# Poker Strategy

**Poker strategy** is a set of choices that describes players' actions in poker. It outlines a plan to maximize the profit in a poker game. The characteristics of poker strategies are influenced by game-theoretic properties of poker, such as imperfect information and the element of chance. Consequently, mixed strategies, methods of deception, and probabilistic considerations are common features of effective poker strategies.

An example for the elaborateness of a well-thought-out poker strategy is the complexity of a decision in a single poker hand, which is reflected in the multitude of factors to be considered, such as the number of opponents, position at the table, opponents' playing styles and their perception of the player's own playing style, previous actions, pot size, stack size and some other circumstances like the stage of the tournament.

Certain strategies proved to be more successful over the long run. This justifies the view that skill plays an important role in poker, which was the object of the debate of the legality of online poker in U.S.

In 2015 it was announced that the heads-up limit Hold'em (a variant of Texas Hold'em with restrictions on the betting structure played by two players) was essentially solved, meaning that a sufficient approximation of a Nash equilibrium was computed, making this the first nontrivial game of imperfect information played by humans to be essentially solved by a computer.

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## Fundamental Theorem of Poker

The fundamental theorem of poker formulates a criterion for deciding whether the most profitable decision has been made which applies to almost all considerations, with a few exceptions. Although it provides the theoretical basis for most poker techniques, due to the fact that we never know other players' hands with 100% certainty, it has limited applications in practice in its original form.

Instead of putting an opponent on a specific hand, a more applicable approach is to try to determine the best play against a range--the set of possible hands he could be holding in this situation. This approach clears some misunderstandings which arise in the use of the fundamental theorem of poker in practice:

Folding the Second-best HandYou hold $6\heartsuit 7\heartsuit$ and the board is $8\heartsuit 9 \heartsuit J\diamondsuit T\heartsuit A\diamondsuit$. Based on the previous actions you can not make any accurate estimates on your opponent's cards. Your opponent moves all in with a stack equal to the pot size and you call his bet. The opponent shows $J\heartsuit Q\heartsuit$. Is your call correct?

While the fundamental theorem of poker implies that calling was the wrong play (if you would have known that the opponent has a higher straight flush you would have folded), folding a straight flush under these circumstances is not profitable in the long run because there is only one hand in opponent's range (presumed to contain more hands) that beats yours. $_\square$

## Odds

The profitability of a play in poker is determined on the basis of the risk vs. reward concept. This concept takes a simple mathematical form in the definitions of various odds and the relations between them. The most common use of odds is the comparison of drawing odds and pot odds.

A drawing hand is a hand that has little value in the current situation but can improve significantly through certain future cards (**outs**) which are said to complete a draw. Drawing odds are defined as the ratio of the probability that a hand will not complete the draw to the probability that a hand will complete a draw, calculated as $\frac{1-p}{p}$, where $p$ equals the probability of completing the draw. They are commonly stated in the form of a fraction $\left(\frac{1}{p} - 1\right):1$.

Flush DrawThe hand $A\diamondsuit J\diamondsuit$ on the board of $2\clubsuit 6\diamondsuit 8\diamondsuit T\clubsuit$ has no immediate value. However, if a card that is a diamond comes, the hand will turn into a flush. The probability that the next card is a diamond, ignoring the possible holdings of opponents, is $p=\frac{9}{46}\approx 19.6 \%$ (there are 9 diamonds out of 46 unseen cards). Drawing odds of completing the flush in this situation are approximately $4.11:1$.

Pot odds are defined as the ratio $\frac{P}{C}$ of the current size of the **pot** $(P)$ to the cost of the call $C$.

Pot OddsYou are playing against one opponent who bets $\$10$ in a pot of $\$20$. The current size of the pot is $\$30$. If you call, you are paying $\$10$ to win $\$30$ so your pot odds are $3:1$.

The expected value of the call with a drawing hand where no future actions are possible is determined by the relation of drawing odds to pot odds:

Expected value of the call with a drawing hand with no future actions:Suppose that a player is holding a hand that has the probability $p\in (0,1)$ to complete a draw, thus becoming the winning hand and the player has to call a bet of size $b$ to win a pot of size $P$. The expected value of the call is calculated as $E = p\times P - b\times (1-p)$ $($if the hand completes the draw the win is $P$, otherwise the loss equals the cost of the call $b).$ The call has positive expected value $E>0$ if and only if

$p\times P - b\times (1-p)>0 \iff \frac{1-p}{p} < \frac{P}{b},$

that is, if the pot odds are numerically larger than the drawing odds.

Calling with a DrawYou hold $9\diamondsuit 6\heartsuit$ and the flop is $A\diamondsuit 7\spadesuit 8\heartsuit$. Your opponent has a stack of half of the pot size and moves all in. He coincidentally flips his cards over, revealing $J\diamondsuit J\clubsuit$. Do you have proper pot odds to call his all in?

Your probability of winning the hand if you call his all in is around 33%, or expressed as odds, roughly 2:1. If you call his half-pot sized all in bet, your pot odds are 3:1. Since you are getting better pot odds than winning odds, the call has a positive expected value. $_\square$

The confrontation of a made hand (a hand with no significant chances of improving) and a drawing hand illustrates the most common application of the concept of odds. A player with a drawing hand compares the odds to determine if the call is profitable while a player with a made hand can adjust the size of the bet to deny favorable odds to a drawing hand, making the call unprofitable.

When future actions are possible, one should take more factors into account, such as implied odds and the potential for bluffing.

## Balance and Exploitation

Important features of a poker strategy are its potential to exploit the opponent and its potential to be exploited. If we observe that our opponent folds too often, we can start bluffing more often to increase our profit by exploiting his play. However, the opponent can adjust to our new strategy by calling more often, thus exploiting our strategy. By increasing the potential to exploit the opponent, we have opened the potential to be counter-exploited. The main flaw of exploitative strategies is that they can be counter-exploited. Strategies which are more difficult to exploit are referred to as balanced strategies. In general, it is recommended to use exploitative strategies against nonobservant players and chose balanced ones otherwise.

The problem of counter-exploitation requires game-theoretic considerations, such as the use of mixed strategies to develop the optimal bluffing/calling frequency:

Balanced StrategiesConsider the following simplified game between two players, A and B:

- The flop, turn and river are dealt and player B has a hand that beats $40 \%$ of all possible hands player A can have.
- Player B cannot draw any conclusion (besides card removal) about player A's hand--he assumes that player A has a random hand.
- Player A can see player B's hand.
- The pot size is $P.$
- Player A has the option to bet an amount equal to the pot or to check and player B can only check (if player A checks), fold or call (if player A bets).
- If both players check, the best hand wins the pot; if A bets and B calls, the best hand wins the pot plus the bets; if A bets and B folds, A wins the pot.
Since A knows B's hand, to optimize his expected value, he should bet always if he has a hand that beats B's hand. The problem is to determine how A should play if he has a losing hand. Player A can choose to bet as a bluff or check. Considering only two options for A--to bet all the winning hands and to bluff always or to check always when having a losing hand--and only two options for B--to check/call always or to check/fold always--the payoff matrix which represents these pure strategies is as follows:

(B, check/fold always) (B, check/call always) (A, bet only winning hands) $(A: 0.6P, B: 0.4P )$ $(A: 1.2P, B: -0.2P )$ (A, bet always) $(A: P, B: 0)$ $(A: 0.8P, B: 0.2P)$ The payoffs are calculated as follows:

For the strategy pair $($(A, bet only winning hands) and (B, check/fold always)$):$ if A has a winning hand $($probability $0.6),$ he will get the pot $P$ since B will fold to a bet; if A has a losing hand $($probability $0.4),$ the action will be check-check and B will win the pot $P;$ hence the payoff is $0.6P$ for A and $0.4P$ for B.

For the strategy pair $($(A, bet only winning hands) and (B, check/call always)$):$ if A has a winning hand $($probability $0.6),$ he will get $2P$ (the pot plus the bet) and B will lose $P$; if A has a losing hand $($probability $0.4),$ he will get 0 and B will get $P$--hence the payoff is $0.6(2P)=1.2P$ for A and $-0.6P + 0.4P = -0.2P$ for B,

The payoffs for the other two strategy pairs are calculated similarly.

None of the strategy pairs is a Nash equilibrium:

The pair $($(A, bet only winning hands) and (B, check/fold always)$)$ is not a Nash equilibrium since A can switch to the strategy (A, bet always) to increase his payoff if B keeps his strategy.

The pair $($(A, bet only winning hands) and (B, check/call always)$)$ is not a Nash equilibrium since B can switch to the strategy (B, check/fold always) to increase his payoff if A keeps his strategy.

Similar reasoning applies to the other two pairs of strategies.

In other words, each of player A's pure strategies can be exploited. In order to protect himself from being exploited, player A should randomize his bluffing; that is, he should bluff with, say, $b\in (0,1)$ of his losing hands. The value of $b$ is chosen so that player B is indifferent between his pure strategies combined in a best-response mixed strategy, in which case B should have the same payoff with strategies (B, check/fold always) and (B, check/call always).

If A bets all his winning hands and $b$ of his losing hands, the payoff for B using the strategy (B, check/fold always) is

$(1-b)\cdot 0.4 P,$

since B will fold to a bet always and win the pot $P$ only when A checks $\big((1-b)\cdot 0.4$ times since he bets all of his winning hands ($0.6$) and part of his losing hands ($b\cdot 0.4)\big).$

Using the strategy (B, check/call always) the payoff for B is

$0.6(-P) + b\cdot 0.4(2P) + (1-b) 0.4 P,$

since B will lose a bet $P$ when A bets his winning hand $(0.6),$ gain $2P$ when A bets as a bluff $(b\cdot 0.4),$ and win $P$ when the hand is checked $\big((1-b)\cdot 0.4\big).$

Since B has to be indifferent between these two strategies, the payoffs must be equal:

$(1-b)\cdot 0.4 P = 0.6(-P) + b\cdot 0.4(2P) + (1-b)\cdot 0.4 P.$

Solving for $b,$ we get $b=0.75,$ so in order to protect himself from being exploited by B, A should bluff with $75 \%$ of his losing hands.

Similarly, if B uses pure strategies (B, check/fold always) resp. (B, check/call always), A can start bluffing all the time or bet only his winning hands to maximize the profit. To prevent this from happening, B has to use a mixed strategy, that is, call $\gamma\in (0,1)$ of A's bets. Using the indifference principle, $\gamma$ has to be chosen so that A is indifferent between his two pure strategies. If B calls $\gamma$ of A's bets, the payoff for A if he bets only his winning hands is

$0.6\cdot \gamma \ 2P + 0.6\cdot (1-\gamma) P,$

and

$0.6\cdot \gamma \ 2P + 0.6\cdot (1-\gamma) P + 0.4\cdot \gamma (-P) + 0.4\cdot (1-\gamma) P$

if he bets always. Equating these payoffs we get $\gamma = \frac{1}{2}$, so B should call half of A's bets.

The mixed strategy pair $($(A, bet all winning hands and bluff with 75% of losing hands), (B, check/call 50% of A's bets)$)$ is a Nash equilibrium of the game--no player can unilaterally change his strategy to exploit the opponent.

Note:A situation which can be approximately modeled by this game would be that B holds $3\clubsuit 2\heartsuit$ on the board $Q\heartsuit A\spadesuit 5\spadesuit 8\spadesuit 2\clubsuit$ and cannot draw any conclusions about A's hand based on previous actions.

The example shows how to apply game theory to find non-exploitative strategies. However, a balanced strategy does not necessarily result in maximizing the profit in practice--the opponent will not necessarily use the potential to counter-exploit an exploitative strategy.

## Examples of Applications of Mathematics

Conditional probability is frequently applied to gain information in poker. Standard examples include the calculation of probabilities to connect with a flop and complete draws and gaining information about opponent's range based on his previous actions. Proper application of conditional probability helps to implement our observations to reach the correct decision and can serve to devise a deceptive play based on the concept of levels of common knowledge.

Is your opponent bluffing?You are faced with a decision to call a bet on the river. Based on your estimates, you are certain that your opponent has either missed his draw (the probability of which you estimate to be 70 %) or has a hand that beats yours (30%). Your hand beats a missed draw. Suppose you know that your opponent will always bet a winning hand and will bluff with a missed draw 20% of the time. Your opponent bets. If this is all the information you have, how can you estimate the probability that your opponent is bluffing?

Let $P(C)$ and $P(M)$ denote, respectively, your estimates of the probabilities that your opponent has a winning hand or has missed the draw, and denote by $P(B|C)$ and $P(B|M),$ respectively, the probabilities that your opponent will bet a winning hand or bet a missed draw. Then we have $P(C) = 0.3, P(M) = 0.7, P(B|C) = 1, P(B|M) = 0.2.$ Using Bayes theorem and $P(B) = P(C)P(B|C) + P(M)P(B|M),$ where $P(B)$ is the probability that the opponent will bet, we find $P(M|B),$ the probability of opponent bluffing, as follows:

$P(M|B) = \frac{P(M)P(B|M)}{P(B)} = \frac{P(M)P(B|M)}{P(C)P(B|C) + P(M)P(B|M)} \approx 0.32.$

For practical purposes of the calculation, one can use a version of natural frequencies representation described in Bayesian theory in science and math.

Out of 100 hands the opponent could be holding, 30 hands beat our hand. The opponent will bet all of 30 winning hands and $0.2 \times 70 = 14$ hands that missed the draw. Out of 44 hands he is betting, 14 of them are bet as bluffs. Hence the probability of him bluffing is $\frac{14}{44} \approx 0.32$.

Deceiving the OpponentSuppose you want to increase your chances of bluffing an observant opponent by faking a tell. You assume that your opponent has some basic knowledge of tells and considers nervousness as an indication of a strong hand. In order to influence his decision you pretend that your hands are shaking. Your opponent notices that your hands are shaking and modifies his estimate of the probability that you are bluffing based on new information.

In some cases, such as heads-up with small stacks, strategic advice based on Nash equilibrium charts is frequently applied in practice.