A bit of game theory

A very fast intro to classic game theory

Games

Historically, game theory developed to study the strategic interactions among rational decision makers (players) who try to maximize their payoffs. Consider the following game of Matching Pennies between two players A and B. Both A and B contemporaneously place a penny on the table. Let H be A’s strategy of playing heads (placing heads up) and T that of playing tails. Similarly, let h be B’s strategy of playing heads and t that of playing tails. If the pennies match, then A wins B’s penny (and keeps his own); if not, B wins A’s penny (and keeps his own). We could represent the game with a tree; however, it is more convenient to use a strategy matrix:

Player B
Player A		h	t
	H	+1;-1	-1;+1
	T	-1;+1	+1;-1

Table 1

Here is how to read the matrix. The top-left column with +1;-1 tells us that when both A plays H and B h, A wins one penny and B loses one; the box immediately below shows that when A plays T and B h, A loses his penny and B wins one. The rest of the matrix is read analogously.

Matching pennies is a zero-sum game in that whatever one player wins must be lost by another player. Since there are only 2 players, one’s wins are the other’s losses: the game is one of pure competition.

Note that in Matching Pennies each player knows:

The payoffs of each player and the strategies available to each player
The fact that each player knows that all the players know this.

A game in which such knowledge is available is a game of complete knowledge. Henceforth, unless otherwise stated, we consider only games of complete knowledge.

Equilibrium

The central notion for studying games is that of equilibrium, namely, a combination of strategies such that each player uses a best strategy, namely one most conducive to maximizing his payoff, given that the other players are trying to do the same. (What counts as a best strategy depends on the type of equilibrium, as we shall see). Solving a game consists in exhibiting an equilibrium. At times, games have more than one equilibrium.

Dominance

The simplest type of equilibrium is dominance equilibrium. Consider the following (table 2) strategy matrix for two persons zero-sum game, where S1 and S2 are A’s strategies and s1 and s2 B’s.

Player B
Player A		s1	s2
	S1	1,1	-2,2
	S2	2,2	0,3

Table 2

Note that A should adopt strategy S2 no matter what strategy B adopts, as in each box S2 has a greater payoff than S1. If by E(Si,sj) we understand the payoff of playing Si against sj, then

Sh strongly dominates Si if and only if E(Sh,sj)>E(Si,sj) for all sj’s.

We indicate this by Sh>Si. So, a strategy Sh has always better payoffs than strategy Si if and only if Sh>Si. In our case, S2 strongly dominates S1, that is, S2>S1. Note that dominance is a relation among the strategies of one player; hence it does makes sense to say (falsely) that S1 dominates S2, or to say that s1 dominates s2 (does it?), but not that, say, S1 dominates s2. When in a game both players have a dominant strategy, then the game has a dominant strategy equilibrium. In the game we are considering, S2 and s2 provide such equilibrium. Notice that an equilibrium need not be fair, as in the game 1,1 or 2,2 are fair outcomes while dominant strategy equilibrium outcome 0,3 is not.

Sometimes games can be simplified by simultaneously eliminating all dominated strategies. For example, consider the game in table 3.

Player B
Player A		s1	s2	s3
	S1	-1;1	2;0	4;-1
	S2	0;3	3;2	5;2
	S3	1;4	4;3	6;1

Table 3

S3> S1 and S3>S2, while s1>s2 and s1>s3. Hence, (S3; s1) is the (unique) solution for this game. Once the two players consider their own strategies and those of their opponents and keep in mind that all the players act only to maximize their own payoffs, (S3;s1) is the only possible outcome.

In the game in table 4, s1 does not dominate s2 and s2 does not dominate s1.

Player B
Player A		s1	s2
	S1	4;-4	2;-2
	S2	3;-3	-4;4

Table 4

However, both players know that S1>S2, and therefore both will reason as follows. As S1>S2, player A will always choose S1 no matter what. Hence, the S2 row can be deleted. But now, s2>s1, and therefore the column for s1 can be eliminated. This leaves only one strategy per player, namely (S1; s2), which is the solution to the game. The solution is reached by dominance iteration, namely by the sequential elimination of dominated strategies. Games that are solvable by the elimination of dominated strategies are dominance solvable. It is a nice feature of dominance solutions that they are unique.

Even when dominance iteration does not lead to only one strategy per player, and therefore to an equilibrium, it is important to apply it whenever possible in order to simplify the game.

It is also possible to simplify games by eliminating weakly dominated strategies, where if a strategy Si has never worse and at least one better payoff than strategy Sj then Si weakly dominates Sj. That is,

Sh weakly dominates Si if and only if E(Sh,sj)≥E(Si,sj) for all sj’s and E(Sh,sj)>E(Si,sj) for some sj’s.

We symbolize the fact that Sh weakly dominates Si with Sh≥Si. The order in which weakly dominated strategies are eliminated may matter to the final solution. For example (table 5),

Player B
Player A		s1	s2	s3
	S1	2;6	1;3	1;6
	S2	0;6	0;3	1;4
	S3	0;6	1;3	0;7

Table 5

eliminating strategies in the order (S3, s3, s2, S2) leads to solution (S1, s1), while following (S2, s2, s1, S3) leads to (S1,s3). So, weak dominance solutions need not be unique. In addition, simplification by iterated weak dominance may lead to elimination of Nash equilibriums (of which more later). However, since a Nash equilibrium that does not use weakly dominated strategies always exists, it is usually reasonable to eliminate them when possible.

Stag Hunt and Prisoners Dilemma

Matching Pennies is a zero sum game: the interests of the players are diametrically opposite. The opposite kind of game is a coordination game, in which an outcome that is considered best by one player is considered best by all the others, as in Stag Hunt. Player A must decide whether to hunt a stag (S) or a hare (H); player B must do the same. Both A and B must pre-commit before they know what the other has decided. The problem is that if one choose S, one will be able to kill a stag only if the other has also chosen S, while if one chooses H, one is assured one will kill a hare. Of course, there’s more meat in a shared stag (4) than in an unshared hare (2), but choosing H is playing it safe. In effect, this is a game of cooperation vs. defection in which cooperation provides higher payoffs but at a higher risk.

Player B
Player A		S	H
	S	4;4	0;2
	H	2;0	2;2

The game is not dominance solvable, but as 4;4 is clearly the best outcome for everyone the real problem is the difficulty in coordinating strategies to get there.

Strategically, the more interesting games are mixed motives games, namely ones that are neither zero-sum nor coordination. The most famous mixed motive game is Prisoners Dilemma. Consider the following story. Two criminals are arrested and the prosecutor has not enough evidence to convict either of a serious crime unless one or both confesses; however, the two criminals do not know this. Hence he tells one of prisoners: “If you confess and the other guy does not, I’ll grant you immunity and you walk free. If the other confesses and you don’t, I shall make sure that you get a heavy sentence. If neither of you confesses I shall settle for misdemeanor charges, with the result that you will pay a small fine and walk free. If both of you confess, I shall charge both with a felony but also argue for shorter sentences than you would get if the other guy squeals and you do not.” Keeping in mind that the game is one of complete knowledge, what should a prisoner do?

Here is the strategy matrix, with S representing “keeping silent” and T “talk”, +10 the utility of walking free, -10 that of a heavy sentence, -6 that for felony charges but with shorter sentence, and +6 that for misdemeanor charges:

Player B
Player A		S	T
	S	+6;+6	-10;+10
	T	+10;-10	-6;-6

Table 6

The game is neither a coordination nor a zero-sum game, but it is dominance solvable: T dominates S; consequently, (T,T) provide a dominance equilibrium. No matter what the other does, it’s better to talk: if you squeal and the other does not, you walk free (+10 payoff); if you squeal and the other does as well, you get a -6 payoff. At all cost you want to avoid keeping silent when the other confesses. Self-interest prevents both from following (S, S) (both keep silent), which would provide a better outcome for the two together. This is why this game is called “Prisoners Dilemma”: apparent private rationality leads to common failure. One can think of Prisoners Dilemma in terms of cooperation (cooperating with the other by keeping silent) and defection (going at it alone by confessing). Notice two things:

1. If the players are self-regarding (only trying to maximize their own payoffs) communication does not solve the problem: even if I know that you will not squeal, it is still in my self-interest to confess.

2. Even finite iteration of the game need not change its outcome. For example, suppose I know that the game will be played 10 times. Then I know that in the tenth round I should confess, independently of what happened in the ninth round. Hence, in the eighth round I should confess because what happens in the ninth does not affect what happens in the tenth, and so on. In other words, backward induction demands that I confess.

Fortunately, most people do not behave as classical game theory suggests; in fact, there’s ample experimental evidence that we tend to cooperate unless we perceive that we are being taken advantage of. In other words, most people do not use backward induction in games. But backward induction may (but it need not!) manage to maximize one’s payoffs only if the other players use backward induction as well. Hence, since most people are conditional cooperators, it often makes sense to cooperate, at least initially, unless the stakes are so high that cooperating against a defector leads to immediate big losses. (Note that in Prisoners Dilemma it may be reasonable for the players to set up an enforcer that compels them to choose cooperation: in some cases limiting one’s options is perfectly rational as it maximizes one’s payoffs).

When Prisoners Dilemma is played an indefinite amount of times between two players, the structure of the reiterated game is different from that of each Prisoners Dilemma round. We shall come back to that later.

The Prisoners Dilemma can provide a rough strategic description of many real life situations. For example,

· a one shot arms “race” between two counties has the same strategic structure: arming dominates over not arming, but of both countries arm they’ll incur severe expenditures they could avoid by not arming

· a one shot tariff confrontation between two countries has the same logic: if A raises its tariffs and B doesn’t, A will improve its trading balance, and if A doesn’t raise tariffs and B does, A will do worse than if both raise tariffs. So, raising tariffs dominates over not raising them. Hence, both countries will raise tariffs, with a decrease in business for both.

Nash equilibrium

Dominance equilibrium in interesting games is rare. However, there are other types of equilibrium. A strategy Si available to A is a best reply to strategy sj if Si’s payoffs are greater than, or at least equal to, those of any other strategy available to A. (B’s strategies are treated in the same way).

A pair of strategies Si and sj form a Nash equilibrium if and only if they are best replies to each other.

In symbols, if by E(Si, sj) we indicate the payoff of playing strategy Si against strategy sj,

E(Si,sj)≥E(U, sj) and E(sj, Si) ≥E(V, Si), for any U, V≠ Si and U,V≠ sj.

When the symbol ≥ is replaced by >, we have a strict Nash equilibrium.

For example, in the following game (table 7)

Player B
Player A		s1	s2	s2
	S1	1,4	0,3	0,0
	S2	-1,0	5,2	1,1
	S3	2,-1	-1,0	1,1

Table 7

there are no dominant strategies. However, S2 is a best reply to s2 and vice versa. That is, if A plays S2, then B’s best reply is s2, and A’s best reply to s2 is S2. Consequently, (S2,s2) constitute a Nash equilibrium. The idea is that player A has no incentive to change strategy S2 as long as B follows s2, and vice versa. Abstractly understood, human conventions are Nash equilibriums; for example, if you drive on the right side of the road, it’s best for me to do the same, and viceversa. Note that

· A dominance equilibrium is a Nash equilibrium as well, but not vice versa.

· A Nash equilibrium need not be fair and need not result in the best payoff for either player. For example, player B would be better off if he followed s1 and A followed S1.

Nash equilibriums are important for two reasons.

· The first has to do with dynamical systems, of which more later, and does not assume the rationality of the players, thus amounting to a significant extension of the theory. For example, imagine fitness strategies competing in a population. Typically, they will outcompete each other until they reach a Nash equilibrium, at which point all of them will be optimal in the sense of being best responses to each other, at least as long as the environmental conditions do not change.

· The second reason involves the assumption of the rationality of the players, a more traditional ground for the theory. If the game is of complete knowledge, the players intend to maximize their payoffs, they are rational and try to predict the moves of their opponents, and all this is common knowledge among them, they can avoid much strife by settling for a Nash equilibrium. Here is why. Imagine a demonic book that told every player which strategy to follow to maximize his payoff given that all the others also maximize theirs. If the book is to be fully authoritative, then it must settle for some Nash equilibrium because otherwise at least one of the players would improve his payoff by changing strategy.

Multiple Nash equilibriums

The existence of Nash equilibriums presents a new problem: many games have two or more Nash equilibriums, and they do not produce the same payoffs. For example, one can see that Stag Hunt has two Nash equilibriums (which are they?) with different payoffs. Another famous example comes from Battle of the Sexes, which goes as follows. Joe and Jill want to go on vacation. Joe can choose to go to the sea (S) or to the mountains (M), as similarly for Jill, who can choose s (sea) or m (mountains). Joe prefers the sea and Jill the mountains. However, they prefer going together rather than going alone (Table 8).

Jill
Joe		s	m
	S	4;1	0;0
	M	0;0	1;4

Table 8

This game has two Nash equilibriums, (S,s) and (M,m), that do not produce the same payoffs, and therefore are not interchangeable, as obviously Joe prefers the former and Jill the latter.

Many situations can be modeled by Battle of the Sexes. For example, two companies A and B want to standardize their products but need to decide what standard to follow, A’s or B’s, or two parties that want to reach an agreement but need to decide on what language to use, and so on. Note that pre-commitment is advantageous in this game: if Joe has already a non-refundable train ticket for the sea, he has an advantage over Jill, another case in which limiting one’s options may be rational. However, Jill might decide to retaliate if she thinks Joe’s pre-commitment amounts to cheating.

How can the players choose among non-interchangeable Nash equilibriums?

In some cases, this can be achieved by appealing to Pareto dominance. An outcome X Pareto dominates an outcome Y just in case for all players X has at least a good a payoff as Y and at least for one player a better one. For example in Stag Hunt (table 9),

B
A		S	H
	S	4;4	0;2
	H	2;0	2;2

Table 9

(S,S) Pareto dominates (H,H), although both are Nash equilibriums. A and B have an identical interest in choosing (S,S) because Stag Hunt is a coordination game. Note, however, that Pareto dominance does not work in Battle of the Sexes, a mixed motives game. Moreover, in a real game (one in which, among other things, players make mistakes) would the Pareto dominant equilibrium be reached if in table 9 the payoffs for (S,H) were (-1000;2), or would A get cold feet, fearing a mistake by B?

In other cases, a common conjecture on how to play the game can be the outcome of cultural practices, human psychological idiosyncrasy, or of having played the game many times before.

Mixed strategies

The games we considered up to now are in pure strategies, meaning that there are no probabilities (different from 1) associated with the strategies. However, a player may decide to play his available pure strategies with different probabilities. If we associate probabilities with the strategies of a player, so that Pr(S1)+Pr(S2)+….+Pr(Sn)=1

(Note the normalization, which tells us that one of the strategies must be used), the game is one of mixed strategies. Pure strategies are then limiting cases, with one strategy played with probability 1 and all the others with probability zero. So, in Matching Pennies, player A might play H with probability 60% and T with probability 40%.

In a game in which A can play S1 and S2, saying that, for example, a mixed strategy amounts to using S1 with probability 40% and S2 with probability 60% can be interpreted in two different ways:

It could mean that A follows S1 40 times out of 100 and S2 60 times out of 100.
It could mean that in 100 games, 40 players in A’s position follow the pure strategy S1 and 60 S2. In other words, instead of referring to the strategy frequencies in the playing of one player, it may refer to the strategy frequencies in a population in which each player plays only one pure strategy.

The second interpretation has more interesting applications than the first, as we shall see.

Determining the EP of a pure strategy against a mixed strategy

Consider Matching Pennies again (table 1), and suppose that player A follows H (playing heads) with probability 1/3 (we shall write Pr(H)=1/3) and T with probability 2/3. We can decide which of B’s pure strategies is a best reply simply by determining the expected payoff of h (heads) and t (tails) with respect to A’s mixed strategy.

Suppose a pure strategy s is played against strategies S1, …,Sn, and that Pr(Si) is the probability that Si is played. Then, the expected payoff of s is:

EP(s)=E(s,S1) Pr(S1) + …..+ E(s,Sn) Pr(Sn).

Or, in more compact notation,

EP(s)=.

That is, if by Si we denote a generic S,

The expected payoff of pure strategy s is the sum of the payoffs of s against each of the Si times the probability that Si is played.

Hence, in Matching Pennies if B plays h, then his expected payoff is

EP(h)= 1/3 (-1) + 2/3 (1) =1/3,

and if he plays t it is

EP(t)= 1/3 (1) + 2/3 (-1) = -1/3.

Now the strategy that has the highest expected payoff is B’s best strategy against A’s mixed strategy. Hence, h (heads) is a best reply against A’s mixed strategy.

Determining Nash equilibriums in mixed strategies

To determine the equilibrium of a game:

Eliminate all the strongly dominated strategies
Equate the expected payoffs of the pure strategies of each player.

The rationale behind (2) is this: I should choose a mixed strategy M against which the pure strategies of my opponent fare the same, which is equivalent to saying that he will have no incentive in changing strategy as long as I keep playing M. Of course, the same reasoning applies to my opponent.

For example, consider the following generic game between A and B from which all the dominated strategies have been eliminated (table 10).

B
A		s1	s2
	S1	a;w	b;x
	S2	c;y	d;z

Table 10

Suppose that Pr(S1)=p and Pr(s1)=q. Then, following (2), we set the expected payoffs of the two strategies at A’s disposal to be equal:

EP(S1)=EP(S2), or

qa+(1-q)b=qc+(1-q).

Similarly,

EP(s1)= EP(s2), or

pw+(1-p)x= py+(1-p)z.

Solving for q and p, one obtains

p=(z-y)/[(z-y)+(w-x)]

and

q=(d-b)/[(d-b)+(a-c)],

which are the probabilities determining the mixed strategies, so that

(pS1, (1-p)S2; qs1, (1-q)s2)

constitute a Nash equilibrium.

For example, in Matching Pennies, which does not have a Nash equilibrium in pure strategies,

p=(-1-1)/[(-1-1)+(-1-1)]=-2/-4=1/2;

similarly,

q=1/2.

Hence,

[(Pr(H)=1/2, PR(T)=1/2); (Pr(h)=1/2, Pr(t)=1/2)]

is the Nash equilibrium for that game. Note that although h (that is, Pr(h)=1) is a best reply it does not produce an equilibrium because if B always plays h, A’s mixed strategy Pr(H)=Pr(T)=1/2

is not a best reply. To have equilibrium, the opposing strategies must be each other best reply.

In Battle of the Sexes, which has two Nash equilibriums in pure strategies,

p=4/5 and q=1/5, so that

[Pr(S)= 80%, Pr(M)= 20%; Pr(s)= 20%, Pr(m)= 80%]

constitutes the Nash equilibrium in mixed strategies.

Expected Payoffs in Nash equilibriums

In Battle of the Sexes, what is Joe’s expected payoff EP in this Nash equilibrium? We need to determine the EP of a mixed strategy. We know how to calculate the EP of a pure strategy (one that is played 100% of the times): the expected payoff of pure strategy s is the sum of the payoffs of s against each of the Si times the probability that Si is played. The extension to mixed strategies is obvious. If a mixed strategy s is constituted by strategies sj, each played with probability Pr(sj), then

the expected payoff of s is the sum of the expected payoffs of each sj times the probability that sj is played.

More compactly:

EP(s)=

So, the EP of Joe’s mixed strategy S (Pr(S)= 80%, Pr(M)= 20%) against Jills’s mixed strategy s (Pr(s)= 20%, Pr(m)= 80%) is:

EP(S)Pr(S) + EP(M)Pr(M)= 4 x (2/10) x Pr(S) + 0 + 0+ 1 x (8/10) x Pr(M)=

4 x (2/10) x (8/10) +0 + 0 + 1 x (8/10 ) x (2/10) =16/25 + 4/25 = 4/5.

Note two things:

1. the EP of a mixed strategy is in fact the average payoff for that strategy

2. since in a Nash equilibrium all the strategies constituting the mixed strategy M of a player pay the same, the EP of M is identical to that of any of its component strategies. For example, EP(S)=EP(M) =4/5, so that the EP of the mixed strategy is: (4/5) Pr(S) + (4/5) Pr(M) = (4/5) Pr(S) + (4/5) (1- Pr(S)) = 4/5.

The EP of Jill’s mixed strategy is the same, as can also be seen by considering that the game is player symmetric in that the game does not change if the two players are switched. However, keep in mind that unless the game is symmetric the expected payoffs of a Nash equilibrium need not be the same, or fair, or even advantageous to all the players.

More on equilibriums

Sometimes, one finds oneself in such a dangerous situation that one should try to minimize one’s losses. A maximin strategy is one that minimizes one’s losses, one, in other words, that will guarantee the maximum of the minima , the best of the bad outcomes. Von Neuman proved the Maximin Existence sTheorem, the first important theorem in game theory:

Every two person finite zero-sum game has at least one maximin equilibrium in mixed strategies.

Note that in a zero sum game, the minimization of a player’s losses directly affects the gains of the other player, as the game is one of pure competition: my gains are your losses and viceversa. Two points are worth noticing:

1. if there is more than one maximin equilibrium, they are interchangeable, providing the same payoffs

2. maximin equilibriums are also Nash equilibriums and viceversa. Hence, finding a Nash equilibrium in a two player zero-sum game is finding the maximin equilibrium.

Keep in mind that (1)-(2) hold only for two-persons finite zero-sum games, and therefore they do not apply to prisoners Dilemma, for example. However, in Matching Pennies if A adopts the mixed strategy [(Pr(H)=1/2, PR(T)=1/2)] he compels B to adopt the best reply [(Pr(h)=1/2, Pr(t)=1/2)], with the result that A’s average payoff will be 0, much better than his worst outcome with only pure strategies, namely, -1. In other words, if A adopts his maximin defensive strategy, B cannot push him below an average payoff of 0. Von Neuman’s existence theorem notwithstanding, often finding the equilibrium can be actualy impossible. For example, as the game of chess is a finite zero-sum game with two players, a mixed strategy exists that guarantees the outcome will be a tie; but as the strategies are complex and in great number, knowing that such a strategy exists is not the same as knowing what it is.

In spite of the great historical importance of maximin equilibriums, Nash equilibriums have broader application because of the Nash Existence Theorem:

Every finite sum game has at least one Nash equilibrium in mixed strategies.

However, such equilibriums (if the game has more than one) are not interchangeable, in contrast to two-players-zero-sum games.

Indefinitely repeated games and the Folk Theorem

Up to now, we have considered one-shot games. However, suppose we repeat a game G an indefinite amount of times. Each round of G is called a “stage” of the repeated game. It turns out that the repeated game has different properties than G; for example, there are a great number of Nash equilibriums of the repeated game that are not Nash equilibriums of G. To understand the import of this we need the notions of discount factor and of signaling.

The discont factor δ is the equivalent in present units of one unit of value to be received one time unit from now.

Hence, when δ=1, the highest value of δ, one values goods to be received one unit of time from now exactly as much as present goods. Hence, when life is uncertain or the future looks grim, the discount factor is low; in general patient agents act on a basis of a high discount factor, impatient ones on the basis of a low discount factor. In monetary terms, when inflation is high, the discount factor is small, and when inflation is low, the discount factor is high.

In determining the expected payoff of a strategy in a repeated game in which defection matters it is important to have reliable signals telling one whether other players have defected or not. A signal is public if all the players receive it (otherwise it’s private), and it is perfect if it correctly reports whether a player has defected or not (otherwise it’s imperfect).

Consider now Prisoners Dilemma with the payoffs listed in table 6. If the players can use mixed strategies, it turns out that any point in the quadrilateral below, where the abscissa represents player 1’s payoff and the ordinate player 2’s, refers to a possible payoff outcome. However, since by defecting each player can guarantee he’ll incur a loss of at most -6, only the points in the quadrilateral ABCD represent strategies with payoffs greater than those resulting from universal defection.

Suppose now that each player is given a list of moves that will result, if both follow their list accurately, in an average payoff of a for player 1 and b for player two, corresponding to point P in the graph. Then, player 1 could set up the following strategy:

“As long as player 2 follows the list, then follow the list as well; however, if 2 deviates then maximally punish him (in this specific case, defect) forever after”.

(This type of strategy is called a “trigger strategy”)

Imagine now that player 2 follows the equivalent strategy. Clearly, the mixed strategies leading to P, constitute a Nash equilibrium (they are best replies to each other), as any deviation from them will result in the application of the trigger strategy which will assure that the deviator (and the other player once the deviator retaliates) will get -6 ever after. Importantly, there are three assumptions at work:

The discount factor is high enough for the players care enough about their future payoffs for the trigger strategy to work.
Defection signals are public
Defection signals are perfect.

Notice three things. First, in any game any player can always play a strategy that inflicts maximum losses to his opponents; similarly any player can follow a maximin strategy, namely, one that minimizes his losses. (Here -6 is the maximin to which the original deviator may be pushed). Hence, the above argument is general. Second, all of this applies to many-players games as well. Third, conditions (2)-(3) are difficult to achieve when many players and the possibility of deception are present. Since P can be any point whatever in ABCD, we have the most basic version of the Folk Theorem:

If (1)-(3) obtain, any point in ABCD (any payoff outcome in ABCD) can be reached by a Nash equilibrium in the repeated game.

So, when the Folk Theorem applies, saying that a certain outcome is the result of a Nash equilibrium is not saying much, as just about any outcome can be the result of a Nash equilibrium. In short, when (1)-(3) apply, Nash equilibriums are plentiful. The theorem can be extended to many cases of public but imperfect information, but we shall not deal with that.