Dishonest Tor relay math question - tor-talk is to lazy

[Stefan Claas]

Hi,

IIRC the assumed number of malicious nodes is much higher.
Then you do not include the assumed number of honest, but
compromised nodes.

How much would your equation help Tor users, in different
locations, if ISPs would hand over to third parties who is
using Tor with port 9050 and 9051, so that third parties could
take further actions in the long run?

Regards
Stefan


On Thu, Oct 7, 2021 at 10:51 PM PrivacyArms <privacyarms at protonmail.com> wrote:
>
> Dear Cypherpunks community,
>
> I came across a post on the Whonix forum recently. Since I am also interested in this question I copied it here:
> https://forums.whonix.org/t/math-behind-honest-tor-nodes/12464
> http://forums.dds6qkxpwdeubwucdiaord2xgbbeyds25rbsgr73tbfpqpt4a6vjwsyd.onion/t/math-behind-honest-tor-nodes/12464
>
> The question (edited):
> How can I calculate how much impact X honest Tor relays have?
> Is it better to calculate with bandwidth consumed (250Gbps), despite the number of relays (~7000)?
>
> Basically, I want to get the mathematical equation to this statement:
> I run X Tor relays at Y Mb/s each and by doing so I secure Z % of the Tor network!
> Starting thoughts:
> - Each “normal” route has three nodes involved: Guard, Middle, Exit
> - I am aware of guard pinning and vanguard protection for middle relay pinning
> - Maybe it is easier to assume an infinite usage time of the network to eliminate guard and vanguard pinning
> - I guess the best is to assume a scenario with 1%, 5%, 10%, etc. dishonest relays
>
> My take on this:
> Tor has approximately 7000 relays.
> If I consider a number of 5% malicious relays, this would be: 350
> My calculation:
> (1/(7000/350))*(1/(7000/349))*(1/(7000/348))
> = 0.000123931
> = 0.0123931%
>
> 1) Is my approach correct?
> 2) Not every relay has the same bandwidth. How could I change the calculation to make it more realistic?
> 3) How can I add the effect of guard fixation?
> 4) How can I include the effect of mid-node fixation by the vanguard?
>
> I would love to hear your thoughts about it and a concrete math equation would be amazing.