Optimal bidding strategy in "The Price is Right"?

Factual Questions
1

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:

  1. All other players will be playing optimally and employing game theory.
  2. 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.

2

The only strategy I am aware of is: Has the previous bidders bid over or under the price of the item?
If under, then a bid of previous bid + 1, places you above the previous bidder.
If the previous bidder is over then a bid of one dollar is sufficient to win.

3

There is a strategy that I’ve developed in my head (which I’ll use if I ever get on the show) that involves playing off the other players’ bids as much as estimating the price of the prize in question.

The way I see it, the fourth person to bid has a distinct advantage over the other bidders, simply because he or she can bid $1 (if they believe everyone else has bid over) or bid best-previous-bid-plus-one-dollar. Intuition would tell me that the first three contestants have a slight advantage in that they can bid the actual retail price and force the later bidders to bid something other than the ARP. However, since the odds of this are slight I still say that the advantage lies with the last bidder.

For example, let’s say the prize in question is a computer system valued at $1850.

If you’re the first bidder, you have no real option but to take your best guess.

If you’re the second bidder, consider first what Bidder #1 bid. Bidding $1 is no help, because Bidder #3 or #4 could bid $2 and put you out of the game. Also, bidding plus-one-dollar would also be foolish, because later bidders could add one dollar to your bid and put you out of the game.

For reasons that I can’t really justify, my strategy, were I in the second seat, would be to estimate my guess and, if it were higher than the first bidder’s, split the difference (unless the first bidder bid ridiculously low). For example, if the first bidder bid $1400 on the prize, and I thought the ARP was $1700, I’d bid $1550.

The third bidder also couldn’t use the $1 or plus-one-dollar bid, lest the fourth bidder price them out. So the bids on the hypothetical computer system at this point are $1400 and $1550. The #1 bidder now only has a $150 window in which she can win the prize. The idea now, if we’re bidder #3, is to either try to close #2’s window without making a small one for ourselves, or try to go ahead with our own best guess. Again, using my half-the-difference-of-the-nearest bid method, if I thought the ARP was $1700, I’d bid $1625 were I in seat #3. This gives #1 a $150 window and #2 a $75 window in which they could win (if neither of them had gone over).

Bidder #4 now has all the power in the world. If I were bidder #4 at this point, I would merely bid $1 (if I thought they had all gone over) or add $1 to whichever bid I thought was closest. Bidder #4 should never EVER make a bid by guessing the ARP.

Of course, Bidder #4 still has several ways to lose, even without guessing. They could be priced out by another bidder who bid the ARP. They could bid $1 but find out that not all of the other three bidders were over. Or, they could add $1 to the bid they thought was closest but find out that they were over.

I think a cool science experiment, if there are any high-schoolers on the board who want to try it, is to get a bunch of friends to bid on, say, 500 different items. They would have to be taught how to avoid foolish bids (which is to say, bidding $1 or plus-one-dollar when not in seat #4), and seat #4 would always have to bid either $1 or closest-bid-plue-one-dollar, and see how big seat #4’s advantage really is.

4

This, I think, is the key to the matter. The first thing we must assume is that each player values the prize in the same way; that is, P1, P2, P3, and P4 all assume the same expected price for the item V and a similar distribution to this (for them) random variable. Call the Bids each player makes B1, B2, B3, and B4, and assume a distribution p(x), representing the probability that V>=x.

If the game had only 2 players, P1 is in a difficult spot; if his bid B1>V, P1’s best strategy is to bid $1; if B1<V, P2’s best strategy is to bid B1+1, a choice that makes it (nearly) impossible for P1 to win. I think then P1’s best strategy is to bid B1=V, giving P2 a 50-50 shot.

BTW on the actual game, if all contestants overbid the game is replayed; P2 could therefore force a situation where he will always win (unless P1 guessed the exact price) or force a replay by just bidding B2=B1+1. I’m laying this aside, and assuming the game ends with no winner if everyone overbids. Otherwise I think the game is trivial: Each player would simply bid $1 more than the last, and P4 could always force a win or replay.

I think the general plan can be applied to X players such that bids end up evenly spaced with respect to the probability distribution p(x) of V, i.e. the highest bid B’ among x players would have p(B’)=1/x, the next highest B’’ would have p(B’’)=2/x, etc. I’ll have to think about this some more, as I can see problems in extending the 2-player argument…