<p>I'm not sure if there is a formal &quot;proof&quot; available as this is implied by basic logic.</p>
<p>Assuming a block is found (aka mined) in future and there are only two mining subgroups &quot;Honest&quot; and &quot;Attacker&quot;. 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.</p>

CC BY-SA 4.0
2022-07-23T19:15:55.100Z
2022-07-23T19:15:55.100Z
Michael Folkson
<p>I'm reading <a href="https://arxiv.org/pdf/1402.2009.pdf" rel="nofollow noreferrer">Analysis of hashrate-based double-spending</a> where the following assumptions are made (on page 5):</p>
<ul>
<li><p>The total hashrate of the honest network and the attacker is constant, say <code>H</code>. Let the honest network have <code>pH</code> hashrate while the attacker has <code>qH</code> hashrate, <code>p</code> + <code>q</code> = 1.</p>
</li>
<li><p>Mining difficulty is constant, and with hashrate <code>H</code> the average time to mine a block is <code>T_0</code>.</p>
</li>
</ul>
<p>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 <code>p</code> (resp. <code>q</code>).
Now, while it is intuitive to grasp why this should be true, I am unable to prove so.
Any help would be appreciated!</p>

<doublespend><hashpower><math>
Modelling random walks for double spending based on hashrate
SyntasticMonoid
# Modelling random walks for double spending based on hashrate

(Asked by: username_0 on 2022-07-23)

I'm reading [Analysis of hashrate-based double-spending](https://arxiv.org/pdf/1402.2009.pdf) 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!

**Tags:** doublespend, hashpower, math


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## 1 answers

### Answer by username_1 on 2022-07-23. Score: 0 | Accepted answer

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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