Poker, unlike chess or checkers a game with incomplete information. Nevertheless, it is beyond not the mathematical analysis, because, with the aid of probability theory and game theory approaches Poker is actually modeled mathematically. The game is incredibly complex and therefore there is a mathematical solution to this game only for limited special cases.
The most profound mathematical results to Texas Hold 'Em are for the heads-up play (two players) before. The complexity of Texas Hold 'Em is much higher if more players are involved. to give
To gain insight into the mathematics behind the poker game, I want to give a very simple example from the literature (from an article by James Swanson). In this example, the strategy of the players are completely determined. At the same time gaining But if we have any idea how difficult it must be the model with 52 cards, and more than 2 players.
"One Card Poker - The Rules
We play with 2 players, and only 3 cards that we ace, 2 and 3 are significant.
The ace is the lowest card and 3, the highest card.
One of the players is the dealer and the other players in the opener.
Both pay an ante of $ 1 and $ 2 so that the pot is large.
Every now a card is allocated. One is left.
Then the opener to decide if he checks or $ 1 is (We therefore have a fixed limit game).
In the case of Bet from the opener, the dealer may now choose only between fold and call.
In a fold of the dealer gets the opener of the pot.
When a call from the opener, there is a "showdown", ie the higher of the two card hand wins the pot.
If the opener can check in, check the Dealer or give a bet of $ 1.
After a bet from the dealer of the opener, in turn, fold or call options, and if he calls then there's the showdown.
Reasonable assumptions
Assumption 1: No one folds the 3, for the win 3 more the pot.
Assumption 2: No one elects to call a bet when he has the ace, as the ace loses all showdown.
Assumption 3: No one checks out the 3 "in position". If the opener checks to you, will give a 3 Dumit always a bet, because otherwise you get anyway, only the smallest possible pot.
Assumption 4: No one makes a bet with the 2, since the Ace will fold and 3 will call. It therefore gives away so that the 3 only money, while against the ace anyway gets only the small pot, because you can not get a call. In real Hold'Em is also present in such situations where one with an isolated Bet all probability only with better hands and this should be so dear.
decisions and probabilities
The dealer has to ask:
"How many times (probability q1) should I place a bet, if the opener checks to me and I have an ace ?
"How often (Wkt. q2) should I use the 2 Callen, when the opener is a bet?"
The opener has to ask: "How often (Wkt. p1) should I embed with an ace"
"How often (Wkt. p2) should I call with a 2 if the dealer gives a bet?"
"How often (Wkt. p3) should I embed a 3?"
The expected value
Using the above probabilities we can determine what the expected value of the dealer and which has the opener. For this we need to think about how the game can run.
Case 1 holds
Opener Ass
Dealer holds two
a) embeds opener and dealer calls. Loses opener gets $ 1, so - $ 1 The probability is p1 * q2.
b) opener embeds and dealer folds. Then win opener $ 2.The probability p1 * (1-q2) is.
Case 2:
opener holds Ass
Dealer 3 holds
The opener loses $ 1 if he bets. This occurs with wkt. p1.
Case 3: Open 2 holds
dealer holds Ass
checked out the opener. With wkt. 1-q1 also checked out the dealer. Then the opener gets second $
With wkt. q1 * p2 embeds the dealer and is called by opener. Then win opener $ 3
Case 4 holds
Open 2
dealer keeps three
opener loses $ 1 if he bet the dealer calls. This occurs with wkt. p2.
Case 5 holds
Open 3
dealer holds Ass
opener wins $ 3 if he checks, and embeds the dealer. This happened with the wkt. (1-p3), q1.
In other cases, he wins $ 2
Case 6:
opener holds 3
Dealer holds two
If opener embeds and dealer calling, then won the opener $ 3 This happened with wkt. p3 * q2. Otherwise Opener $ wins 2nd
Each of the 6 cases is equally likely.
The opener loses out with wkt. Q1 = 1 / 6 * (p1 * p1 + p2 + q2) a dollar.
The opener wins $ 2 with wkt. Q2 = 1 / 6 * (p1 * (1-q2) q1 +1 +1 (1-p3) * q1 * q2 p3 +1).
The opener wins $ 3-wkt. Q3 = 1 / 6 * (q1 * p2 + (1-p3) * q1 + p3 * q2).
His post-ante EV (expected value after the Ante) is so-Q1 Q2 Q3 +2 * +3 *.
wkt. 1-Q1-Q2-Q3 losing the opener but his ante. With this in mind we have an EV of Q1-Q2 +2 * +3 * Q3-1.
By inserting the expressions for Q1, Q2 and Q3, we get an EV of
1 / 6 * (p1 * (1-3 * q2) + p2 * (3 * q1-1) + p3 * ( q2-q1) q1).
Substituting q1 = q2 = 1 / 3, we obtain independent of p1, p2 and p3 an EV of -1 / 18 for the opener. This strategy is called "indifferent" and it is also the best that can have the dealer. In 1 / 3 of the cases he was bluffing with the ace and in 1 / 3 of the cases he should be with the 2 to call.
against other strategies can achieve the opener by modification of p1, p2 and p3 advantage.
Also for the opener, there is an indifferent strategy, namely its EV can also write as follows: 1 / 6 * (q1 * (3 * p2-p3-1) + q2 * (p1 * p3-3 ) + p1-p2).
We now choose arbitrarily p3, then the strategy of the indifferent form that p1 = 1 / 3 * p3 and p2 = p1 +1 / 3. Again, we get an EV of -1/18.
How to play the best?
Based on this optimum, we must ask ourselves how the strategy against a specific Opponents should look like, does not play to optimum.
We are looking to maximize the opener and the EV. We know q1 and q2.
If q1> 1 / 3, then we maximize the EV, as we choose p2 = 1.
That is, if the dealer is bluffing too often with his ace, then we should call the 2 always.
If q1 <1/3, p2="0">
That is, if the dealer is bluffing too little with the ace, we should call 2 is not a bet, but always fold.
If q1 = 1 / 3, so there is however no matter how we react with the 2 on a bet.
This is somewhat counterintuitive result:
If the opponent plays optimally, it does not matter how you play. But if he does not play optimally, you must consider your strategy at his stats, just adjust q1 and q2. One game after the "optimal" strategy, then the EV would not maximize.
Conclusion:
A determined for the purposes of both players best game no perfect strategy for a player. Rather, it depends on the Stastistik of others, the strategy which we can maximize his EV.
This fact shows the importance in the real Hold'Em so-called "Reads" and statistics about the game the opponents are. One can certainly a relatively "optimal" standard game to learn and succeed, but a profit maximization can only be achieved by adapting to the opponents play.
