# Park Place Is Still Worthless: The Game Theory of McDonald’s Monopoly

McDonald’s Monopoly begins again today. 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. Here’s the post, mostly in its original form:

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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, 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 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!)

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Now time for an update. What might not have been clear in the original post is that McDonald’s Monopoly is a simple illustration of a matching problem. Whenever you have a situation with n individuals who need one of m partners, all of the economic benefits go to the partners if m < n. The logic is the same as above. If an individual does not obtain a partner, he receives no profit. This makes him desperate to partner with someone, even if it means drastically dropping his share of the money to be made. But then the underbidding process begins until the m partners are taking all of the revenues for themselves.

In the book, I have a more practical example involving star free agent athletes. For example, there is only one LeBron James. Every team would like to sign him to improve their chances of winning. Yet this ultimately results in the final contract price to be so high that the team doesn’t actually benefit much (or at all) from signing James.

Well, that’s how it would work if professional sports organizations were not scheming to stop this. The NBA in particular has a maximum salary. So even if LeBron James is worth \$50 million per season, he won’t be paid that much. (The exact amount a player can earn is complicated.) This ensures that the team that signs him will benefit from the transaction but takes money away from James.

Non-sports business scheme in similar ways. More than 100 year ago, the De Beers diamond company realized that new mine discoveries would mean that diamond supply would soon outstrip demand. This would kill diamond prices. So De Beers began purchasing tons of mines to intentionally limit production and increase price. Similarly, Apple and Google once had a “no compete” informal agreement to not poach each other’s employees. Without the outside bidder, a superstar computer engineer would not be able to increase his wage to the fair market value. Of course, this is highly illegal. Employees filed a \$9 billion anti-trust lawsuit when they learned of this. The parties eventually settled the suit outside of court for an undisclosed amount.

To sum up, matching is good for those in demand and bad for those in high supply. With that in mind, good luck finding that Boardwalk!

### 14 responses to “Park Place Is Still Worthless: The Game Theory of McDonald’s Monopoly”

• j. pallare

After researched, my dream went down the drain , I got extremely excited because I got one Park Place at Mc D . At El Paso Texas. Anyways my ticket is for sale if anybody likes to purchase it jpallare@cp.Epcc.edu

2. nick janicki

Has Park Place ever been the rare game piece in the McDonalds game? Please tell me it was at least once. Approximately 17-20 years ago I got Boardwalk and just figured it was a common piece and threw it away. I didn’t go to Mcdonalds often and just assumed they changed the rare pieces annually.

3. nick janicki

Thank you. I’ve tried to look if up before, however I can only find information as far back as 2005. I was never going to keep buying food just to find out the answer back then. Today, with that kind of info on line it would be a different story.

4. lstiffler

Excellent article! Thank you.

5. Nuwind

I have a boardwalk and am looking for someone who wants to split 70/30 me acquiring 70% and them the 30%. Since the boardwalk is so rare I see no issue in this this of course would be a pretax arrangement each party would be responsible for their own taxes.

• Cris

I have Park place and I’m good with 30% 🙂

• Shelby

I have one and good with 30 🙂

• Shelby

You can email me at shelbyacampbell@gmail.com I have a park place and 30 is good

• Ryan

hahaha, I’ll do 10%…

6. mrtravel

I will sell you a Park Place for \$10. Credit cards accepted.

• Ryan

I’ll sell mine for \$1.50!!