Conflict in Loons

You probably came up with ideas like these:

1. Fight only if the resource is worth it.
2. Fight only if you already own the resource, to repel an intruder.
3. Fight only if you’re likely to win.

Game Theory

Let’s use game theory to make a model of #1. To keep things simple, we’ll consider just two strategies now, fight and not-fight. As a shorthand, we’ll call the fighting strategy Hawk and the non-fighting strategy Dove. Later, we’ll consider additional strategies.

For our model, we need to figure out the payoff to each strategy when it is used against each of the other strategies. That is, what does a hawk get if it meets another hawk? What does a hawk get if it meets a dove? What does a dove get when paired with either a hawk or a dove?

Then we’ll put these outcomes into a payoff matrix:

Each cell in a payoff matrix shows the average payoff to the strategy on the left when meeting someone playing the strategy on top.

Before we can assign payoffs, we need to decide on the relevant variables. Our hypothesis, “fight only if the resource is worth it,” implies that we can assign a value to the resource and a cost to fighting. To keep things simple, we’ll just have one variable for each of these:

• v is the value of the resource, so a winner gets a payoff of v.
• c is the cost of being injured when losing a fight, so the loser of a fight gets a payoff of −c.

(You may wonder if the winner also pays a price for fighting, since a winner could still be injured, and at least uses up some energy fighting. That’s a good point, and we’ll come back to it later when we add to the model. For now, we’ll assign a cost only to the loser of a fight.)

What are the payoffs in terms of v and c? Enter your choice and then check your answer with the ✓ button before going on to the next one. We start with the simplest payoff and work up to the most complex. As you check each answer, the payoff will be explained just below the question and the correct payoff will appear in the matrix to the right.

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Keep in mind that these are the average payoffs to each player. That’s why the dove-dove payoff is v/2, and that’s why the hawk-hawk payoff is ½v (chance of winning × value of winning) − ½c (chance of losing × cost of losing). The ½ in these payoffs reflects the probability of each outcome (win or loss). Because probabilities determine the payoffs in most games, see the tutorial on Probability in Payoffs before continuing (the link opens in a new window so you won’t lose your place in this tutorial).

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Using the Payoff Matrix

We now have a game-theory model of conflict, but what does it tell us? It’s not obvious from the matrix whether it’s better to fight or not, or whether it might be best to fight in some cases but not in others. For that, we need numbers for v and c.

Let’s try some arbitrary values, say v = 8 and c = 4, so the value of the resource is double the cost of fighting. That gives us:

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With these values of v and c, it’s better to be a hawk. We see this by looking at each possible opponent (columns labeled at the top) and considering each of our options (rows labeled at the left of the matrix). If you meet a hawk, you get 2 if you fight but 0 if you don’t, so it’s better to be a hawk. If you meet a dove, then you get 8 if you fight but 4 if you don’t, so again it’s better to be a hawk. With these values of v and c, it’s always better to be a hawk, and over time, the whole population would end up playing the hawk strategy.

What if we increase the cost of losing a fight, so v = 8 and c = 10?

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Let’s analyze this as before, starting with the Hawk column. If you meet a hawk, you get −1 if you fight but 0 if you don’t, so it’s better to be a dove. However, if you meet a dove, you get 8 if you fight, but only 4 if you don’t, so now it’s better to be a hawk.

If almost everyone you meet is a hawk, you’re better off as a dove, since 0 is a better payoff than −1. But if almost everyone you meet is a dove, you’re better off as a hawk, since 8 is better than 4. Thus it’s not clear which strategy is better overall. Since you don’t know in advance which strategy your opponent will use, it wouldn’t even help if you could switch between strategies.

In this case, whether it is best to fight or concede depends on the proportion of hawks and doves in the population. Recall that game theory is a way of analyzing competitive situations in which the outcome of a strategy for one player depends on what strategy other player uses. This is just that kind of situation.

Let’s go back to the payoff matrix with v = 8 and c = 4, where it is always better to be a hawk:

When we look closely, this seems odd. If everyone is a hawk, everyone gets an average payoff of 2. But if everyone were a dove and peacefully shared resources, they’d get an average payoff of 4, clearly a better result! Something about the situation forces this unfortunate outcome on everyone.

Why can’t everyone be a dove? Suppose the whole population consisted of doves. What would happen if one individual had a mutation that made him play the hawk strategy? That hawk would always meet doves and obtain the hawk-dove payoff of 8, compared to the dove-dove payoff of 4 that everyone else got. With this higher fitness, the hawk mutation would spread in subsequent generations, to the point that eventually the whole population would play the hawk strategy.

Now suppose that the entire population consisted of hawks and a dove mutant appeared. That dove would always meet hawks and get a payoff of 0, compared to the 2 that everyone else got. The dove mutation would die out in subsequent generations, and the population would remain hawks.

This is why the optimal situation, a population of doves, can’t exist. It is unstable, easily invaded by a hawk mutant.

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Mathematical Simulation

Mathematical simulation can help clarify situations like these. The simulation below models the hawk-dove game with the payoff matrix that we just developed. The graph shows how proportions of the two strategies change over time; the area taken by each color indicates the proportion of its strategy (red for Hawk and blue for Dove). You can set the values of v and c with the controls at the right of the graph, and you set the initial relative proportions of Hawk and Dove with the controls below the graph. (You may want to open the game in a window so that you use it without having to scroll back here.)

Try different proportions of Hawk and Dove to satisfy yourself that a population of Doves is unstable and would be taken over by a Hawk mutation. For example, start with an all-Dove population (set the Hawk proportion to 0 and the Dove proportion to 100), and then introduce the Hawk strategy by setting the initial proportion of Hawks to 1.

Next, try changing the values of v and c to explore their effect on the ultimate outcome of the game. Do you see any patterns in the strategy that dominates depending on these values? What happens when v > c? When v < c?

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Evolutionarily Stable Strategy

An evolutionarily stable strategy (ESS) is defined as a strategy that cannot be invaded by another strategy. As we just saw, Dove, even though it would be the overall best strategy, is not evolutionarily stable. If you haven’t yet done so, you should now go through the tutorial on Stable Strategies before continuing (the link opens in a new window so you don’t lose your place in this tutorial).

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You can now calculate the ESS proportion of hawks in the Hawk-Dove game, in terms of resource value (v) and fight cost (c):

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We solve this using basic algebra. First, we need to represent the proportions of hawks and doves. We’ll call the proportion of hawks p, which means that 1−p is the proportion of doves.

What is the total payoff for a hawk? It gets the hawk vs. hawk payoff whenever it meets another hawk, and it gets the hawk vs. dove payoff whenever it meets a dove. Since the probability of meeting a hawk is p and the probability of meeting a dove is 1−p, the total payoff to a hawk is p multiplied by the hawk-hawk payoff plus 1−p multiplied by the hawk-dove payoff:

• Payoff for hawks = p(payoff for hawk vs. hawk) + (1−p)(payoff for hawk vs. dove)
• or
• Payoff for hawks = p(v/2−c/2) + (1−p)(v).

Similarly for doves:

• Payoff for doves = p(payoff for dove vs. hawk) + (1−p)(payoff for dove vs. dove)
• or
• Payoff for doves = p(0) + (1−p)(v/2).

Our goal is to find a value of p that makes the total payoffs for hawks and doves equal, so we set the two total payoffs equal and then solve for p (proportion of hawks) in terms of v and c:

• Payoff for hawks = Payoff for doves
• (p)(v/2−c/2) + (1−p)(v) = p(0) + (1−p)(v/2)
• pv/2 − pc/2 + v − pv = v/2 − pv/2
• pv − pc/2 + v/2 − pv = 0
• v/2 − pc/2 = 0
• v = pc
• p = v/c

So the proportion of hawks should be v/c. When v ≥ c, hawk is a pure ESS, as in the first example, where v = 8 and c = 4. In our second example, where v = 8 and c = 10, we expect an equilibrium with 80% (v/c) playing the hawk strategy. Now go back to the Hawk-Dove simulation and try a few different values of v and c to verify that the proportion of hawks always stabilizes at v/c regardless of the initial proportions.

What does this 80% mean? Does 80% of the population always play the hawk strategy and 20% always play dove? Does each individual act like a hawk 80% of the time and as a dove 20% of the time? It could mean either of these. It could even mean that in each encounter, everyone is 80% aggressive (whatever that might mean). The model just predicts 80% “hawkishness” in the population without specifying how it is achieved.

The ESS calculation in the Hawk-Dove game makes sense intuitively as well as mathematically: The amount of fighting for a resource should increase as the value of the resource increases or the cost of fighting decreases.

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Building on the Model

Now let’s build on the hawk-dove model. Recall that we simplified things initially by having no fight cost to the winner. That’s clearly unrealistic in the case of loons, which use a lot of time and energy in their fights. How would you change the payoff matrix to take into account the cost of fighting to both the winner and the loser?

There are at least two ways to approach this.

First, we could say that the only cost of fighting is the use of energy and exposure to predators while fighting, in which case the cost is equal for the winner and loser. Adding this cost only affects the hawk-hawk payoff, since there’s still no cost when hawks and doves interact, since they don’t fight, nor is there a cost when two doves meet, since they don’t fight either.

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If the winner and loser have the same cost, then the winner gets v−c and the loser gets −c, for an average hawk-hawk payoff of ½(v−c)+½(−c), or v/2−c. That’s effectively the payoff matrix we had before; all it does is double the magnitude of cost. If we substitute 2c for c in the original matrix and in the ESS calculation, we now get an ESS of v/(2c).

Another way to more realistically model the cost of fighting is to say that there is an energetic cost of fighting (c) for both winners and losers plus a special injury penalty for losers. Then we need a new variable, d for damage. Again, this applies only if there’s a fight, so it only affects the hawk-hawk payoff.

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This can be calculated in several different ways. We can say that each hawk wins half the time (v/2), loses half the time −(d/2), and always bears the fight cost (−c). Or we could say that the winner gets v−c (value minus cost of fighting), but the loser gets −c−d (cost of fighting and cost of damage). So the average payoff is ½(v−c)+½(−c−d), or v/2−(2c+d)/2. All of these answers are algebraically equivalent.

What is the ESS of this new game? Calculate it algebraically as before and check your answer below.

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Neither of these ways of adding to the fight cost changes the game much. All they do is change the cost parameter from c to 2c or to (2c+d), so we might as well keep things simple by combining all the costs in a single variable, c. There are other ways we could make costs more realistic and complex, such as adding energetic costs to a hawk frightening a dove, or to a dove scurrying away from a hawk. Neither of these would alter the outcome much (you can verify this by modifying payoffs in the Hawk-Dove simulation). It is usually best to stick with a simpler model whenever possible.

Adding New Strategies

Recall that we chose the most basic strategies, fight or don’t fight, when we started. We considered two other options as well:

2. Only fight if you already own the resource, to repel an intruder.
3. Only fight if you’re more likely to win, if you’re bigger than your opponent.

How would we add strategy #3 to this mix and test it against the two original strategies? First, we need a 3×3 matrix to accommodate the new strategy (fight-if-bigger is usually called Assessor, because it assesses its opponent):

Now there are three strategies: always fight regardless of one’s own size or the size of one’s opponent (Hawk), never fight regardless of size (Dove), or fight if one is bigger and back off if one is smaller (Assessor). Nothing changes about interactions between hawks and doves, so we only need to fill in the new payoffs that involve Assessor.

What are the payoffs for Assessor? We can simplify this by realizing two things. First, we are looking for the average payoff. On average, an assessor is larger than its opponent half the time and smaller half the time. Second, when the assessor is larger, it acts like a hawk, and when smaller, it acts like a dove, and we already know the payoffs for hawk and dove. Keep in mind another assumption implicit in this model, which is that the larger animal always wins. If you get confused when calculating these payoffs, you may want to refer to the Probability in Payoffs tutorial again.

Let’s go through these one-by-one. What are the payoffs in terms of v and c? Enter a choice and then check your answer before going on to the next payoff. As you check your answers, correct payoffs will appear in the matrix to the right.

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How can we determine the ESS in a 3×3 matrix like this one? If we have numeric values, we may be able to find a pure ESS (if it exists) by the methods given in the Stable Strategies tutorial. For example, if v = 8 and c = 10, what is the ESS?

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Let’s analyze this as before. If we just look at each column, we see that Assessor does best against either Hawk or Assessor, while Hawk does best against Dove. So it isn’t clear yet whether there is a pure ESS. However, if we notice that Dove is never the best strategy against any opponent, then we can remove Dove from the matrix. Now when we consider the resulting Hawk-Assessor matrix, it’s clear that Assessor is a pure ESS because Assessor always does better than Hawk. In fact, Assessor is a pure ESS for all values of v and c, since v/2 is always a better payoff than v/2−c/2.

You can verify this with a simulation. Open the Hawk-Dove-Assessor game below and add the Assessor strategy to it, using the above matrix. To add a strategy to the game, click the “Configure Game” button at the bottom-right of the simulator, select “3×3 Symmetric” as the game type, and name the third strategy “Assessor”. Save your changes. Now you can enter the missing payoffs into the matrix and test Assessor against Hawk and Dove with different values of v and c.

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Explaining Loon Fights

Mathematically modeling the evolution of a behavior is not an end in itself, but a way to produce testable hypotheses. Our Hawk-Dove-Assessor model predicts that loons should all play the Assessor strategy, assuming that they are capable of assessing one another. (If they are unable to assess one another, then the Hawk-Dove model predicts that the amount of hawkishness should depend on v/c.)

How well do these models explain the main features of loon fights? From the videos, we know that territories are required for reproduction and thus extremely valuable, while fights are rare but intense and sometimes fatal.

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If all loons play the Assessor strategy, there should theoretically be no fights at all. When an intruder enters a resident’s territory, the two loons should simply assess each other’s size (or other measure of fighting ability) and the lesser one should leave without a fight. Before long, all territories would be held by the largest strongest loons. This might explain why fights are so rare. It also turns out that loons that don’t have territories often fly over occupied territories, which may be a way to assess and challenge a territory owner before deciding to actually fight.

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Even if all loons play the Assessor strategy, there will be times when an intruder and an owner are closely matched in size. When each loon thinks it is larger, both will be agressive and fight. That might explain why, when fights do occur, they are intense and long-lasting. If two combatants are evenly matched, it may take a long time to result in a winner.

While the Hawk-Dove-Assessor game goes a long way toward explaining the loon situation, there is another twist to the story. The next video reveals a surprising feature of loon fights.

When fights are fatal, it is always the territory owner who is killed. Furthermore, fatal fights occur only between males. This raises two points that are hard to explain with the Hawk-Dove-Assessor game. First, why is it always the owner who is killed? Why does he not concede when the intruder is larger? Is the rare fatality merely bad luck, or does an owner take on extra risks, defending his territory beyond the point that it is rational to do so? Secondly, why is it only male, and not female, territory owners that are killed defending their territory? Another bit of data addresses both points.

Now we know that a territory is more valuable to its male owner than to a male intruder. Owners have often held a territory for several years and have found the best nesting site on it. Surprisingly, when a new male takes over, he has to go through the process of finding a good nesting site again, apparently without input from the resident female. This can take several seasons of trial-and-error, during which breeding attempts might fail. Because female owners don’t keep special knowledge about the territory, the value of the territory to a female owner is no greater than it is to a female intruder. That may be why fighting among females is less intense and never fatal.

A final point, not mentioned in these video clips, is that male owners are repeatedly challenged by intruders flying over. As a result, the size and strength of a male territory-owner declines significantly over the course of the breeding season. Even when fights don’t occur, fending off intruders prevents a territory-owning male from feeding as much as he otherwise would. Frequent fly-overs by territory-less males may be a way of assessing whether an owner has deteriorated enough to be worth challenging.

To make a model that takes these facts into account, we would use an asymmetric game. All of the games we’ve looked at so far are symmetric. That means that all players have the same role, the same set of strategies, and the same payoffs for playing each strategy. An asymmetric game, on the other hand, is one in which the players have different roles and thus different payoffs for playing different strategies. To expand our model, we would have different payoff matrices for owners and intruders, with the owner’s payoffs taking into account the greater value that the territory holds for him.

This concludes the tutorial on conflict and symmetric games. You may now try the practice questions. If you wish, you can continue by creating a model for asymmetric conflict in the Conflict II tutorial.

For Cornell BioNB 2210 students: download the Game Theory Quiz here (the same quiz is posted at the end of Conflict I, Conflict II, and Cooperation; we urge you to complete all three tutorials before answering the quiz).