Thread by @murchandamus

@Mario_Gibney 1/ Yeah, lemme try to draw it out. Let's assume it takes 6s for a block to reach other miners and compare 60s vs 600s block intervals for miners with 5% vs 30% of the hashrate. 6s is roughly the situation after Compact Blocks: dsn.kastel.kit.edu bitcoin/#propagation

@Mario_Gibney 2/ The propagation delay is both an advantage and disadvantage: it gives the author a headstart for the next block, but also gives time for a competing block to emerge which may oust the author's block. Let's assume half the remaining HR works on the competing block.

@Mario_Gibney 3/ We can calculate the chance of finding one or more blocks in a time interval by starting with 100% and deducting the chance of finding 0 blocks. The chance to find 0 blocks in a time frame where we'd expect λ blocks is `p(0|λ) = exp(-λ)*(λ^0)/(0!) = exp(-λ)*1/1 = exp(-λ)`.

@Mario_Gibney 4/ Given a 600s block interval, we'd expect 0.01 blocks to be found in 6s by 100% of the hashrate, `0.05×0.01 = 0.0005` blocks to be found by the 5% miner, and 0.0030 blocks by the 30% miner.

@Mario_Gibney 5/ Given a 60s block interval, we'd expect 0.1 blocks to be found in 6s by 100% of the hashrate, `0.05×0.1 = 0.005` blocks to be found by the 5% miner, and 0.030 blocks by the 30% miner.

@Mario_Gibney 6/ As noted above, we'll simplify to expect half the remaining hashrate to work on a competing block whenever one is found. In case of the 5% miner this is 47.5% of the HR, and it's 35% for the 30% miner.

@Mario_Gibney 7/ The author wins either when they find another block during the headstart or when their collaborators find the next block, and loses when both a competing block is found and it sires the successor: `P_loss = (1-e^(-λ_comp))×comp/2×e^(-λ_self)`

@Mario_Gibney 8/ (That's a simplification, because we don't cover the case where both sides find at least one block in the headstart time window, but the competitors find more blocks.–I think we can get away with it.)

@Mario_Gibney 9/ At 600s blocks for the small miner:
`P_loss = (1−e^(−0,0095))×0,475×e^(−0,0005) = 0.0045`. I.e. the 5% miner mines at a handicap of 0.45% and has an effective HR of `5%×(1-0.0045)=4.98%`.

@Mario_Gibney 10/ At 600s blocks for the big miner:
`P_loss = (1−e^(−0,007))×0,35×e^(−0,003) = 0.0024`.
I.e. the 30%-miner mines at a handicap of 0.24% and has an effective HR of `30%×(1-0.0024)=29.93%`.

@Mario_Gibney 11/ At 60s blocks for the small miner:
`P_loss = (1−e^(−0,095))×0,475×e^(−0,005) = 0.043`.
I.e. the small 5%-miner mines at a handicap of 4.3% and has an effective HR of `5%×(1-0.043)=4.79%`.

@Mario_Gibney 12/ At 60s blocks for the big miner:
`P_loss = (1−e^(−0,07))×0,35×e^(−0,03) = 0.023`.
I.e. the bigger 30%-miner mines at a handicap of just 2.3% and has an effective HR of `30%×(1-0.023)=29.3%`.
At the same handicap as the smaller miner it would be only 28.7%.

@Mario_Gibney 13/ We see that going from 600s blocks to 60s blocks, a 5%-miner goes from a 0.45% handicap to a 4.3% handicap, but a 30%-miner goes from a 0.24% handicap to a 2.3% handicap.

@Mario_Gibney 14/ Also, the stale block rate would increase from one every ~100 blocks to about one in every ~10 blocks; increasing hashrate not contributing to security, the chance of multi-block reorgs, and easing majority and selfish mining attacks.

@Mario_Gibney 15/15 In effect, an individual miner contributing to a 5%-pool would make about 0.2% more money by joining a 30%-pool at 600s blocks, but the same move would gain 2.1% more money at 60s blocks.

@Mario_Gibney Gonna run off for lunch before the first mathematician looking at my napkin math corrects me. 😅

The link in the first tweet was supposed to be dsn.kastel.kit.edu/bitcoin/#propa…