The Prisoner’s Dilemma is considered as one of the greatest situations in game theory, and like many problems it seems very simple yet the math and logic behind it is very interseting. The situation simply goes:
Two partners in crime are arrested. The police put them in separate rooms and offer them the same deal:
If you both betray each other, you both get 2 years.
If you betray your partner and they stay silent, you go free and they get 3 years.
If you both stay silent, you both get only 1 year on a minor charge.
So what do you do? Now you would think that a huge part of this has to do with your relation to the other person and how strong your bond and loyalty is. While this does play a factor if you were entirely focused on a shorter sentence this social aspect is irrelevant.
Let E(silent) be the expected value of prison sentence if you stay silent, and let E(betray) be the expected value of prison sentence if you betray your partner.
Also let P(this is the variable, depending on the other person’s loyalty) be the probability that your partner stays silent, meaning 1-P is the probability that they betray you.
Now let’s split into cases on if you stay silent or if you betray:
Case 1: You stay silent–> if they stay silent as well you get 1 year, if they betray you get 3 years
E(silent)=1*(P)+3*(1-P)=3-2P (if you are confused where this formula is coming from, it is simply just multiplying the outcome by the probability of it happening, this is the most straightforward way of calculating expected value)
Sentence for you: 3-2P years
Case 2: You betray–> if they stay silent you get 0 years, if they also betray you get 2 years
E(betray)=0*(P)+2*(1-P)=2-2P
Sentence for you: 2-2P years
Conclusion: E(silent) is 3-2P years, while E(betray) is 2-2P years. As you can see you will always get 1 year less on your sentence if you betray!
So the moral of the story is to never lie to police(and always betray your friend, just kidding), but like always don’t blame the math!
DON'T BLAME THE MATH 