The two player one is trivial, but more players is still debated.
For any # of players of equal ability, your chance of first is equal to your fraction of the total chips. So in your example, one player has 1/3 chance of winning. His "fair" share of the prize money is thus 2nd+1/3*(1st-2nd).
For more players, I have a scheme that is correct for fairly reasonable assumptions (
Computing tournament finish place probability) . Others have made various approximations to get reasonably close (to these) results.