1

I'm reading Analysis of hashrate-based double-spending where the following assumptions are made (on page 5):

  • The total hashrate of the honest network and the attacker is constant, say H. Let the honest network have pH hashrate while the attacker has qH hashrate, p + q = 1.

  • Mining difficulty is constant, and with hashrate H the average time to mine a block is T_0.

The author writes that if the honest network and attacker start mining blocks at the same time then the probability that the former (resp. latter) mines a block first is p (resp. q). Now, while it is intuitive to grasp why this should be true, I am unable to prove so. Any help would be appreciated!

1 Answer 1

0

I'm not sure if there is a formal "proof" available as this is implied by basic logic.

Assuming a block is found (aka mined) in future and there are only two mining subgroups "Honest" and "Attacker". If they have equal hashpower (p=0.5, q=0.5) then they are searching the solution space at the same rate and each has a 50 percent chance of finding this next block. If Honest has twice the hashpower of Attacker (p=2/3, q=1/3) then Honest is searching the solution space at twice the rate of Attacker and hence has twice the probability of finding the next block. Similarly with p=0.8, q=0.2 Honest is searching the solution space at 4 times the rate of Attacker and has 4 times the probability of finding the next block.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.