Can repeated runs of a Las Vegas algorithm equate to a better Las Vegas algorithm? Let's say I have a Las Vegas algorithm $L$ for some problem $P$, which runs in $n^3$ steps with 50% likelihood. My friend asks me for an algorithm for $P$ that runs in $n^3$ steps with 75% likelihood. My intuition says, "No problem! I'll just run $L$ once, and if it doesn't return after $n^3$ steps, I'll terminate it. Then, I'll run it again. The likelihood that it fails to return after $n^3$ steps twice in a row is only 25%."

Is my intuition correct? I know that if it's running twice, that's technically $2n^3$ and that would be subsumed by big-O notation. Maybe the problem that was posed to me was not carefully worded enough...
Monte Carlo and Las Vegas algorithms are equivalent and this question is more naturally asked in terms of the former so I will pretend you asked about Monte Carlo.

Any Monte Carlo algorithm can be run multiple (say, a constant number of) times to amplify the probability of success while slowing down the algorithm. And in fact you can achieve any fixed probability of success $&lt; 1$ while slowing down by a constant factor.

Where you "cheat" is that you adopt the convention of ignoring the constant factor in run time while simultaneously caring about the constant probability of success. As the probability can be freely adjusted by multiplying run time by a constant, caring about its particular value is equivalent to caring about the constant factor in run time so big O is inappropriate in this situation.