0

If I am running my own node, and using a mixer to anonymize my BTC, will it not be connected anyway because the transactions are to/from my own node?

I mean I am sending the BTC to the mixer from my node, and then getting it back to another address, sure, but I am accessing that address from my own node...

Even if I am using TOR, is there not something that identifes the nodes and tells observers that these two addresses have something in common, the node on my machine?

Improve this question
3
  • 1
    Are you using a custodial mixing service? (a very bad idea). Or are you using a wallet that implements a coinjoin/payjoin protocol? (not a bad idea).
    – chytrik
    Commented May 5, 2021 at 5:19
  • In this case a mixing service. I tried coinjoining but it was rather complicated to setup and make work... Anyway, I'm not mixing large sums.
    – CodeOrElse
    Commented May 5, 2021 at 22:30
  • Those services provide terrible privacy assurances (if any at all), and expose you to massive counterparty risk (lots of examples of them running away with customer coins). I would absolutely not recommend using them, even for smaller sums.
    – chytrik
    Commented May 5, 2021 at 22:37

1 Answer 1

Reset to default
1

Technically the others nodes you are connected to would see you sending the transaction to them but they wouldn’t know if it is your transaction or someone else’s and you are just relaying it.

There might be a scenario where someone has strategically placed lots and lots of nodes into the network and monitors when certain transactions arrive to each of them and might be able to narrow down where the transaction originally came from to a small set of nodes.

Using TOR would limit this potential deanonymization by obfuscating your IP (which is the identifying piece of info)

Improve this answer
0

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.