. Every time the network finds a block the problem changes cause they are all a chain. A share isn't a piece of a puzzle it's a lottery ticket. And we are all pooling our lottery tickets in hope one of us wins with the agreement we all split the proceeds. So this %CDF isn't coverage of a problem space it's just a probability that a winning lottery ticket will be made for the latest link of the chain.
The real trick is that we are all working on low difficulty attempts so we can make lots of them in the hopes that by chance we'll hit one that is above the required difficulty.
Or at least this is what I'm pretty sure is the case usual disclaimer applies I'm not an expert.
We aren't all doing part of a big problem, and a round is not just one block. Every time the network finds a block the problem changes cause they are all a chain. A share isn't a piece of a puzzle it's a lottery ticket. And we are all pooling our lottery tickets in hope one of us wins with the agreement we all split the proceeds. So this %CDF isn't coverage of a problem space it's just a probability that a winning lottery ticket will be made for the latest link of the chain.
The real trick is that we are all working on low difficulty attempts so we can make lots of them in the hopes that by chance we'll hit one that is above the required difficulty.
Or at least this is what I'm pretty sure is the case usual disclaimer applies I'm not an expert.
I'm not an expert on the btc protocol per se, but I do understand the mathematics involved, and this is a very good and succinct description for lay people. However, the %CDF is not really a probability, but a measure of how many blocks of this difficulty one would "normally" expect to have been found after this many lottery tickets (shares) have been cast.
EDIT - odd as it may seem, the actually probability that share N finds the block is exactly the same as the probability that share 1 will find it.
