Game Theory Article
by LearnedfromTV on 10/25/08
This is a rough draft of an article I'm working on, the first in a handful of game theory articles. Some of them will deal with naked ace theory in PLO, and the scenario this one covers is similar. Important Disclaimer; the work here and in future articles is my own, but I should reference Chen and Ankenmann's Mathematics of Poker, which is the bible of poker game theory and covers, in some form, anything I'll write about.
This article will cover a simple contrived multi-street nuts-or-nothing betting scenario that can be solved using game theory.
The rules are simple. Pot-limit structure where you must bet pot. The game: two opponents Player A and Player B each hold a single spade. Pot contains 1 unit. Player A has and is known by B to have, the Ten of Spades. Player B's spade is unknown.
One street version
Since B has perfect information about A, the game begins with a checking to B, who then bets pot or checks behind.
So B has a 4/12 chance of beating the T, and an 8/12 chance of being beaten; if there were no betting, B's range has 33% showdown equity, A's has 66%.
In order to find the equilibrium solution for the play of this game, we must find the frequencies A can choose that will make B indifferent between his two options, and vice versa.
That is, B needs to bluff with a frequency that makes A indifferent to calling; and A needs to call with a frequency that makes B indifferent to bluffing.
We know B bets every time he has J,Q,K,A. To bet only those and check all losers would be expoitable; A would never call, and would claim the pot the 66% of the time B checks behind.
If we let B bluff x of the remaining 8 times, we find that if A calls he wins 2 units x/(x+4) of the time, and loses 1 unit 4/(x+4). So his total profit is determined by (2x-4)/(x+4), which equals 0 when x = 2
This makes sense. Since he's laying 2-1, he should have a vbet:bluff ratio of 2:1.
What is A's optimal response?
His goal is to make B indifferent to bluffing. The pot size is 1 unit and B's bluff costs 1 unit, so A makes him be indifferent to bluffing by calling 50% of A's bets.
What is the overall equilibrium strategy:
A bets all his winners and 1/4th his losers (say 9 and 8, it doesn't matter), and A calls half the bets.
12 cases:
1/2 the time B checks; A wins 1 unit, B 0 units
1/12 of the time, B bluffs and A calls; A wins 2 units, B loses 1
1/4 of the time, B bets and A folds; A wins 0 unit, B wins 1 units
1/6 of the time, B value bets and A calls; A loses 1 unit, B wins 2 units
So A wins 1/2 + 1/6 + 0 - 1/6 = 0.5 units
And B wins 0 - 1/12 + 1/4 + 1/3 = 0.5 units
A and B both have 50% equity. B gains 16.66% equity from the street of betting + the information advantage
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Two street version:
For simplicitly, call the first street the "turn" and the second street the "river".
Let's assume (for now)that B will always value bet the first street (ignoring that it is possible that the most profitable strategy would include checking the turn with the nuts some % of the time, in order to be more likely to gain a river bet)
Because B has perfect information, A's options are to check-fold the turn, check-call then check-fold, or check-call twice. B will always value bet his winners on both streets. When he has a loser he can check twice, bluff once and then check, or bluff twice (He can't check and then bluff, because he doesn't have a river valuebetting range after checking)
B's optimal bluff:value bet frequency on the first street is the strategy that will make A indifferent to calling or folding, given that the result of calling is being placed in the river situation where A will often face another bet.
If A check-folds he nets 0 units; if he check-calls he will be placed in the river situation described above. The pot will now be 3 units, but the exact same equilibrium strategy determined above will hold, provided that B has enough bluff hands in his range; that is, provided he bluffs at least 2 hands on the
turn. B will bet all his winners and 2 of his losers (2:1 value bet:bluff ratio), and A will call half the time.
B will bluff the turn with some number x of his 8 bluffing hands. If called, he will then check behind x-2 of those hands. So A's river outcomes are
(x-2)/(4+x) he wins 2 units (the 1 unit in the pot before the turn, and the unit B bet on the turn)
1/(4+x) he wins 5 units (the 1 unit in the pot before the turn, the unit B bet on the turn, and the 3 units B bets on the river)
2/(4+x) times he loses 4 units (the 1 unit he calls on the turn and the three units he calls on the river)
3/(4+x) he loses 1 unit (the 1 unit he called on the turn before folding the river)
So his total outcome is [2(x-2)+5-8-3]/(x+4). For him to be indifferent between calling and folding to the turn bet, we need to set this equal to 0 (the outcome if he folds the turn) and solve for x.
2x - 4 + 5 - 8 - 3 ----> 2x - 10 = 0 -----> x = 5
So B's strategy is to bluff 5 of his 8 losers on the turn, along with betting the 4 winners, and check behind the other 3.
A's goal is to make B indifferent to bluffing or checking the turn. If we let A's turn calling frequency = y, B's outcome for bluffing the turn is to win 1 unit (1-y)% (when A folds) and to enter the river game y% (when A calls).
In the river game, A calls half the time.
If B bluffs the turn and A folds, B gains 1 unit. If B bluffs the turn and A calls, the outcome is dependent on the river action. When he checks behind he loses 1 unit, when he bluffs and gets called (half the time) he loses 4 units, and when he bluffs and succeeds (half the time) he gains two units.
So if A calls turn y% of the time, the net outcome of B's bluff is
1-y +1 units
3y/5 -1 units
y/5 -4 units
y/5 +2 units
(1-y) - 3y/5 - 4y/5 + 2y/5 = (1-y) - y = 1 - 2y; y=1/2.
So, add it all up and we get the following outcomes:
1/4 of the time B checks turn; A +1 unit, B 0 units
3/8 of the time B bets and A folds the turn, A 0 units, B +1 unit
1/8 of the time A calls the turn, B checks the river, A +2 units, B -1 unit
1/12 of the time A calls twice and loses; A -4 units, B +5 units
1/24 of the time A calls twice and wins; A +5 units, B -4 units
1/8 of the time A calls the turn and folds the river; A -1 units, B +2 units
So A's total win is 1/4 + 0 + 1/4 - 1/3 + 5/24 - 1/8 = 0.25
B's total win is 0 + 3/8 - 1/8 + 5/12 - 1/6 + 1/4 = 0.75
In other words, A's share of the pot goes from 66% with no betting to 50% with one street to 25% with 2 streets.
I'll leave the 3-street version for another time.
This article will cover a simple contrived multi-street nuts-or-nothing betting scenario that can be solved using game theory.
The rules are simple. Pot-limit structure where you must bet pot. The game: two opponents Player A and Player B each hold a single spade. Pot contains 1 unit. Player A has and is known by B to have, the Ten of Spades. Player B's spade is unknown.
One street version
Since B has perfect information about A, the game begins with a checking to B, who then bets pot or checks behind.
So B has a 4/12 chance of beating the T, and an 8/12 chance of being beaten; if there were no betting, B's range has 33% showdown equity, A's has 66%.
In order to find the equilibrium solution for the play of this game, we must find the frequencies A can choose that will make B indifferent between his two options, and vice versa.
That is, B needs to bluff with a frequency that makes A indifferent to calling; and A needs to call with a frequency that makes B indifferent to bluffing.
We know B bets every time he has J,Q,K,A. To bet only those and check all losers would be expoitable; A would never call, and would claim the pot the 66% of the time B checks behind.
If we let B bluff x of the remaining 8 times, we find that if A calls he wins 2 units x/(x+4) of the time, and loses 1 unit 4/(x+4). So his total profit is determined by (2x-4)/(x+4), which equals 0 when x = 2
This makes sense. Since he's laying 2-1, he should have a vbet:bluff ratio of 2:1.
What is A's optimal response?
His goal is to make B indifferent to bluffing. The pot size is 1 unit and B's bluff costs 1 unit, so A makes him be indifferent to bluffing by calling 50% of A's bets.
What is the overall equilibrium strategy:
A bets all his winners and 1/4th his losers (say 9 and 8, it doesn't matter), and A calls half the bets.
12 cases:
1/2 the time B checks; A wins 1 unit, B 0 units
1/12 of the time, B bluffs and A calls; A wins 2 units, B loses 1
1/4 of the time, B bets and A folds; A wins 0 unit, B wins 1 units
1/6 of the time, B value bets and A calls; A loses 1 unit, B wins 2 units
So A wins 1/2 + 1/6 + 0 - 1/6 = 0.5 units
And B wins 0 - 1/12 + 1/4 + 1/3 = 0.5 units
A and B both have 50% equity. B gains 16.66% equity from the street of betting + the information advantage
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Two street version:
For simplicitly, call the first street the "turn" and the second street the "river".
Let's assume (for now)that B will always value bet the first street (ignoring that it is possible that the most profitable strategy would include checking the turn with the nuts some % of the time, in order to be more likely to gain a river bet)
Because B has perfect information, A's options are to check-fold the turn, check-call then check-fold, or check-call twice. B will always value bet his winners on both streets. When he has a loser he can check twice, bluff once and then check, or bluff twice (He can't check and then bluff, because he doesn't have a river valuebetting range after checking)
B's optimal bluff:value bet frequency on the first street is the strategy that will make A indifferent to calling or folding, given that the result of calling is being placed in the river situation where A will often face another bet.
If A check-folds he nets 0 units; if he check-calls he will be placed in the river situation described above. The pot will now be 3 units, but the exact same equilibrium strategy determined above will hold, provided that B has enough bluff hands in his range; that is, provided he bluffs at least 2 hands on the
turn. B will bet all his winners and 2 of his losers (2:1 value bet:bluff ratio), and A will call half the time.
B will bluff the turn with some number x of his 8 bluffing hands. If called, he will then check behind x-2 of those hands. So A's river outcomes are
(x-2)/(4+x) he wins 2 units (the 1 unit in the pot before the turn, and the unit B bet on the turn)
1/(4+x) he wins 5 units (the 1 unit in the pot before the turn, the unit B bet on the turn, and the 3 units B bets on the river)
2/(4+x) times he loses 4 units (the 1 unit he calls on the turn and the three units he calls on the river)
3/(4+x) he loses 1 unit (the 1 unit he called on the turn before folding the river)
So his total outcome is [2(x-2)+5-8-3]/(x+4). For him to be indifferent between calling and folding to the turn bet, we need to set this equal to 0 (the outcome if he folds the turn) and solve for x.
2x - 4 + 5 - 8 - 3 ----> 2x - 10 = 0 -----> x = 5
So B's strategy is to bluff 5 of his 8 losers on the turn, along with betting the 4 winners, and check behind the other 3.
A's goal is to make B indifferent to bluffing or checking the turn. If we let A's turn calling frequency = y, B's outcome for bluffing the turn is to win 1 unit (1-y)% (when A folds) and to enter the river game y% (when A calls).
In the river game, A calls half the time.
If B bluffs the turn and A folds, B gains 1 unit. If B bluffs the turn and A calls, the outcome is dependent on the river action. When he checks behind he loses 1 unit, when he bluffs and gets called (half the time) he loses 4 units, and when he bluffs and succeeds (half the time) he gains two units.
So if A calls turn y% of the time, the net outcome of B's bluff is
1-y +1 units
3y/5 -1 units
y/5 -4 units
y/5 +2 units
(1-y) - 3y/5 - 4y/5 + 2y/5 = (1-y) - y = 1 - 2y; y=1/2.
So, add it all up and we get the following outcomes:
1/4 of the time B checks turn; A +1 unit, B 0 units
3/8 of the time B bets and A folds the turn, A 0 units, B +1 unit
1/8 of the time A calls the turn, B checks the river, A +2 units, B -1 unit
1/12 of the time A calls twice and loses; A -4 units, B +5 units
1/24 of the time A calls twice and wins; A +5 units, B -4 units
1/8 of the time A calls the turn and folds the river; A -1 units, B +2 units
So A's total win is 1/4 + 0 + 1/4 - 1/3 + 5/24 - 1/8 = 0.25
B's total win is 0 + 3/8 - 1/8 + 5/12 - 1/6 + 1/4 = 0.75
In other words, A's share of the pot goes from 66% with no betting to 50% with one street to 25% with 2 streets.
I'll leave the 3-street version for another time.
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