Game Theory 101: The Hawk-Dove Game

[@bradsmith6019]

This video makes sense until 6:08, at which point new things are introduced with zero explanation.

[@directordissy2858]

For future watchers - see his video on the mixed strategy algorithm. He covers this earlier in the series and the math he starts at 6:08 is straightforward by this point in the series.

[@ValentineRogue]

​@@directordissy2858 A quick summary of the math: In any two-player-two-option game, to find the nash equilibrium (pure- or mixed- strategy), we set the expected utility of the opponent (EU_2, what player 2 will get out of the game) for each option (EU_2(H) and EU_2(D)) equal to each other and solve for the proportion of options player 1 should select when playing the game multiple times (sigma_1(H) and sigma_1(D), where, since there only only 2 options, sigma_1(H)=1-sigma_1(D)). So from player 2's perspective, then replacing sigma_1(D) with 1-sigma_1(H) EU_2(H)=sigma_1(H)*(V/2-C)+sigma_1(D)*(V)=sigma_1(H)*(V/2-C)+(1-sigma_1(H))*(V) and EU_2(D)=sigma_1(H)*(0)+sigma_1(D)*(V/2)=sigma_1(H)*(0)+(1-sigma_1(H))*(V/2) The objective of player 1 is to make player 2's actions predictable, so they should act in a way such that, from player 2's perspective, EU_2(H)=EU_2(D), so that player 2 have no incentive to pick one over the other. This means that player 1 can reasonably expect that player 2 will have sigma_2(H)=sigma2(D)=0.5, and act accordingly. Because neither sigma_2 is equal to 1, this is a mixed-strategy nash equilibrium. If player 1 instead somehow knows ahead of time what player 2 will choose, sigma_2(H)=1 and sigma_2(D)=0 or vice versa, resulting in the pure-strategy nash equilibrium. So there are three cases for the exogenous variable: V/2-C>0, V/2-C=0, and V/2-C0 and V/2-C=0, player 1 maximizes their expected utility by always playing hawk, because sigma_1(H)*(V/2-C)+(1-sigma_1(H))*(V)=sigma_1(H)*(0)+(1-sigma_1(H))*(V/2) always results in sigma_1(H)=V/2C, and for V/2-C>0 and V/2-C=0, V/2C is greater than or equal to 1. Player 1 can assume player 2 has come to the same conclusion because the game is symmetrical, so sigma_1(H)=sigma_2(H)=1, resulting no mixed-strategy nash equilibria, only the pure-strategy nash-equilibrium of (Hawk, Hawk) In case V/2-C

[@anatyy2]

Thank you you helped me study for my uni exam!

[@ehatipo4598]

did you graduate?

[@lubababinthhalim3840]

Can the cost of conflict be divided among the players so that when they both play Hawk strategy, they both get a payoff of (v-c)/2. When you write (v/2)-c that means each incurs a different cost of conflict? Or the cost is not divided but each incurs the cost fully?

[@Gametheory101]

@Dexmonicus ...which is technically covered by this game, when V = C!

[@akawumpus]

I guess this is also an important game in Evolutionary biology. Don't understand why yet, but thanks for the quick explanation.

[@AnthonyLopez-di5vv]

because the payoffs can represent fitness, which can help explain evolvability of a phenotype/strategy in a population. have a look at Maynard Smith's classic "Evolution and the Theory of Games"

[@alexanderrahl7034]

It's also an example of aggression and agreeableness. Animals in the wild rarely fight to the death, or even to severe injury, because any injury can be life threatening. The hawks in this case, are aggressive, they will beat the dove every time. In this way, the hawk's aggression beats the Dove's so the dove leaves. Hawk gets +1, and both live. In this way, it's preferable to be the hawk, the Predator, over the prey. However, in the event of 2 hawks, the fittest survives. So it's best to be the strongest hawk overall. Extrapolated to humans, we are the apex Predators, and we have our own internal hawk and dove dynamic within our species.

[@belleviewable]

yes I would also like to know the answer please!

[@Pallerim]

Did you say 'peer' or 'pure' strategy @ 4:17 ? Since you've only mentioned 'pure strategy' before i guess that's it, but i'm still in the learning process, so i don't know if there is such a thing as 'peer strategy' :)

[@ValentineRogue]

This is 11 years late, but he says pure-strategy nash equilibrium. Pure because the sigma for one of the options for one of the players is 1, whereas in a mixed-strategy bash equilibrium, none of the players have any sigma of 1.

[@Gametheory101]

@Pallerim Pure.

You were saying Heads and Tails instead of Hawk/Dove.. haha.. it was on purpose? But thank you!

[@kiemdoan]

What about V/2-C is 0? Would there not be weak dominance?

[@cremeog7362]

5 years late but I'm pretty sure that the population doves would go to zero over how ever many iterations dependent on the base fitness.

[@ApplepieFTW]

probs for coming back to this comment FIVE YEARS LATER DAMN

[@ValentineRogue]

The V/2-C=0 case is similar to the V/2-C>0 case, because EU_2(H)=EU_2(D) always results in sigma_1(H)=V/2C, which indicates there are no mixed-strategy nash equilibria. The pure-strategy nash equilibria are determined by setting each sigma_2=1. For V/2-C>0 and sigma_2(H)=1 -> sigma_2(D)=0, EU_1(H)=V/2-C>0 and EU_1(D)=0, so sigma_1(H)=1 and sigma_1(D)=0, and sigma_2(D)=1 -> sigma_2(H)=0, EU_1(H)=V>0 and EU_1(D)=V/2>0, so sigma_1(H)=1 and sigma_1(D)=0. Then the only pure-strategy nash equilibrium for V/2-C>0 is (Hawk, Hawk), because if player 1 should always pick Hawk, so Player 2 should also always pick Hawk to maximize their EU. For V/2-C=0 and sigma_2(H)=1 -> sigma_2(D)=0, EU_1(H)=V/2-C=0 and EU_1(D)=0, but then EU_1(H)=EU_1(D)=0, and sigma_2(D)=1 -> sigma_2(H)=0, EU_1(H)=V>0 and EU_1(D)=V/2>0, so sigma_1(H)=1 and sigma_1(D)=0. Since the game is symmetrical, player 1 and player 2 come to the same conclusion that if the other player plays only Hawk they get nothing, but if the other player plays Dove they should play Hawk. This means the nash equilibrium is based on whether each player has faith the other player won't screw. We have to assume bad faith, so the nash equilibrium is still (Hawk, Hawk). In some strange universe where we assume good faith, the pure-strategy nash-equilibrium becomes (Dove, Dove), but either player betraying that good faith even once to play Hawk and get the EU of V instead of V/2 produces bad faith, resulting in the bad faith nash equilibrium of (Hawk, Hawk). Ultimately the EU of both players is 0, so the game has no utility, so the real solution is to not play (or to just play for fun).

[@hadikhaled1410]

Hello , i have a question regarding this topic , i am willing to have a mathematical analysis including this topic . i am an Ib student doing his math Standard level exploration , and i found this interesting. But i need help to know if the aim of my topic is suitable and i would like to inquire about some calculations including this. Firstly, do you think that exploring which strategies are more successful to win a hawk and dove, in addition can we determine whether there is an equillibrium or not . Lastly what kind of mathematical tests can i use on game theory. Thank you

[@AkifIsmail-r2i]

The Hawk Tuah game

[@dompatrick8114]

thanks G

[@stillamitchinmybook6320]

Huh? Lol no but really this is cool

[@AkifIsmail-r2i]

The Hawk Tuah Game

[@LissusMurataj]

was looking for this comment