The Colonel Blotto problem is a zero-sum game about how to best position resources. While Colonel Blotto games are described in a military context, I will explain in future articles some of its useful applications in sports, advertising, elections, and many other areas.

Today I want to highlight a specific Colonel Blotto game that is convenient to solve mathematically.

A simple example of a Colonel Blotto game

Colonel Blotto is planning his attack for the next day. He wishes to capture 2 different locations, and he has 4 regiments of troops.

The opposing commander, Colonel Lotso, also wishes to attack those 2 locations, and he has 3 regiments.

To make the problem precise, it is necessary to ascribe payoffs to possible outcomes.

Let’s say that if both send troops to the same site, the person who sent more troops will win the site.

Specifically, if one Colonel sends x troops, and other sends a lesser number y, then the Colonel who sent x regiments will win. The payout will be equal to y + 1: the Colonel gets y points for defeating that number of enemy troops, plus he gets 1 point for securing the location.

For example, if Colonel Blotto sent all 4 troops to one site, and Colonel Lotso all 3 to the same site, then Colonel Blotto would win the battle and get 4 points because he defeated 3 of Lotso’s regiments and got 1 point for securing the location.

If each sent all of their troops to different sites, however, then both would gain a location but not defeat any enemy troops. Each Colonel would get a location, but neither gains an advantage, so the net payout to each would be 0.

If both Colonels play strategically, how should they distribute their troops?

Solution to the Colonel Blotto game

Colonel Blotto has 5 strategies at his disposal: he can send 4 troops to either location, he can send 3 to one location and 1 to another, or he can send an equal number of troops 2 to each location. We will abbreviate these in set notation (4, 0), (0, 4), (3, 1), (1, 3), and (2,2).

By similar reasoning, Colonel Lotso has 4 strategies at his disposal that can be abbreviated (3, 0), (0, 3), (2, 1), and (1, 2).

We will write out the 5 x 4 matrix for the set of strategies and diligently calculate the payout for each battle. Since this is a zero sum game, we will write the payouts in terms of Colonel Blotto, and note that Colonel Lotso’s matrix would have the opposite values.

(The matrix displays net payouts to each site, so careful accounting is required to calculate some of the payouts. For example, consider the strategy (4,0) versus (2,1). Blotto wins the first site and gets a total of 3 points. But Lotso wins the other site and gets 1. So the net payout to Blotto is 3-1 = 2).

We can use a handy zero gum game solver to figure out the optimal strategy.

Colonel Blotto will play (4,0) with probability 4/9, he plays (0, 4) with probability 4/9, and he will play (2,2) with probability 1/9. He never plays (3, 1) or (1, 3).

Colonel Lotso does not have a unique mixed strategy. He can actually play a variety of mixed strategies. The symmetric solution, as explained in Introduction to game theory, is that he will play both (3,0) and (0,3) with probability 1/18, and he will play both (2,1) and (1,2) with probability 4/9.

Interpreting the solution

When both are playing their optimal strategy, Blotto can expect a payout of 14/9. This makes sense that Blotto has a positive payout because he has 1 extra regiment and should be favored.

The solution indicates that Blotto should concentrate his troops to specific sites and occasionally split his troops, just to make sure that Lotso cannot steal a location easily.

Lotso responds by doing the opposite. Lotso cannot win against Blotto in the numbers game, so Lotso has to spread his troops out and hope to secure an undefended location with 1 regiment. Occasionally Lotso will deploy all his troops to one site or the other, just so that Blotto cannot win by spreading his troops evenly all the time.

Let’s you and I play this very simple game and analyze the best strategy.

Imagine we are playing this game in a college experiment. We each have a chance to win money depending on how we play.

Here are the rules:

–You and I each secretly play “A” or “B”

–If we both pick “A,” then we each get $4

–If one person picks “A,” and the other “B”, the person picking “A” gets $1 and the person playing “B” gets $3

–If we both pick “B,” then we each get no money and leave with $0

Here is the matrix of payouts:

The game is played once. What option would you pick?

Analyzing the game

This game is a no-brainer: it is a dominant strategy to pick “A” and both of us should get $4.

Verifying this is an easy task. Each person thinks about the best response to the other player’s move. If the other player picks “A,” then it’s best to also pick “A” to get $4 rather than “B” to get $3. If the other player picks “B,” it is also better to pick “A” and get $1 rather than “B” to get $0.

The best strategy is to pick “A,” regardless of what the other person is doing. Both players should easily cooperate and get $4.

There is no sensible reason to pick “B.” And yet, that’s exactly what researchers found people doing over half a century ago, in a similar game played with pennies rather than dollars.

The results were astounding: more than 50 percent ended up playing the strategy “B”!

(The experiment is referenced in The Survival Game regarding this 1960 article)

We could be tempted to chalk up the result to the small payouts, or maybe people did not understand the rules. It is possible that people did not take the game seriously.

But the researchers also raised another possible, biological explanation that’s worth investigating.

The green-eyed monster of jealousy

As explained above, both players maximize their payout when they pick “A.” It should be obvious that picking “A” is the best thing to do. Except, perhaps we are thinking about the problem with the wrong motivation.

In game theory, economics, or business, we often choose the option that brings us the highest profit in absolute terms. All things equal, we would rather have $1,000 than $100.

But people do not always think in absolute success. They can sometimes think in terms of relative success: the goal is not to maximize payout, but rather, in the researchers’ words, “to maximize the difference between one’s self and the other player.”

There is further experimental evidence of this idea, as explained on Michael Shermer’s blog:

Would you rather earn $50,000 a year while other people make $25,000, or would you rather earn $100,000 a year while other people get $250,000? Assume for the moment that prices of goods and services will stay the same.

Surprisingly — stunningly, in fact — research shows that the majority of people select the first option; they would rather make twice as much as others even if that meant earning half as much as they could otherwise have. How irrational is that?

…In this case, relative social ranking trumps absolute financial status.[emphasis mine]

The point is that relative thinking, and jealousy of other people’s success, factors into how people think about money.

So let’s use this idea to re-analyze the game.

The destructive nature of relative thinking

The original payout matrix accurately reflected the total payouts for each player. It was apparent that “A” was the obvious choice.

But imagine that people were thinking in terms of relative rather than absolute payout. That is, they only cared about how much more they earned than the other player.

The game is now transformed into the following payouts:

–If we both pick “A,” then we each get $0 more than the other person

–If one person picks “A,” and the other “B”, the person picking “B” gets $2 more than the other player

–If we both pick “B,” then we each get no money and also leave with $0 more than the other person

Here is the matrix of relative payouts:

The payouts of this game are completely changed from the original one. Notice how the original game–which was non-zero sum and mutually profitable–is now suddenly a zero sum, and competitive, game.

The strategy becomes competitive too: in this game, it is a dominant strategy to pick “B,” meaning both players are expected to leave with nothing.

Rather than cooperating for mutual gain, both players “happily” end up with nothing to avoid letting the other person gain in stride.

Get over your money jealousy

This outcome is sadly not just theoretical: people gleefully act towards mutual destruction out of money jealousy.

I went over many such examples in a previous article. In that article, I made a plea for people to be more calm and not worry so much about other people’s success.

Here’s my closing advice from that article. I hope it will inspire people to be less jealous and look for mutual gain:

I have my own personal analogy to get over jealousy. I think about success as filling up water flowing from an ocean. Each of us has a different size glass that represents a personal level of achievement. There’s really no point worrying if your neighbor has a bigger glass than you since there is more than enough water to go around. If you want to get more, then focus on what you can do. Success will come from building your own glass and filling it, not from shattering what your neighbor has. It’s time to put the green eyed monster of jealousy to rest.

One of my favorite topics related to game theory is the subject of auction theory.

The results from auction theory are very interesting, but I have yet to cover much about it because the math can be quite intimidating.

Today’s problem is still challenging, but it should be within the reach of a math enthusiast. Here is the puzzle:

Alice wants to auction off a rare collector’s item. She knows the item is worth somewhere betweeen $500 and $1,000, but she has had trouble finding interested buyers.

A company offers to find interested participants at the rate of $10 per bidder. (So they’ll find one bidder for $10, and ten bidders for $100)

How many bidders should Alice tell the company to find?

A couple of points:

–Assume the bidders have valuations randomly drawn from the uniform distribution on [500,1000]

–Suppose Alice holds an eBay style auction and she will sell the item for a price equal to the second highest valuation of the bidders*

(*this is a standard result in auction theory, though technically it’s for one bid above the second highest valuation. An example: if bidders had valuations of $500, $600, and $700, the person who values the item at $700 would win the auction. The price he would pay in an eBay style auction with dollar bid increments is $601–just enough to outbid the person with the second highest valuation)

The puzzle is about two conflicting forces: Alice wants more bidders to bring her higher bids, but she faces a tradeoff in the cost of acquiring bidders.

Can you figure out the optimal number of bidders?

The answer is written below.

Spoilers below!
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Answer to the puzzle

Alice wants to maximize her expected auction profits. The equation for profits for n bidders is something like this:

Profit(n)= E(revenue n) – Cost(n)

The cost part is easy to figure out. Alice pays $10 per bidder, so her cost is 10n.

The harder part is figuring out the expected revenue for n bidders. What we want to know is the following. If we take n draws from a uniform distribution, what is the expected value of the second-highest draw?

This question is actually part of a larger topic in probability called order statistics. One can explicitly solve for the expected value of any distribution using the technique outlined in this lecture (pdf).

I will not go through the math here. But I will mention the order statistics for the uniform distribution are easy to visualize. What happens is that if you take n draws from the uniform distribution, the expected value of the n draws can be visualized as n points being evenly spaced on the interval.

Here is a picture to illustrate what I mean:

So the n points separate themselves along the interval. So you divide the interval into n+1 segments, and the points will be at the fractions 1/(n + 1) along the way for the minimum, then 2/(n + 1) along the way for the second lowest point, etc., until the maximum draw which has an expected value of n/(n + 1).

By this logic, the second highest draw is expected to be at (n – 1)/(n + 1) along the way from 500 to 1000. This means the second highest valuation is expected to be:

500 + 500 * (n – 1)/(n + 1)

This is our formula for expected revenue. So we can substitute this expression back into the formula for profits:

Profit(n)= E(revenue n) – Cost(n)
Profit(n) = 500 + 500 * (n – 1)/(n + 1) – 10n

Now we need to solve for the profit maximizing point. Let’s just use a shortcut and plug this into WolframAlpha:

We can see that the profit maximizing point is at n = 9 bidders, and Alice can expect $810 of profit.

The lesson is that more bidders is not always optimal: you capture much of the expected revenue from the first few bidders, and then the returns are diminishing (unless some bidder is a big outlier and you can extract money from him).

Extension: suppose Alice earned the highest valuation

As an extension, let’s imagine that Alice somehow was able to extract the highest bidder to pay his entire valuation. This is not an assumption used in theory, but let’s say it happens because of some irrational bidding war.

In that case, Alice would expect to earn slightly revenue (the term (n – 1)/(n + 1) becomes n/(n+1)), meaning her profit function is:

Profit(n) = 500 + 500 * n/(n + 1) – 10n

How will that change the number of bidders?

Here is the answer:

So Alice will only need to acquire 6 bidders, but she will earn nearly $870. This is 3 fewer bidders than above and she gets about $60 more.

This is, of course, exactly what we would expect: if Alice can extract more money from the bidders–the highest valuation instead of the second–she does not need as many bidders and she can earn more out of it.

This is common sense, but it’s useful to check the theory matches intuition.

But today I am thrilled to say there is finally a great game theory introduction that I can recommend. The e-book is called:

Game Theory 101: The Basics & Extensive Form

The book covers the basics of game theory, including the Prisoner’s Dilemma, mixed strategy equilibrium, and it also covers extensive form games (game trees) in which players move in sequence, like the ultimatum game. There are tons of diagrams and lengthy discussions to help you understand the concepts.

One of the remarkable things is how cheap the book is. This ebook which has over 100 pages is selling for a mere $2.99 on Amazon (there is also a lite version for $0.99 called Game Theory 101: The Basics, but I would suggest the $2.99 version as it is more comprehensive and suited for readers of this site).

Very important: while the book says it’s available for Kindle, you don’t need a Kindle to read it. You can read the book on your PC, Mac, iPhone, Android phone, or virtually any device by downloading an appropriate Kindle reading app.

About the author William Spaniel

The book is written by William Spaniel, a name that may be familiar to some of you. William Spaniel has created an impressive series of Youtube videos with the same brand name of “Game Theory 101.” He has also been a long time reader of this blog, and he has contributed with many comments and a great idea for how to find cheap gas using game theory (a bit more on that in the interview below).

I really respect William for writing the book and creating the videos, and even more so, it’s impressive he is doing this while he’s a student pursuing a PhD in political science at the University of Rochester. I got a chance to interview William Spaniel and I picked his brain about why he wrote the book and created the videos. I also got his opinion about why game theory textbooks are often not very good.

I hope you enjoy the interview, and definitely be sure to check out the ebook Game Theory 101.

What got you interested in game theory?

WILLIAM SPANIEL: A paper called Ethnic War as a Commitment Problem. Most people think I am an economist, but I am actually a political scientist by trade. Unlike economics, political science does not have any core tenants that all undergrads at every department must know before they can graduate. I think this leads to a watered-down field with mostly murky theories. But the concept of a commitment problem was refreshingly clear. I figured if game theory could create such focus in the otherwise directionless world of political science, I should learn more about it.

What motivated you to make Youtube videos?

I had played around with YouTube a bit prior to the creation of Game Theory 101. I noticed there were no good comparable options out there. When I started creating lesson plans to teach my roommate the field, I figured I would throw a few short clips on YouTube. It took off past my wildest expectations. Two years later, I have almost 100 videos, a partnership with YouTube, a website (gametheory101.com), and a book series.

Why did you decide to write a book?

My friend Chris pestered me endlessly until I wrote the book. This is the same friend from the car ride where we used the secretary problem to find cheap gas.

I get a lot of emails and comments with strange requests from this blog. What’s an odd/funny request you’ve gotten?

Every December and May/June, I get a few desperate emails from students offering hundreds of dollars to complete their take-home finals for them. I am still unsure how to tell them “no.”

Most game theory textbooks I have read are not very good. Why is that?

There are two reasons. First, every other game theory textbook I have read is written as a reference manual, not a teaching book. Every serious game theorist needs one of these on their bookshelf, but they are completely inaccessible if you do not already understand most of what they discuss. I do not understand why everyone writes this way. Maybe publishers will not publish anything less rigorous. Perhaps professors simply write in the way they were taught. Regardless, that is just how the genre is.

Second, print textbooks have restraints that digital textbooks do not have to worry about. When you take any sort of math class, the first thing that a professor stresses is that math is best learned through repetition. Yet textbooks often show a single example to introduce a new concept! How are you supposed to learn through repetition when you see one example and have 30 practice problems to contend with? But more examples require more paper and ink to publish, which publishers will not allow.

The digital market has no comparable restrictions—megabyte transfers are dirt cheap in comparison to traditional printing and shipping. Thus, I can cram multiple examples of every concept, show picture-by-picture how to best look at a game, and calculate all solutions line-by-line. A traditional publisher would have a heart attack if he saw such a manuscript, yet I can get away with selling a book less than a dollar.

You have a basic book for $0.99 and another one for $2.99. What’s the difference?

The Game Theory 101 textbook is an ongoing project. My $0.99 book is what you might think of as the first chapter. It covers how to solve basic matrix or strategic form games, like the prisoner’s dilemma, matching pennies, and chicken. The $2.99 book is the first two chapters. It includes everything in the first book, plus extensive form games. I recommend anyone who is taking an actual game theory class or who wants a deeper understanding of the field to buy the larger book. If you only have a casual interest in game theory or just have a micro economics test with a splash of game theory, the first one is the best bet.

Why split the book at all? One of the neat things about digital publishing is that you can sell exactly what the consumer needs. The average microeconomics or game theory class covers just a fraction of the material in a normal game theory textbook. Why should you spend money for the part of the book that you will never use?

I am in the process if writing two more chapters. When I finish them, I will package them in a similar way. I understand that it might be a little daunting for the consumer to see so many options, but I think the money they will save makes it worthwhile.

Thanks for the interview. Be sure to check out Game Theory 101: The Basics & Extensive Form, it’s a great read, and stay tuned as William Spaniel releases more chapters.

The article is about three friends who agree to share a cab, and the possible ways they can split the costs.

I highly recommend you read the article.

The thing I liked most is the article describes various fair division methods. As I have described before in my article about splitting restarant bills, fair division is not just a mathematical concept. Fair division depends on social norms and how people perceive fairness.

Therefore, it is useful to understand many methods of fair division and have them in your toolkit. Below I will describe some of the fair division methods mentioned in the article about splitting cab fares.

The details of the cab ride

The situation is a common one: three friends agree to share a cab to different destinations, and they need to split the costs fairly. How can you do that?

More specifically, let us consider the following situation:

Let’s say that passenger A’s usual fare would be $1, passenger B’s is $5 and passenger C’s is $9. If all three share a cab (and assuming A and B are allowed to hop out on the way to C’s destination, without incurring any special fees), the total bill would be $9 — rather than the $15 they’d have to pay, total, to ride alone. How should they divide up the cost of the shared $9 ride? Or, put another way, how do they share the $6 of total savings?

Think about how you might solve this problem if you were splitting a cab with your friends.

Division methods

The author of the article, Carl Bialik, tackled the question by asking several economists how they would solve the problem.

He additionally came up with his own solution of how to divide the trip.

Here is a table that summarizes the different methods. Below I will explain the logic in more detail.

Method 1: Proportional by time in cab

The author of the article took a stab at the problem before asking the economists. He suggested that each person should pay proportionally to the time they spent in the cab. That is, the first leg to passenger A’s house should be shared by all three, then the next leg to passenger B’s house should be divided by B and C, and finally the rest of the way should be paid by passenger C.

The logic here is that you are paying for the time you are in the cab. Under this method, passenger A pays 33 cents, passenger B pays $2.33, and passenger C pays $6.34.

The advantage of this method is it is simple to implement. After a passenger leaves the cab, the fare can be read, and each person immediately knows how much to pay.

The downside is that passenger C ends up saving less percentage-wise. In this example, passenger C only saves 30 percent compared to its normal fare, versus passenger B who saves 53 percent and passenger A who saves 67 percent. It seems a bit unfair that passenger C does not exactly share in the savings. One could argue passenger C should save the most as his trip home is most inconvenienced by having to drop off A and B.

Method 2: split the savings proportionally

Another way to think about the problem is to consider the savings surplus. Had each person gone home separately, the trip would have cost $15 = $1 + $5 + $9. By sharing the cab, the passengers only pay $9, which is a savings of $6.

This method frames the problem as how to split up the $6. The proposal is to split the $6 proportionally, using each person’s individual ride cost out of $15 for the ratio.

Since passenger A would have normally paid $1, passenger A gets a 1/15 share of the $6 savings, or 40 cents. Similarly, passenger B gets a 5/15 share of the $6 savings, or a $2 savings. Finally, passenger C gets the remaining 9/15 share of the $6 savings, or a $3.6 savings.

Ultimately, the ride is split with passenger A paying 60 cents, passenger B paying $3, and passenger C paying $5.40.

Sharing the savings proportionally makes a lot of sense, and this is a preferred method in many legal settings like in bankruptcy situations.

The downside is the answer is a bit harder to compute. The passengers need to know how much the final ride costs before any split can be made.

Method 3: Bargaining solutions

Another way to split the cab ride is by using a procedure known as the Nash bargaining solution. This is a precise mathematical formulation in which each player seeks to maximize his profit from agreeing to a deal, knowing that he can walk away and break the deal.

The amount a person saves depends on how important he is to the deal. Suppose that it is a busy night, and cabs are preferring to pick up groups of 3 passengers rather than individual fares. In that case, each person is vital to catching a cab and equally important.

The Nash bargaining solution, in this case, is that each person gets an equal share of the $6–each gets $2.

This is quite unrealistic as it means passenger C pays $7, passenger B pays $3, and passenger A profits by $1 for his role. The odd solution is because we assumed that 3 passengers were needed in order to hail a cab, and that means each person has equal bargaining power.

A more realistic assumption is that passengers can form subgroups, or coalitions, and continue to bargain. If passenger A is being annoying, then passenger B and C could just split off and take their own cab. In that case, the two would still end up paying $9, which means a saving of $5 plus the satisfaction of not having to travel with passenger A.

This means passengers B and C could argue $5 of the savings belong to them, split as $2.50 per person. They could then come to A and offer the remaining $1 of surplus from all three riding be split equally as 33 cents per person. This leads to passengers A paying 67 cents, passenger B paying $2.17, and passenger C paying $6.18.

Bargaining solutions are extremely important in game theory as they depend on how powerful each player is.

Method 4: Equal division of the contested sum

There is another method that is one known as an equal division of the contested sum.

The idea is you look at which portions are contested, and then you split that evenly. This method has historical importance and I have written about it before: how game theory solved a religious mystery.

I will not repeat the details of the algorithm here. The interesting part in this case is this method produces a exactly the same result of the bargaining solution.

Extension: my method to split cab rides

The problem leaves out an important detail in cab rides. Normally fares are quoted as some base cost, plus an amount that depends on mileage. Additional passengers usually cost extra too.

In the city of Chicago, the cab fares as of this writing are:

• The flag pull or initial charge is $2.25 for the first 1/9 mile.
• The additional fraction of a mile charge is $.20 for each additional 1/9 mile.
• Every 36 seconds of time elapsed is $.20.
• The flat fee for the first additional passenger over 12 and under 65 is $1.00.
• Each additional passenger after first passenger, over 12 and under 65 is .$50.

For three passengers, the initial cost is $2.25 and the two extra passengers add $1.50 to the fare.

From my perspective, the base amount of $3.75 should be borne by all passengers: each person should pay $1.25 minimum to ride. (You can also add in other surcharges if there are any, like tolls or fuel surcharges)

From there, I would split the cab fare proportionally based on time in the cab.

So my method is two steps:

1. Split the initial and extra passenger charges equally
2. Split the remaining charges proportionally

For instance, let’s say three people end up with a $12 fare with tip, and the passengers normally would have had $5, $10 and $15 cab rides.

The first step is to split up the initial and passenger charges. As explained above, that means each person pays $1.25 for a total of $3.75.

The second step is to split up the remaining $8.25 proportionally. Riding separately, the passengers would have racked up $30 in cab fares. Thus, the proportions are 5/30 for the first passenger ($1.38) , 10/30 for the second passenger ($2.75), and 15/30 for the third passenger ($4.12).

Therefore, the passengers pay $2.63, $4, and $5.38. To make it easier, people would probably round that to $3, $4, and $5.

How do you split cab rides?

Personally my cab splits are pretty easy. My friends and I usually take a cab to the same destination, and so we just split everything evenly.

The issue is when you are going to different destinations, as in the problem above. How would you split a cab fare when going to different destinations?