The Prisoner's Dilemma

Story Time!

Two members of a notorious gang are arrested and imprisoned. Upon capture, they are immediately placed in confinement with no means of communication with one another. The authorities are having a hard time gathering evidence to convict and accuse these criminals of their main charges. They would like to have both of them sentenced to a year in prison on a lesser charge, but at the same time, the authorities offer each prisoner to a possible deal and bargain. Each can either 1) Betray the other by testifying that the other committed the crime, or 2) to cooperate with the other by remaining silent. The offer stands as follows: 

1. If A and B both betray each, they will both serve 2 years in prison. 
2. If A betrays B but B remains silent, A will be set free and B will serve 3 years in prison (and vise versa)
3. If A and B both remain silent, both of them will serve 1 year in prison only for the lesser charge

Where clearly 0 > 1 > 2 > 3 .... right? 

One thing that makes this game so interesting is the Prisoners inability to communicate with each other. If they were able to, then you would assume that they would both remain silent. With each prisoner being allowed to remain silent, the two could both decide to do so and earn the lowest possible jail time for the both of them.

** But where is the fun in that? This game would be wayyyyy too boring if they could talk. **

Oh Shaggy, just take responsibility sometimes! 

So because they are not allowed to communicate, lets start with some scenarios!

From the perspective of Prisoner A, Prisoner B can either cooperate or defect. If:

1. Prisoner B has decided to cooperate. Then Prisoner A has the choice of also cooperating and giving him a one year sentence, or to defect, setting him free. Logically, Prisoner A should choose to defect, as it means he spends the lesser time of the two options in prison. 
2. Prisoner B has decided to betray. Then Prisoner A has the choice of once again cooperating and thus serve three years, or to also defect, giving him 2 years of jail time. Looking at this logically, Prisoner A chooses to defect, which would give him less time in jail. 

Prisoner B faces the exact same situation in both scenarios and would rationally come to the same conclusion, and this is how the Nash Equilibrium (which is the theory that states that ones strategy isn't influenced by knowing what the other player will choose) is used in this dilemma.

After using the Nash Equilibrium, we come to the conclusion that the best route for each Prisoner is to actually defect, which would get them both 2 years in prison. The interesting thing is that although this choice was the most rational option, there is a better possible payoff if they both decided to cooperate. 

Lets Play Again! Lets Play Again!

In most dilemmas, there are many assumptions that seem to take place, which is actually the main reason why the less desirable choice ends up being the most "rational" choice in the Nash Equilibrium. But lets playing this one more time....

The Iterated Prisoners Dilemma  is when the game is played more than once in succession, while of course, remembering the previous moves of the other player. Because they remember the previous moves, they can change their own choices based on that info. 

For this version, we say that 2R > T + S where R is cooperation, and T and S are the outcomes when each prisoner doesn't choose the same option. So, a mutual decision of cooperation always yields a better outcome than an alternation (one choosing defect and one choosing betray). 

Lets say this game is played X amount of times.  Once again, cooperating would be the most logical for both players to choose and its the most fair, and it would allow a binding trust between the two players to decide to cooperate for each game. However, defecting gives that player the best possible option if you have "earned the other players trust" and want to betray him, but will also motivate to not make anymore future "nice" (cooperating) moves. Sadly, there is no strategy to how these multiple games should play out, but analyzing the game from the last "turn/game", we see that the iterated dilemma doesn't differ from the original. 

So consider the very last turn of X. There are no more turns after this round, so you (in theory) want to look out for yourself because the worry of losing the other players trust isn't a factor anymore. Being the evil person we all naturally are, we would choose to defect as there is no retaliation later, and would earn the highest payout reward. Of course, monkey see, monkey do, as this process is copied by the other prisoner. Using this same logic, it would also make the most sense to defect on the second to last turn as well, since the last turn is already set in terms of what move you will do. Trickling this down all the way down to the first turn, where you will once again defect, the strategy is the same for both the iterated and regular versions of the Prisoners Dilemma.

Lets Lock it up! (Get it?!? Does that play-on-words work?Let me know...)

The Prisoners Dilemma is a great practical use of Game Theory and with the use of the Nash Equilibrium, we get to see the strategy that would yield the "safest" outcome for both players. There have been many variations to this game which allow for the players to grab a new twist to the concept, but all of the theories and applications of Game Theory and Nash Equilibrium remain the same. 

Hope you enjoyed!

Jon 

References: 

https://cdn.andertoons.com/img/toons/cartoon4053.png

https://en.wikipedia.org/wiki/Prisoner%27s_dilemma

https://31.media.tumblr.com/90fe57665c665cf9aebbb1f34aeae151/tumblr_inline_n04negmVba1qzo1my.gif