While that is reasonable to do when there are only two individuals like the example, good luck buying all Park Places in reality. This would force the Boardwalk controller to split the million dollars. Note 3: An enterprising Park Place owner could purchase all other Park Place pieces and destroy them. However, Park Place itself is not creating any value here-it’s purely the transaction cost. That is because it is (slightly) costly for a Boardwalk owner to seek out and solicit bids from more Park Place holders. Note 2: In practice, we might see Park Place sell for some marginally higher value. (With the right tie breaker, it could also be $0.02.) Either way, this is not good for owners of Park Place. Note 1: If money is discrete down to the cent, then the winning bid could be $0 or $0.01. Moral of the story: Don’t get excited if you get a Park Place piece. 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. What is going wrong for the Park Place holders? Supply simply outstrips demand.
#Mcdonalds monopoly hardest pieces get full
Despite requiring another piece, your Boardwalk is worth a full million dollars. So both bidding $0 is the unique Nash equilibrium. This sell price is larger than the expectation. That player can profitably deviate to 3z/8 and win outright. Regardless of the tiebreaking mechanism, one player must lose at least half the time. The final case is when both players bid the same amount z > 0. Thus, this is a profitable deviation and bids x and y are not an equilibrium.
He still wins the piece, but he pays less for it. 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. Thus, both bidding 0 is a Nash equilibrium. So neither player can profitably deviate. (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. It is clear that both bidding $0 is a Nash equilibrium. For example, if one person wrote $500,000 and the other wrote $400,000, you would buy it from the second at $400,000.Īssume that sell prices are continuous and weakly positive, and that ties are broken by coin flip. You will complete the transaction at the lowest price. Each of them must write their sell price on a piece of paper. (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. Imagine you have a Boardwalk piece and you corral two Park Place holders into a room.
Yet, it is easy to show that this intuition is wrong. (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.)Įven with that disparity, you might think Park Place maintains some value.
However, they only one or two Boardwalks available. I do not have an exact number, but I would imagine there are easily tens of thousands of Park Places floating around. 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. It helps to know how McDonald’s structures the game. Not close to nothing, but absolutely, positively nothing. That’s worth about $500,000, right?Īctually, it is worth nothing. As always, if you collect Park Place and Boardwalk, you win a million dollars. Here’s the post, mostly in its original form:
#Mcdonalds monopoly hardest pieces get update
With that in mind, I thought I would update my explanation of the game theory behind the value of each piece, especially since my new book on bargaining connects the same mechanism to the De Beers diamond monopoly, star free agent athletes, and a shady business deal between Google and Apple.
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