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Title: basic probability Post by: ignacio_buendia on August 04, 2020, 11:06:53 AM Is it correct to state the following?
If the difficulty of the protocol requires a digest with 5 leading zeros, than I have the chance of ((1/16^5)+(1/16^6)+...(1/16^32)) to find a valid proof of work with 1 attempt - regardless of any variables? Title: Re: basic probability Post by: j2002ba2 on August 04, 2020, 01:47:46 PM No. The zeroes are independent. The probability of having 5 (hexadecimal) leading zeroes is 16-5.
Title: Re: basic probability Post by: BrewMaster on August 04, 2020, 02:15:06 PM since the difficulty is defined as an integer comparison not a "leading zero count" thing, if you want to compute the probability of finding a block at first attempt then you must find the probability of finding a number that is in range from 0 to target from a bigger range of numbers that go from 0 to 2^256.
and if i'm not mistaken that probability is calculated by dividing target by 2^256. |