McDonald’s Monopoly is back. As always, if you collect Park Place and Boardwalk, you win a million dollars. I just got a Park Place. That’s worth about $500,000, right?
Actually, as I show in my book on bargaining, it is worth nothing. Not close to nothing, but absolutely, positively nothing.
It helps to know how McDonald’s structures the game. Despite the apparent value of Park Place, McDonald’s floods the market with Park Place pieces, probably to trick naive players into thinking they are close to riches. I do not have an exact number, but I would imagine there are easily tens of thousands of Park Places floating around. However, they only one or two Boardwalks available. (Again, I do not know the exact number, but it is equal to the number of million dollar prizes McDonald’s want to give out.)
Even with that disparity, you might think Park Place maintains some value. Yet, it is easy to show that this intuition is wrong. Imagine you have a Boardwalk piece and you corral two Park Place holders into a room. (This works if you gathered thousands of them as well, but you only need two of them for this to work.) You tell them that you are looking to buy a Park Place piece. Each of them must write their sell price on a piece of paper. You will complete the transaction at the lowest price. For example, if one person wrote $500,000 and the other wrote $400,000, you would buy it from the second at $400,000.
Assume that sell prices are continuous and weakly positive, and that ties are broken by coin flip. How much should you expect to pay?
The answer is $0.
The proof is extremely simple. It is clear that both bidding $0 is a Nash equilibrium. (Check out my textbook or watch my YouTube videos if you do not know what a Nash equilibrium is.) If either Park Place owner deviates to a positive amount, that deviator would lose, since the other guy is bidding 0. So neither player can profitably deviate. Thus, both bidding 0 is a Nash equilibrium.
What if one bid $x greater than or equal to 0 and the other bid $y > x? Then the person bidding y could profitably deviate to any amount between y and x. He still wins the piece, but he pays less for it. Thus, this is a profitable deviation and bids x and y are not an equilibrium.
The final case is when both players bid the same amount z > 0. In expectation, both earn z/2. Regardless of the tiebreaking mechanism, one player must lose at least half the time. That player can profitably deviate to 3z/8 and win outright. This sell price is larger than the expectation.
This exhausts all possibilities. So both bidding $0 is the unique Nash equilibrium. Despite requiring another piece, your Boardwalk is worth a full million dollars.
What is going wrong for the Park Place holders? Supply simply outstrips demand. Any person with a Park Place but no Boardwalk walks away with nothing, which ultimately drives down the price of Park Place down to nothing as well.
Moral of the story: Don’t get excited if you get a Park Place piece.
Note 1: If money is discrete down to the cent, then the winning bid could be $0 or $0.01. (With the right tie breaker, it could also be $0.02.) Either way, this is not good for owners of Park Place.
Note 2: In practice, we might see Park Place sell for some marginally higher value. That is because it is (slightly) costly for a Boardwalk owner to seek out and solicit bids from more Park Place holders. However, Park Place itself is not creating any value here—it’s purely the transaction cost.
Note 3: An enterprising Park Place owner could purchase all other Park Place pieces and destroy them. This would force the Boardwalk controller to split the million dollars. While that is reasonable to do when there are only two individuals like the example, good luck buying all Park Places in reality. (Transaction costs strike again!)