For those who don’t know the game: four contestants bid on a prize of some sort. The contestant who bids closest to the actual value without going over wins. Thus if a pair of skis was worth $348, and the players bid 500, 200, 300, and 350 respectively, the 3rd player would win.

What is the game theoretical optimal strategy for playing such a game? Assume the following:

- All other players will be playing optimally and employing game theory.
- Unlike the actual show, there is no special prize for hitting the number exactly on the head.

My thoughts below; please correct me if I’m wrong:

The key for each player is to divide the chances of winning such that it is optimal for each subsequent player to intrude on a zone other than the one he has already “reserved.” For a four person game, this would be quadrants.

Thus, the first player should bid such that he thinks there is either a 25%, 50%, or 75% chance that the product is priced greater than his bid.

P2 should now bid so as to divide the remaining “zones” into 25% and 50%.

Let’s assume P1 bid 75% and P2 25%. There is now a 50% zone between these two bids. P3 should bisect it exactly and bid such that there is exactly a 50% chance of the product being more than his bid, and a 50% chance of it being less. Again, any other number between P2 and P1’s bids would hurt him to the benefit of P4, because whatever “chunk” of the 50% P3 took or left, P4 could grab the lion’s share by outbidding either P2 or P3 by exactly 1 dollar.

P3 could also bid either $1 or P1’s bid + $1, taking a 25% chance for himself but leaving a 50% hunk for P4 to grab up by bidding P2+$1. The way for all earlier players to avoid such a situation as this is to divide the zones as slightly off from quadrants as possible, taking a little bit smaller zone for themselves in order to assure that subsequent players have a slight incentive to not screw them. In other words, P1 should actually bid 75+x%, where x is an arbitrarily small number, so that there is no chance that a player could play optimally by choosing P1+1 as a bid.

After P3’s bisection, there is now a 25% chance each for P1, P2, and P3 to win, and a 25% chance that they have all overbid. P4 should thus either outbid one of the previous contestants by 1 dollar, or bid 1 dollar. As discussed above, by bidding ever so sightly differently from true quadrisection, each previous player can guarantee that the proper strategy is for P4 to bid $1, rather than outbid any of them by $1.

Each player thus ends up with a 25±x% chance of winning.

Any other bidding strategy is suboptimal.

As an example, lets say that P1 bids such that there is a 60% chance the product costs less than his bid, and a 40% chance that it costs more. P2 should then bid 70%. This leaves a 30% zone above P2’s bid, and a 60% zone below P1’s bid. P3 then bisects the larger zone by bidding 30% (+x), making P4 bid $1, giving P2, P3, and P4 a 30% chance of winning, at the expense of P1, due to his poorly chosen bid.

I have a feeling I’m overlooking some complexities here, but I’m not certain. The logic seems solid, but I’m no game theoretician.