Variant of The Price Is Right

Let there be four contestants in a game similar to "The Price Is Right". They simultaneously write down bids for an object they don't know, the bids can range from 1 to 1000 USD. The object's value will be 75% of the average of the four bids. The bid nearest to the value will win, and it does not matter whether the bid is too high (unlike in The Price Is Right). Is there an optimal play?

I think there is none, because if there was, then all four contestants would bid the same amount, giving them each a chance of 25% to win. However, for every single contestant it would be advantageous to make a different bid, thus this cannot be a Nash equilibrium.

• Is there not a Nash Equilibrium at bids of one dollar? – JP McCarthy Mar 3 '15 at 12:07
• If they all bid the same, it is not advantageous for any ONE person to change their bid because that person will never be closest to the average. – Michael Burr Mar 3 '15 at 12:22
• @MichaelBurr but they have to get closest to 75% of the average... therefore this is not a Nash Equilibrium. E.g. if they all bid €100 then if one contestant drops to €99 then they win as they are closest to €75. – JP McCarthy Mar 3 '15 at 14:18
• You're right. That makes sense. – Michael Burr Mar 3 '15 at 14:19
• So is all four contestants bidding 1 USD the only Nash equilibrium? – user220658 Mar 3 '15 at 14:43

If the goal is to guess a fraction $p$ of the average where $p \in (0,1)$, as we have here, then, as several have commented, the unique Nash equilibrium is (generally -- there are exceptions in some specifications of the game where there is rounding and such) for everybody to bid the minimum bid. This is sometimes called a "race to the bottom."