Game Theory and Bluffing
March 5th, 2006 by MarkLast night, I went back to Taibei to play chess with my old co-workers. It’s kinda scary, really. They’ve all been playing regularly, Martin’s been reading books on strategy, and I fully expect to get left way behind in terms of ability shortly. Mike W. was there, though and the topic turned to gambling. As many of my friends know, Matt and I got really interested in poker, especially Texas Hold ‘em. Matt wrote a program to analyze the strength of various opening hands against different numbers of opponents. I wrote a Perl program to help train myself to group various opening hands in terms of strength based upon where one is sitting. For example, if you’re in a 10 person game, sitting to the dealer’s left and holding an unsuited Ace-Queen, it’s time to fold. If you’re holding the same cards and sitting to the dealers right, you’ll definitely want to pay to stay in and see the flop.
Game Theory
Then game theory and optimal bluffing came up. Sometimes, there are situations in which a consistent strategy will always fail, and yet a somewhat random strategy, or a mixed strategy will prevail. It sounds irrational, but it is true. Mike asked me to explain it, and I wasn’t able to do so very clearly. So, I’ll have another go at it here. First, there are a few important distinctions to make, though.
Optimal Strategies vs. Exploiting Strategies
There are two kinds of strategies used in poker. In optimal strategies, the opponent is assumed to be strong and adaptive. Optimal strategies are evaluated based on how well they would fare against an optimal opponent. Exploiting strategies are designed to exploit a weak opponent as fully as possible. For example, if you play against a timid opponent who never calls, you can win money from him more quickly by bluffing every time (a strategy designed to exploit his weakness) than you could by using an optimal strategy. The bluffing strategy I’m about to describe is an optimal strategy that will work even if opponents know you do it.
An Example in Which Random Betting is Optimal
Imagine that you are playing a poker game in which the first four cards are dealt out face up, and the last card is dealt face down. After each new card is dealt, each player may bet and raise once. In this game, you and your opponent have just been dealt your final cards, there are $40 in the pot and he has bet $10. The maximum bet is $10.
| Opponent’s hand: | 3♣ | 3♠ | 6♦ | 8♣ |
| Your hand: | K♠ | J♠ | Q♦ | 10♥ |
There are 8 cards on the table. Your last card, which your opponent has not seen, is one of the 44 remaining cards in the deck. Of those remaining cards, any of the Kings, Jacks, Queens or Tens would give you a bigger pair than your opponent’s threes, and any of the Aces or nines would give you a straight. In other words, of the remaining 44 cards, 20 will give you a winning hand and 24 will give you a losing hand. However, you are still in the stronger position if you use an optimal bluffing strategy. Consider these three cases:
You Never Bluff
In 24 cases out of 44, you have the weaker hand and you fold. In the other 20 cases you bet $10. Your opponent, being a strong player, recognized that you do not bluff and never calls your bet. You win the $40 pot in 20 games out of 44. Since half of the money in the pot was yours to begin with, you earn $20×20 = $400. In the 24 games in which you fold, you lose $20×24 = $480. In the long run, if you employ this strategy, you’ll lose $80 every 44 times you play this way.
You Always Bluff
In 20 cases out of 44, you have the stronger hand and bet $10. Your opponent knows you always bluff, so he calls. You win $40 from the pot, plus his $10 from calling. Since $20 of the pot is your money to begin with, you win $30×20 = $600. In the 24 games you lose, you lose your $20 in the pot, plus a $10 bet each time. That’s a $30×24 = $720 loss. In the long run, if you employ this strategy, you’ll lose $120 every 44 times you play this way.
You Use Game Theory to Bluff an Optimal Amount
It is possible to use your opponent’s pot odds to determine how often to bluff. In this case, always bet on the 20 winning cards plus four of the 24 losing cards, and you’ll have the edge. It doesn’t matter how you determine when to bluff, as long as it’s random (at least to your opponent’s perspective). You could say, I’ll bet if I draw a winning card OR a two of any suit. You could ask your friend to give you a random number and then divide it by 24 and only bet on losing cards if the remainder were under 4. Anything random will work. Bet on the 20 winning cards, plus 4 losers. Unless you give tells or your opponent can crack your “randomization” scheme, there is no strategy he can employ that will give him the edge.
Your Opponent Folds When You Bet
You bet on your 20 winners, plus 4 of the losers. Your opponent folds every time, so you win $20 from the pot 24 times for a total of $480. You fold on 20 of the 24 hands in which your last card was a loser, losing $20×20 = $400. In the long run you’ll win $80 every 44 times this situation comes up against an opponent who folds.
Your Opponent Calls You
You bet on your 20 winners, plus 4 of the losers. Your opponent calls each time. On the 20 hands in which you really do have the stronger hand, you win the $20 he put in the pot, plus the $10 call. That’s a total of $30 x $20 = $600. On the four hands in which you bluff and lose, you
lose $30 x 4 = 120. On the 20 hands you fold, you lose $20 x 20 = $400. In the long run, you’ll win $80 every 44 times this situation comes up against an opponent who calls.
Conclusion
Don’t underestimate math geeks! By utilizing game theory, it is possible to construct a mixed strategy that can win in some situations when any consistent strategy would fail.
The optimal bluff is calculated based on the pot odds your opponent would if you bet. Advantage is maximized when the odds that your bet is a bluff are equal to your opponent’s pot odds.

March 6th, 2006 at 4:41 am
Game theory has always been dear and true to my heart. I’ve done many small experiments with various games, including Texas Hold ‘em, and Bridge. For the interested, you can look at some of my poker tables on the probabilistic outcomes of various starting hands in Hold ‘em. I’m hoping to clean this up at some point, but for now, you have to look at raw tables. Similary, I’ve got a bridge table showing the probability of making a given number of tricks in No Trump, with a given number of Goren High Card Points within the partnership.
But, back to the discussion at hand… I was thinking that it would be an interesting project to create a game that demonstrates game theory without the complicated rules of poker. I’ll call it (for lack of a better name) “Matt’s Game Theory Game”. Here’s how it goes:
The game uses a deck of 101 cards, each containing a whole number between zero and 100. Each player is dealt 1 card face down. The player to the left of the dealer posts a blind bet. Play continues as in normal poker, with each player given the option to fold, call, or raise. Highest card wins.
The advantage to this game is that you can accurately compute the odds of winning, just based on the card you hold. For example, if you’re playing just one opponent (heads up), the percentage chance of winning (against a random card), is equal to the number on your card. If you hold a zero, you have no chance of winning a show-down, a 100 will always win, and a 50 wins half the time.
The probabilities of winning show-downs only stays easy against one opponent. Against two opponents, you expect to win half the time with a 67; against three, you win half the time with a 75 ( = 100 * N / (N + 1), where N is the number of opponents).
Anyways, I’ll see if I can find some time to put up an algorithm that can optimally play this game “heads up”. I’m still deep into learning Ruby, so that might be where I start.
November 24th, 2007 at 6:24 am
Actually, you missed a couple other cases: If their down card is a 3, 6, or 8, it will beat your hand if you get a 10, J, Q, or K. This does in fact change the odds.
Furthermore, you don’t consider an opponent who also optimally bluffs (overcalls), who in this case will win on average.
November 24th, 2007 at 7:27 am
It’s the last bet. You’re right about the winning cards the opponent could be holding being too numerous, though.
Since two-pair and three-of-a-kind each lose to a straight, your opponent has 8 cards out of 44 possible cards that will lead to a win against 12 of the wins that I wrote out above. I suppose I could make the hands more like this:
8♣ 8♦ 6♦ 3♦
K♦ Q♦ J♦ 10♦
That would be lead to a loss without using game theory bluffing: Out of 44 times, you’d win 12 (any A, 9 or ♦) times, and 12 times out of 44 (i.e. you get a high pair), you’d beat all but 8 of your opponent’s possible cards (any 6, 8 or 3). Overall that would be (12+12*(36/44))/44 times, which is a losing proposition with any fixed strategy, and yet within striking distance for an optimal bluffer. It would probably be a confusing example for a non poker-player, though.
February 11th, 2008 at 11:51 am
oooooook
this is definatly confusing for a non-poker player
February 24th, 2008 at 3:41 am
[...] I started doing randomized bluffs, but far too loosely. I continued to bet and play good hands, but I also played every single hand [...]