And I’m an astrophysicist who, mea culpa, originally got the definition of the Nash equilibrium wrong, and therefore thought it actually applied to the TD. Shrug. One can argue that $2 is the Nash equilibrium to the TD (I’d argue it isn’t, but we can run with it). However, unless you manage to come up with a way that players A and B can submit different bids (remember, players A and B are perfect rational players playing a symmetric game), then they must submit the same bid…
The Nash equilibrium argument claims that player A will go: if player B bids $2, then my best bid is $2, so I’ll bid $2 because the same is true for player B. But that is just a statement that all bids like ($4, $2) give player A a smaller payoff than ($2, $2). We know that all bids ($X, $2) for X =/= 2 are forbidden by symmetry. So the Nash equilibrium statement is that: Player A will submit a bid of $2 because his payoff in the ($2,$2) case is better than than in a large set of impossible cases that the system can never see. A rational player doesn’t give weight to impossible scenarios, so a rational player will ignore the Nash equilibrium because it isn’t applicable to symmetric games.