how to know how long it will take to complete a block?

[D4rkAng3l]

Hello to everybody,

if I know the difficult of a coin and the khash\s speed of my hardware can I know how many time need to close a block in solo mining?

Exist a formula to calculate it?

Tnx

[S4VV4S]

http://www.bitcoinx.com/profit/

[D4rkAng3l]

http://www.bitcoinx.com/profit/

Is this the value that I have to look "Average generation time for a block (solo)" to know how time I need to close a block? or what?

[S4VV4S]

http://www.bitcoinx.com/profit/

Is this the value that I have to look "Average generation time for a block (solo)" to know how time I need to close a block? or what?

Yeap

[flounderella]

Hello to everybody,

if I know the difficult of a coin and the khash\s speed of my hardware can I know how many time need to close a block in solo mining?

Exist a formula to calculate it?

Tnx

Blocks are not "completed" per se, they are found ... i.e. there is no progress bar, each hashing is independent of the next.

[greenlion]

Take the current target and divide by 115792089237316195423570985008687907853269984665640564039457584007913129639935.

That number represents the probability of any given hash solving a block.

You now have a very standard binomial probability problem where n represents the number of hashes your hardware is capable of generating over the amount of time you're considering.

If you call the above calculated number P, then the probability of producing a hash that qualifies for the target difficulty after n hashes is going to be 1 - (1-P)^n

[JesseCoin]

Take the current target and divide by 115792089237316195423570985008687907853269984665640564039457584007913129639935.

That number represents the probability of any given hash solving a block.

You now have a very standard binomial probability problem where n represents the number of hashes your hardware is capable of generating over the amount of time you're considering.

If you call the above calculated number P, then the probability of producing a hash that qualifies for the target difficulty after n hashes is going to be 1 - (1-P)^n

Thanks for this