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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 havepHhashrate while the attacker hasqHhashrate,p+q= 1.Mining difficulty is constant, and with hashrate
Hthe average time to mine a block isT_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!
asked yesterday
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.
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