In this variance, however, the value of the prizes are randomly determined within a defined range ($1 to $3000, both inclusive).
Logically, the person in the 4th and final seat will bid $1 over another bid (or just $1) so that they have the best chance of winning. (This tactic is used all the time in the real game, sometimes even by the 2nd or 3rd players.)
Player 2 bids $1800
Player 3 bids $650
So, player 4, knowing the prizes have a range of 1 to 3000:
bidding $1 gives a 649/3000 chance of winning,
bidding $651 gives a 649/3000 chance also,
bidding $1301 gives a 499/3000 chance, and
bidding $1801 gives an 1200/3000 chance.
So player 4 would logically bid 1801.
We'll assume the first 3 contestants bid randomly between $1 and $3000, and the 4th contestant always bids $1 more than another player (or just $1) to give him/herself the best odds of winning.
As a logician, is there an optimal strategy of bidding so that you have the best chance of winning as the first, second, or third contestant?