What economists call game theory, psychologists call the theory of social situations, which is an accurate description of what game theory is all about. Basically, it is the process of figuring out the results of an ‘interaction’ between two or more ‘players’. Furthermore, the name ‘Game Theory’ is a bit of a misnomer. The interaction might be a chess game, stock market or even full-out war. What’s common in all these interactions is that they have certain rules and conceivable outcomes.

However, there’s one assumption that Game Theory allows for. It assumes that the participating players are rational decision-makers and do whatever they can to maximize their own payoff. The number of players in a game can theoretically be infinite, but most games are framed in the context of two players. This theory was developed extensively in the 1950s by many scholars. Game theory was later explicitly applied to biology in the 1970s, although similar developments go back at least as far as the 1930s. Since then, it has been brought to wide attention through movies like *A Beautiful Mind,* based on Game Theory pioneer John Nash, and *The Imitation Game*, based on famous mathematician Alan Turing.

However, to understand it better, we need to look at some of the most famous examples of the application of Game Theory.

**The Monty Hall 3-door Problem**

This is a widely popular head-scratcher that gained wide popularity in 1963 through the TV reality show *Let’s Make a Deal*. It was hosted by Monty Hall in the USA, hence the name. The contestant was asked to agree to a proposition that the host made. The proposition was as follows.

- The Game show set is equipped with three identical doors (A, B and C).
- Each door has an ‘object’ on the other side.
- The object might either be a goat or an expensive racing car.
- Behind two of the three doors is a goat, leaving only one door with the car.
- The contestant is asked to choose a door.
- The host then opens one of the two remaining doors. Since he knows the placement of goats/car beforehand, the door that he opens always has a goat behind it.
- The contestant is then asked if he/she wants to ‘stick’ with their choice of the first door or ‘switch’ to the last remaining door.

**Question**: Should the contestant ‘stick’ or ‘switch’?

**Solution**: They should always switch!

Assuming that the contestant actually wants to win the Car, he/she wants to maximize the chances of the winning it. The goat symbolizes a failure according to Game Theory. The contestant has a 1 in 3 chance of winning the car when they choose the first door, since there is no additional information available to them, and the doors are all equally likely to have the car behind them. Let’s assume they choose Door A.

It’s only when the host reveals the ‘failure’ Door B does their chance of winning a car change. Since Doors B and C also had a success probability of 1/3, they had a combined probability of 2/3. When Monty Hall revealed B to be a ‘failure’, the combined probability condenses onto door C. Now, Door C has a 2 in 3 probability of hiding the car! Over a large number of games, the contestant would almost always win if they ‘switched’, even though the choices were similar in the beginning. Monty Hall changed the chances without ever touching the ‘success’ door!

To understand this better, assume that there were 100 doors in the same type of situation. After the contestant chose Door 1, the rest of the 99 doors have a combined 99/100 probability of having a car, which seems like much better odds. Then Monty Hall opens 98 of the remaining 99 doors to reveal goats behind each of them. The 99/100 probability then condenses onto Door ‘100’. It seems pretty evident that the contestant should switch, since Door 100 has a 99/100 chance, while Door 1 has 1/100 chance.

**The Prisoner’s Dilemma**

This situations is mostly used in Game Theory 101 courses to explain how the scientific process is embodied in Game Theory. The setup is as follows.

- Two suspects, A and B, are held in separate interrogation rooms after they are caught committing a minor crime.
- The police want their confession for a major crime.
- Each suspect is given an opportunity to confess about the major crime.
- If Suspect A stays ‘quiet’, he/she goes to prison for 1 year and vice versa.
- If Suspect A ‘defects’ or cuts a deal with the police, while incriminating B, he/she escapes jail time and vice versa.
- If Suspect A stays ‘quiet’, while B makes the deal with the cops to incriminate A, he/she gets a 3-year jail sentence.
- If both of them cut a deal with the police, they both go to prison for 2 years.

This data is represented in table form.

**Question:** How should the suspects act?

**Solution**: If the suspects act in their self-interest, they would end up serving 2 years each. The smartest move is actually to stay quiet, which rarely happens in real-world scenarios.

Firstly, we can observe that the best option for them as a team is to stay quiet, since the combined jail time is 2 years. However, if they incriminate each other, they would escape jail time entirely. Also, since the other person has no idea about the second person’s choice, he/she should always stick to staying quiet, since it leads to the minimum amount of jail time.

The Prisoner’s Dilemma is very similar to the act of stockpiling nuclear warheads by powerful nations. This assures mutual destruction if even one of them ‘defects’, although the best thing to do would be to get rid of their nuclear arsenal entirely.

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