Let there exist a block A that we haven't mined, let N represent the decimal value of the share of the total mining hashrate we have (so if we had 50%, N = 0.5), and represent the miner payments for A as X, and the expected MEV payments as Y. If we mine the next two blocks on top of A, we have a (N x 100)% chance of mining each independently, so the expected payoff is N x (4 ETH + 2Y). If we attempt a time bandit (and, as 0x9116 originally assumed, quit if the next block B is mined):
1. With 1 - N probability the next block B is mined before we can uncle A and replace it with A'. We're then back at square one, where we just look to mine the next two blocks fairly. The expected payoff in this case is (1 - N) x N x (4 + 2Y).
2. With N x (1-N) probability A' is mined but B is then mined before we can mine B'. A' gets uncled to B, getting a 1.75 ETH reward, and you look to mine the block after B. The expected payoff in this case is N x (1 - N) x (1.75 + N x (2 + Y)).
3. With N<sup>2</sup> probability, we mine both A' and C'. The expected payoff in this case is N<sup>2</sup> x (4 + X + Y).
We then add together the expected payoffs to get the total expected payoff for the time bandit scenario. We then compare it to our expected payoff for honest mining, which is N x (4 + 2Y) to find the bound on X for a reorg to be worth it.