3 ms·
This is where your mistake is: they can't be using the other blocks as well, because they are erasing their own (uncensored) transactions by doing so. Let's sa
by simiones 4y ago
This is where your mistake is: they can't be using the other blocks as well, because they are erasing their own (uncensored) transactions by doing so.
Let's say that MiningPoolA controls 51% of hashpower, and MiningPoolB controls 49%. MiningPoolA is refusing to mine transactions from/to some wallet W.
Time T1:
MiningPoolA: OldChain -- Block1(no W)
MiningPoolB: OldChain -- Block1'(W receives 1BTC) still in progress
Any client will accept the chain with Block1 (no transactions from/to W)
Time T2 - if MiningPoolB tries to compete
MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W)
MiningPoolB: OldChain -- Block1'(W receives 1BTC) -- Block2' (parent=Block1') still in progress
Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up
Time T2 - if MiningPoolB decides to accept MiningPoolA's chain, but add the transaction in the second block:
MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W)
MiningPoolB: OldChain -- Block1(no W) -- Block2' (parent=Block1, adds transaction W receives 1BTC) still in progress
Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up
If a mining pool had 51+% of hashpower, they would always be mining the longest chain, no one would be able to compete with them and publish another block (in principle, at least; in practice, since mining is not entirely deterministic, someone else will occasionally win the lottery and propose a new block faster).