[Thythe]

Scenario 1: 2 players, the nash equilibrium (NE) would be both players bid 500. Also, one player bidding 499 and one bidding 500 would also be a NE since neither would have a rational reason to leave this particular point.

Scenario 2: 3 players - N Players. A variety of (500, 500, 500, 499, 500, 500) etc pairs would exist that are NE. I'm pretty sure that only one player can bid 499 and still have it be NE. So that would mean that the number of NE for any particular game of this type and N players would be N+1 solutions. The overall idea is the same, though, and at best, the winners makes $(1/(N-1)) or $0.

Scenario 3: Non rational players. Anything goes I guess. I would make some guess as to the players non rationality. The less rational they seemed, the lower I would bid. I would suspect that the very least rational might try to bid something like $50, so maybe I would bid $100. In real life, though, I think I would bid around $400 (against 1 opponent), suspecting that they would bid lower than that and not be rational enough to raise their bid.