Bluffs from a game theory perspective

Introduction

In this article

• The mathematical background
  
• Optimal strategies
  
• The Nash equilibrium
  

Bluffs are an essential part of poker play and every good player should have them in their repertoire. Players make a huge mistake if they never bluff, but also if they bluff too much. How do you find the right balance? How can you tell when it's a good idea to bluff, how do you know when and how often you should bluff against certain players in order to play profitably against them?

This article deals with the mathematical background of bluffs and uses a game theoretic approach to illustrate possible strategies.

Should I bluff?

You are on the river. The board is .

You are pretty sure that your opponent has a made hand. From his actions you think he probably has AA, KK or AK. You hold .

A busted draw. The pot is $100, and you both have $100 each. You act first. If you check, you lose. (Either your opponent checks behind you and wins at the showdown, or he bets, and you don`t have enough money to try to bluff him then.)

Let us suppose that your opponent has read you correctly, and before the river card he put you on a draw. For the sake of simplicity, suppose also that he thinks that the probability of a busted flush (or gutshot) draw is 80%, and he gives 20% to the possibility that you hold 87 and with the river card you made your straight.

If you had a straight with 87, you would bet for value. But unfortunately this time you missed. You have two options: you give up and check, or try to buy the pot with a bluff. Let us suppose that when you bluff, you go all-in with your $100, which is a pot sized bet. How often should you bluff?

How can we answer this question?

For this we need some mathematics. But don`t worry, it won`t be too complicated. Just a little algebra from elementary school, and some common sense. Oops, almost forgot – and game theory, but that will fit in the `common sense` part.

Translating the question to mathematics

We`ll need some notations. P will stand for the pot size, while B is your bet size. Let`s denote the probability (seen by your opponent) that you have the winning hand by q. So in our example P = $100, B = $100 and q = 0.20. If you don`t like our assumption of 20%, you can change this number, the calculation remains the same, just the result changes. Similarly, you could change the pot size or the size of your bet too. So our notations will also generalise the problem.

We need to denote somehow our bluffing frequency, and our opponent`s calling frequency when we bet. Let the first be x, the latter y. So x means the probability of our bluffing. For example, if we bluff with 30% probability (x = 0.3), our bluffing frequency will be just 30% in the long run. Similarly, y means the probability of the event that our opponent calls our bet.

So our original question just asks what the optimal value of x is.

The expected payoff and pure strategies

To start with, let`s go back to the very beginning. What is our goal in poker? To win money. More precisely, to win as much money as possible. Whenever we make a decision, we ask ourselves(or at least we should ask), which action has the largest expected payoff.

In our example, if you had the winning hand, you would bet for value. When you bet, your opponent either calls your bet (y part of the time), or folds his hand (1 – y part of the time). When he calls, you win the current pot, plus the money with which your opponent calls your bet: P + B. When he folds, you just win the current pot: P. (`Current pot` always refers to the original pot size, before you bet.) Thus, if you had the winning hand, your expected payoff Ew (w stands for `winning`) would be

Ew = y(P + B) + (1 – y)P.

For the losing hand (your busted draw), things are getting a bit more complicated. Here you either bluff (x part of the time), or give up the hand and check (1 – x part of the time).

If you bluff, your opponent either calls your bet (y part of the time), or folds his hand (1 – y part of the time). When he calls, you lose your bet, so your net payoff is negative: – B. When he folds, you win the current pot: P. So when you bluff, your expected payoff is given by these two cases, and it is

(1 – y)P – yB.

If you check (just give up), you don`t win anything, so in this case your expected payoff is zero. With these your expected payoff El (l stands for `losing`) with your losing hand is

El = (1 – x)0 + x[(1 – y)P – yB]

(just combine the above cases with the corresponding probabilities).

Of course the first part is zero, and we can omit it, giving us

El = x[(1 – y)P – yB].

If you know that your opponent never calls (y = 0) – e.g. he is disconnected – your expected payoff would simply be:

El, y=0 = xP

To maximize your expected payoff you have to choose x = 1, which means you should always bluff.

On the other hand, if you know that your opponent always calls (y = 1), your expected payoff becomes

El, y=1 = – xB

To maximise this now you have to choose x = 0, which means you should never bluff. (Just remember the mantra `Never bluff a calling station!`)

In the language of game theory, we just calculated your best response, knowing your opponent`s strategy in two special cases. But these were the extremes, the so-called pure strategies. In real life your opponent will be more unpredictable. He will call your bet with a certain probability (not 0 and not 1). In game theory we would say he has a mixed strategy.

Optimal strategies

It is interesting that your opponent can choose a calling frequency y, when your expected payoff is the same whatever strategy (that is, whatever x) you chose. Let`s denote this calling frequency by yopt (as in a sense it is the optimal value of y).

It is quite easy to compute the value of yopt. If you are interested in the details, you find them in the appendix. Here is the final result

yopt = P/(P + B).

In our example, P = B = $100, so yopt = 1/2. If your opponent calls exactly half of the time, you can`t outsmart him. When your opponent plays according to the strategy y = yopt, your expected payoff will be

El,y=yopt = x[PB/(P + B) – PB/(P + B)] = 0.

(Just put what we got for yopt into the general equation we obtained for El). As x doesn`t appear in the equation (actually nothing appears, just the zero), whatever strategy (whatever x) you choose, you can`t increase or decrease your expected payoff.

It is interesting that yopt depends only on the size of the pot and the size of the bet. It is independent of q, for example. This shows that yopt is not the optimum of y in a universal sense. If for example q = 1, i.e. your opponent is sure that you have the winning hand, he won`t call you half of the time – in fact, he won`t call you at all. Then he would choose the strategy corresponding to y = 0. We will see later, in what sense yopt is optimal.

Similarly, you can chose x in such a way, that whatever strategy (whatever y) your opponent chooses, his expected payoff will be the same. As before, we will denote this special value of x with xopt. However, it is a bit more tricky to get the value of xopt. Omitting the details, finally we get

xopt = qB/[(1 – q)(P + B)].

(If you are interested in the details, to compute the above expression first you have to obtain {see "the expected payoff seen by your opponent" in the appendix}. With that, {see "obtaining the value of xopt" in the appendix} is easy.) If you bluff this often with your losing hand, {see "the expected payoff seen by your opponent" in the appendix} will be

Eop = (1 – q)P – qPB/(P + B).

As there is no y in this formula, he simply cannot change his expected payoff.

In our example P = B = $100 and q = 0.2, so xopt = 1/8. If you bluff with a 1/8 probability, your opponent can`t outsmart you, even if he is very observant and knows your strategy (that is, knows that x = xopt). If you bluff more often or less often, your observant opponent can exploit your strategy by finding the corresponding best response. So if you are facing a very good opponent, xopt ensures the optimal strategy for you.

How often will such a very good opponent call your bet? That is given by yopt. If you are playing according to the strategy where x = xopt, he could actually choose whatever strategy he likes. We saw that he cannot increase (or decrease) his expected payoff. But if he doesn`t use the strategy where y = yopt, you, as an observant player, could exploit his mistake, choosing optimal response to his given strategy. The only case when you can`t exploit his strategy is when he chooses y = yopt – now whatever strategy you play, your expected payoff will remain the same. But be aware, if you choose other strategy than xopt, your opponent can exploit your play by adjusting his strategy accordingly.

So now we see in what sense xopt and yopt are optimal: they provide unexploitable strategies. In game theory such a pair of strategies (xopt, yopt ) are called Nash Equilibrium. This is a very important concept in game theory and economics (yes, he is the same Nash as the hero of the movie `A beautiful mind`, winner of the 1994 Noble Prize in Economics!). Now we see that it plays a central role in the theory of poker too.

Expected payoff in some special cases

Let`s finish with two charts showing your expected payoffs in some special cases. The first chart gives your payoff for your busted draw (losing hand), calculated for the given example:

Your opponent`s expected payoff is just the negative of yours, plus the $100 already in the pot. (He gets your net loss in the betting round, plus the current pot. Of course when you win the pot, your net loss is – $100 and he gets nothing. Or take the simplest case: when your expected payoff is zero, your stack remains intact – the same as at the moment when the river card arrived. Thus the pot can`t be yours, but someone has to get it – so it will be your opponent`s. In the language of game theory we could say that our example game – because we are already at the river – is not a zero-sum game.)

While you know when you are bluffing and when betting for value, your opponent doesn`t. So for him (and probably for you too), the following chart is more instructive. This gives your expected payoff for the mixture of your winning and losing hands. In similar situations you`ll be ahead 20% of the time, and behind 80% of the time. Thus your overage expected payoff is qEw + (1 – q)El. (Again, the negative of this plus the $100 in the current pot gives your opponent`s expected payoff, Eop.)

Conclusion

When you are facing a good opponent, your best choice is to use the strategy given by the Nash equilibrium: xopt. In this case your opponent will play the strategy given by yopt. If not, he is making a mistake (so he is not that good a player), and you can exploit his mistake by finding the best response to his strategy. If he tends to call you too often, bluff less. If he calls too rarely, bluff more. If you can guess his actual calling frequency, you can compute how much less (or more) should you bluff by maximising your expected payoff, just as we did above.

Appendix

If y = yopt, your expected payoff will be the same for all values of x. Let`s take x = 0 first. Here you can`t win anything, which is reflected in the formula:

El, x=0 = 0.

Now take x = 1. Than our equation for El gives

El, x=1 = (1 – yopt)P – yoptB.

As El, x=0 has to be the same as El, x=1, we get

(1 – yopt)P – yoptB = 0,

which gives us

(1 – yopt)P = yoptB,
P – yoptP = yoptB,
P = yopt(P + B),

and finally

yopt = P/(P + B).

{the expected payoff seen by your opponent}, {your opponent`s expected payoff}:

Now let`s look at the hand through the eyes of your opponent. First we write up his expected payoff, Eop. As – unlike you – he doesn`t know whether you are ahead or behind, his expected payoff will depend also on q, so it is a bit more complicated:

Eop = – qyB + q(1 – y)0 + (1 – q)[xy(P + B) + x(1 – y)0 + (1 – x)P].

The first term corresponds to the case when you have the winning hand, you bet and he calls, losing his bet. The second term stands for the case when you have the winning hand and he folds, winning (and losing) nothing. The rest corresponds to the cases when he is ahead. The first part in the square brackets stands for the case when you bluff and he calls, winning the current pot plus your bet. The middle part corresponds to the case when you bluff and he folds, winning/losing nothing, while the last term stands for the case when you give up and he wins the pot (either by checking after you, or betting, forcing you to fold).

Omitting the terms that are zero we get

Eop = (1 – q)[xy(P + B) + (1 – x)P] – qyB.

If your opponent knew that you never bluff (x = 0), what would be his best response? He would never call you of course. This can be seen from the above equation, if we write 0 in place of x.

Eop x=0 = (1 – q)P– qyB.

To maximise this, we have to choose y = 0 (never call).

On the other hand, if your opponent knew that you always bluff (x = 1), the best response is not so obvious. For x = 1 we get

Eop x=1 = (1 – q)y(P + B) – qyB = y[(1 – q)(P + B) – qB].

If

(1 – q)(P + B) – qB > 0,

y = 1 (always call) will maximize you opponents expected payoff.

If

(1 – q)(P + B) – qB < 0,

he should choose y = 0 (never call) instead.

(1 – q)(P + B) – qB < 0

means

(1 – q)(P + B) < qB,
P + B – qP – qB < qB,
P + B < q(P + 2B),

finally giving

q > (P + B)/(P + 2B).

In our example, when P = B = $100, if q > 2/3, your opponent should never call your bet (even if he knows you always bet; and therefore in such a case you should always bluff), while if q < 2/3, he should always call (if he knows that you always bluff). Note that this special value of q also only depends on the pot size and the bet size.

If x = xopt, your opponent`s expected payoff will be the same for all values of y. As before, let`s take first y = 0. Our equation for Eop gives

Eop y=0 = (1 – q)(1 – xopt)P.

Now take y = 1. This time we get

Eop y=1 = (1 – q)[ xopt (P + B) + (1 – xopt)P] – qB.

As Eop y=0 has to be the same as Eop y=1, we get

(1 – q)(1 – xopt)P = (1 – q)[ xopt (P + B) + (1 – xopt)P] – qB,

which gives us

qB = (1 – q) xopt (P + B)

((1 – q)(1 – xopt)P appears on both sides, hence cancels), so finally we get

xopt = qB/[(1 – q)(P + B)].

1 When we work with probabilities, we usually write 0.2 instead of 20%, 0.5 instead of 50%, etc. In this setting the probability of an impossible event is 0 (0%), while a `sure thing` has probability 1 (100%). The probability of anything else has to be between 0 and 1.