Here's my attempt at solving the Blast/Naaru metagame. TL;DR: Given those two choices, the equilibrium is to run Mind Blast at least 50% of the time (including 100%), and Light of the Naaru the rest of the time; when running Mind Blast as P1, always assume your opponent is running Mind Blast too.

To start the analysis, I think there are only three potentially-viable strategies: run Naaru, or run Blast and, as P1, either hold it until fatigue or play it on turn 2. Although I haven't quite proven it, intuitively I don't think there's any point as P1 to skip playing Mind Blast on turn 2 but not hold it all the way to fatigue; you already give up your certain win in the mirror by passing turn 2, so you might as well go all-in and play as if P2 is running Naaru.

Call these three strategies Bp (Blast-play), Bh (Blast-hold), N (Naaru). Each strategy has a 50% winrate against itself (only depending on which seat you draw). Bp wins 100% vs Bh and 50% vs N. Bh wins 100% vs N. Usual 2-player game theory can then solve for all equilibriums via linear programming, using this payoff matrix:

` Bp Bh N`

Bp 0.5 1.0 0.5

Bh 0.0 0.5 1.0

N 0.5 0.0 0.5

It's not too hard to reason out the equilibrium without LP, using the fact that the optimal winrate must be exactly 50%, because of symmetry. Bp wins >=50% vs everything and 100% vs Bh, so Bh can't be played or always playing Bp would get better than 50% winrate. But if Bh is not being played, both Bp and N have 50% winrate not matter what proportion they appear. However N must be played no more than 50% of the time or else Bh would have >50% winrate vs the field.