Suppose the repeated prisoners' dilemma is played for 2 rounds. How many pure strategies does each player have?

Question

Suppose the repeated prisoners' dilemma is played for 2 rounds. How many pure strategies does each player have?

The answer $2^4 \times 2 =32$

But I Cannot see this answer. Please help me to explain this answer.

Answer 1

According to the standard game theoretic definition, the answer is in fact $32$.

To see this, it is helpful to recall the standard meaning of a 'strategy'. Informally, a (pure) strategy in a repeated game specifies what action you would take following every possible history that might arise (the document that you linked provides details). The idea is that you might want to condition your actions on past behaviour; so it makes (some!) sense to allow your strategy to depend on the history in this way.

Consider now your example (2 players and 2 actions) and suppose that everyone observes the actions taken in the first round. Clearly,

• Each player has $2$ choices about what to do in the first round;
• As a result, there are $2 \times 2 = 4$ possible histories;
• So a pure strategy needs to specify five things: what do to in the first round, and what to do following each of the possible histories;
• Since there are two choices in each these five circumstances, this means that there are $2^5 = 32$ possible pure strategies.

As an aside, I will note one slightly strange feature of the standard definition (which may have led to the question): it forces one to specify what one would do in situations which one knows for sure will never arise. For instance, even if I choose action A in the first round, the standard definition of strategy insists that I specify what I would do in the second round had I chosen action B! To be honest, I am not entirely sure what strategies are defined in this rather strange way. However, my suspicion is that using this definition was adopted since it allows one to easily convert sequential games into normal form, thereby allowing for application of Nash's (1950) existence theorem.

Answer 2

In my view, the correct answer is $2^3 = 8$.

Each player has 2 options in the first round. In the second round, each player has $2$ options, but their choice is permitted to depend on which choice the other made - thus, there are $2^2$ choices for the second round.

Edit:

The above answer sounds intuitive, but it is actually subtly wrong because it does not permit one to discuss pure strategies as a special case of strategies in general.

A strategy for a game is a function which takes as input all previously-made decisions and outputs a probability distribution for the next decision.

Note that player X must allow previous decisions made by player X to factor in to player X's next decision, because in a strategy, we do not necessarily know in advance what player X's previous decisions would be (as they are probabilistic).

A pure strategy is a special case of a strategy in which all outputs of the strategy function are deterministic.

In other words, a pure strategy has to tell us what we should do if we made choice A in round 1, even if we know that we will always make choice B in round 1.

Thus, to make our first choice, we have 2 options. And to make our second choice, we must make one 2-option choice for each of the 4 possible outcomes of round 1 (even though in fact, two of those outcomes occur with probability 0). This gives the answer of $2 \cdot 2^4 = 32$.