Nash Equilibrium and Bluffing/Calling Frequencies

Introduction

In this article

• An introduction to game theory
  
• The Nash Equilibrium and its implications
  
• Betting and calling frequencies

Game
theory is a branch of mathematics that analyses certain kinds of conflict
situations or so-called games. In this context, the term 'game' is used for
situations in which multiple participants are competing for resources, with each
participant following a strategy, (possibly with cooperative aspects) and there
is the possibility of profit.

A central concept in game theory is the so-called Nash equilibrium, which
describes a state of strategic equilibrium between players, each player knowing
the best response to the opponent's action and no player being able to increase
their profit by making a one-sided change to their strategy.

This article gives you insight into game theory, Nash equilibriums as solution
strategies, and the application of Nash equilibriums through the example of
betting and calling frequencies. Some fundamental knowledge in the area of
matrix theory is required for the comprehension of this article.

Short Introduction to Game Theory

A game in the mathematical sense is comprised of the following:

• a set of players
• a set containing all of a
  player's (pure) strategies, for each player
• a function that
  associates each strategy profile (that is, each player selects a strategy)
  with a payout tuple, which determines the payout for every player. This
  function determines the outcome of players selecting a certain strategy
  and playing.

The
rest of this article will deal with the games that have two participants - a game can
then very easily be represented by two m x n matrices A and B.
Player 1 then has m strategies (S1,...,Sm) and player 2 has n strategies
(S'1,...,S'n). If player 1 chooses the strategy Si and player 2
chooses S'j, then the payout is Aij for player 1 and Bij for
player 2.

payout player 1:

The players can also use mixed strategies: this means that several pure
strategies are played with certain probabilities, which, of course, add up to
100%. A mixed strategy can be represented by a vector p, with pi
representing the probability of strategy number i being played.

If player 1 plays the mixed strategy p element of Rm and player 2
plays the mixed strategy q element of Rn, then you obtain the
payouts pAq and pBq. (For pure strategies, this leads to the
corresponding entry in the respective matrix.)

For player 1, strategy p is called the 'best response' to player 2 playing strategy q iff p produces maximal payout for player 1, that is, iff the following
is the case:

pAq >= p'Aq for all strategies p' of player 1.

By analogy, for player 2 strategy q is called the best response to player 1 playing
strategy p if the following is the case:

pBq >= pBq' for all strategies q' of player 2.

A pair of strategies (p,q), with p being player 1's strategy and q that of player 2, is called a 'Nash equilibrium (NGG)' iff p
is the best response to q and q is the best response to p.

It can be proven that there is at least one NGG in every game (although not
necessarily in pure strategies).

E
An NEQ doesn't have to be Pareto optimal (an NEQ is pareto optimal if it is impossible
for the NEQ to be changed so that one of the two players receives a larger
payout without the other player receiving less than before.).
On the other hand, a pareto optimal pair of strategies isn't necessarily an NEQ. A
well-known example for such a pair of strategies is the prisoners' dilemma, but we will not go into details on this subject. Other equilibrium concepts
that are interesting but will not be discussed here include evolutionarily
stable strategies and correlated equilibrium.

If both players only have two pure strategies, then there is a very simple
method for determining which strategies would produce the NEQ - the above A=B=I2
example can be used here.

Strategy p of player 1 has the form (a,1-a) with a
element of [0,1], while player 2's strategy has the form (b,1-b)
with b element of [0,1]. Now we will search for player 1's best response
to strategy (b,1-b) of player 2: we can do this by comparing the
payouts of the strategies (1,0) and (0,1) (i.e. the payouts of
pure strategies in each of the two rows).

If player 1 chooses the first row of the matrix, then his payout is b,
and if he chooses the second row, then his payout is 1-b. For b >
1-b, which is equivalent to b > 0.5, the best response is
the first row, in other words a = 1. Correspondingly, for b < 0.5 the
second row is best, in other words a = 0. For b = 0,5 player 1 receives the
payout of 1. Therefore, every strategy is a best response.

We repeat this procedure for player 2. Due to the symmetry in the example we
get b=1 for a > 0.5, b = 0 for a < 0.5 and any b
element of [0,1] for a = 0.5.

NEQs are defined as pairs of strategies, with each strategy being the best
response to the other strategy. It is clear that in the illustration the
intersections of the sets show where the NEQs are.

NEQs of this game are ((1,0),(1,0)), ((0,1),(0,1)) and ((0.5,0.5),(0.5,0.5)).

Betting/Calling Frequencies

Let's
examine the following situation: Player 1 (out of position) and player 2 (in
position) are on the river and the strength of player 2's hand is pretty clear.
Player 1 knows whether he is ahead and player 2 knows that player 1 knows this.

If player 1 can beat player 2's hand, then player 1 will of course make a value
bet (Player 2 will always check behind, as he knows that player 1 knows who is
ahead.). The answer to the question of how often player 1 should bluff of
course depends on the opponent. For example, he of course shouldn't bluff
against a calling station, but should bluff against a weak opponent. For player
two, the decision on whether to call or fold to a bet from player 1 is also dependent on the opponent.

This situation can be modeled as a game:

• Player 1 has the (pure)
  strategies bluff and refrain from bluffing. A mixed strategy
  (a,1-a) would then be a strategy where player 1 will bluff with the
  probability a when betting (and not a strategy where player 1 will bluff
  with the probability a when he has a bad hand. More on this in the
  third section.)

• Player 2 has the (pure)
  strategies call and fold. A mixed strategy (b,1-b)
  would be a strategy where player 2 will call with the probability b
  when player 1 bets.

The best responses to the opponent's strategies are relatively easy to
determine, as will now be shown (let player 1's bet size be x*potsize):

To be able to make a profitable call, player 2 has to be ahead in x*Pot /
(x+2) * Pot = x/x+2 of cases. This means that player 2's best response to
the strategy (a,1-a) is a fold (that is, (0,1) or b=0) if a .

If a>x/x+2, then player 2's best response is a call (i.e. (1,0)
or b=1). If a=x/x+2, then every strategy (b,1-b) with b
element of [0,1] is the best response.

For player 1 to be able to bluff profitably, the bluff needs to be successful
in x*Pot/(x+1)*Pot = x/x+1 of cases, i.e. player 2 must fold in at least
x/x+1 of cases.

This means that player 1's best response to (b,1-b) with b>x/x+1
is to never bluff (i.e. (0,1) or a=0). Player 1's best response
to b is to always bluff (i.e. (1,0) or a=1), and all
strategies are the best response for b=x/x+1.

We obtain precisely one NEQ: ((x/x+2),1-x/x+2),(x/x+1,1-x/x+1)).

Practical Implications of the Nash Equilibrium

Let's
examine the situation from player 2's point of view. Let x = 1 be true,
i.e. player 1 makes a potsize bet. This makes the (0.5,0.5) the NGG. If
player 2 happens to be calling 50% of the time and folding 50% of the time,
then player 2 isn't exploitable in this situation, i.e. player 2's opponent
can't make a decision that will improve his expected value.

Each of his strategies is an equally
good response to our strategy.
However, the strategy (0.5,0.5) is, of course, not always the best
response to player 1's strategy. Ideally, we would always have the best response. If, for example,
the opponent is a rock, then it is quite easy to predict his strategy and chose
the best response. In this situation, you wouldn't even think of calling 50% of
the time against a rock.

However, good opponents will continually change their strategy and try to adapt
to our style, while we will ourselves be trying to adapt to the adjustments the
opponent makes, etc. Ideally, we will always be one step ahead,
thinking one level above our opponent. Then we will
be able to predict the opponent's strategy fairly accurately, i.e. we can judge
whether the opponent is likely to bluff at this point in the game (the term
'game' is used in the every day sense here).

Unfortunately, we will also sometimes be confronted with opponents who are
better at this than us, i.e. who are thinking on a higher level than us.
Against such opponents it's best to play a strategy that isn't exploitable. This
way the opponent can't profit from the fact that he can read us better than we
can read him. If we switched cards with our opponent, then we would win exactly
the same amount that the opponent would have won from us, as we can again
choose the strategy that is in the NGG.

Closing Remarks

a)
If, for example, you have decided to perform a certain action in 50% of cases,
then whether you actually perform it or not each time should be as random as
possible, and not predetermined by an easily detectable system. There are many
things you can use as simple random generator, for example the seconds hand of
your watch or the river card, calling if the value of the card is even. This
kind of random generator may be predictable too if it is too obvious and your
opponents are too attentive.

b) As has already been stated in the second section, player 1 using the strategy (a,1-a) meant that the probability of player 1 bluffing is a in general, and not that the probability of player 1 bluffing is a if he has a bad hand. Therefore, knowing the value a
doesn't immediately tell us how often we should bluff if we are player
1. You need to know how often you would actually have the hand you want
to represent with a bluff.

For example, let's imagine a board of JsTs4 2, that you are obviously
on a draw (straight or flush), and that player 2 has a made hand. If
the river is a 6s, completing possible flush draws, and you hold KQ,
then, to determine how often to bluff, you should consider how often
you would have a flush.

(As9s-As5s, As3s, As2s, 9s7s, 8s7s, 8s6s, 7s6s, 7s5s, 6s5s, 5s3s) is an
example for a realistic range, amounting to 14 different hands. You
will hold a busted straight draw just as often (7 combinations each of
KQ and 98, assuming that you would have played OESD+FD differently
earlier on).

If you now want to bluff with a probability of 1/3, i.e. bluff in one
third of cases, then you need 7 different hands to bluff with (7/(14+7)=1/3). Therefore, if you hold a busted draw, you should bluff in 50% of cases: (7/14 = 1/2).

If you can eliminate 98 in this situation (maybe due to the preflop action), then you have no flush in only 33% of cases, which means that you should bluff every time you don't have a flush.

c) If you are in player 2's position and expect that player1 will bluff
with a frequency that corresponds roughly to the NEQ, then you should
always call if you have a *blocker* in your hand. A blocker is a card
that is part of one of the hands the opponent is representing, thus
making it less likely that the opponent actually has what he is
representing.

If, for instance, you have AsJd in the example from point b), then 7 of
the 14 flush hands are impossible, as they contain As, and are thus
*blocked*. If the opponent uses the method described in b), then he
will calculate an incorrect bluffing frequency for his busted draws
because he will think that he would hold a flush more often than is the
case. This will lead him to bluff in 50% of cases instead of 33%.

The situation is similar if the river is a 9 instead of a 6 and you
hold QQ. If the opponents bet, then they would be representing KQ and
half of all KQ combinations can be eliminated.