Home » iPokerVIP Strategy » Poker Maths » Game Theory in Poker

Game Theory in Poker

As we touched upon in the previous introductory article, Game Theory is a field of mathematics in which among other things, various games are analyzed and solved. Solving a game means coming up with a strategy for each player such that no player can improve his outcome by changing his strategy. Such a strategy is often referred to as a game theory optimal (GTO) strategy. It is a theorem that every 2 player zero-sum game (such as any form of heads up poker) has a GTO solution, and for poker (with any given stack size) this can be seen because there are a finite number of hands, streets, bet sizes, ranges and so forth, so in theory a computer could analyze every possible situation and recursively pick the optimal play. In practice, however, all but the simplest games have not been solved, because the computational complexity of most games is too high to be solved by brute force methods.

Since the GTO solution for essentially every commonly played poker variant is unknown (Chinese Poker is an exception), how can Game Theory help us? Well, we can analyze and solve various simplified games and situations, and learn a little bit about the mechanics of the game. One of the most useful examples is this:

Player A and Player B are playing HU NLHE, and are on the river. Player A acts first, then player B. Now, suppose that Player A always ends up in this situation with the nuts or complete air, and player B always shows up with a bluff-catcher (a hand that loses to the nuts but beats air), and that both players know this.

Let's look at what actions each player should take. If Player A checks, Player B will also check. This is because if he bets, Player A will fold his air and will at least call with his nuts, thereby losing him money. Therefore, Player A should always bet his nuts, since Player B will never bet for him, and he may call and earn Player A a bet. But if Player A only bets his nuts, then Player B should always fold. This strategy cannot be optimal, since Player A could then always bet his air too, and since Player B will always fold, he would make more money. Therefore, Player A must bluff with some of his air. Let's fix a bet size for now, say that Player A can only bet pot, and suppose that when Player A bets, he has air x of the time. When Player B calls he wins a pot sized bet (as well as the pot) when Player A is bluffing, and loses a pot sized bet when Player A has the nuts. Player B's EV of calling versus folding will therefore be

x(2p) – (1 – x)p = 3xp – p.

If this is positive, then Player B should always call, and if it’s negative, then he should always fold. Neither is optimal, since Player A should then never bluff and always bluff respectively. Therefore, 3xp - p = 0 and x = ⅓. Similarly, if Player B bluff-catches y of the time, the EV of bluffing for Player A will be

-yp + (1 – y)p.

Setting this to 0, we have p – 2yp = 0 or y = ½. Now, since the EV of bluffing is the same as the EV of giving up, Player A cannot improve his strategy, and likewise since the EV of bluff-catching is the same as the EV of folding, Player B cannot either. Therefore this strategy is optimal for both players. Let's assume now that Player A bets b of the pot instead. Our arguments are the same, but our EV calculations will be a little different. Now the EV of calling is

x(1 + b) – (1 – x)b = x(2b + 1) – b.

So,

x = b/(2b+1 )

. Similarly the EV of bluffing will be

y(-b) + (1 – y) = 1 – (b + 1)y.

So,

y=1/(b+1)

Notice that setting b = 1 yields the above results. Now, using this, how much does Player A win with any given bet size? Well, since Player B will call exactly often enough so that Player A will always break even on his bluffs, Player A will only make money when he is betting the nuts. Player B will call b⁄(2b+1) of the time, so when Player A has the nuts, he will earn an extra b-sized bet b⁄(2b+1) of the time, so he makes

b^2/(2b+1)

This function is increasing; hence Player A maximizes his profit by betting as big as possible.

So, now that the math is done, what have we learned? In a situation where one player is always polarized and the other player always has a bluff-catcher, the first player should bet as much as he can with his nuts, and balance this a certain amount with his air, and the second player should call a certain amount and fold a certain amount. We also noticed that as soon as either player deviates from this strategy, the other should start exploiting this deviation; for example if Player B doesn't call quite enough, Player A should always bluff, and so forth. Finally, we learned that even very simple games such as the above one are quite hard to solve.

Applying Game Theory Concepts to Poker

While we cannot employ a perfect GTO strategy in a full-fledged game of poker, we can employ some aspects of toy games like the one above to our game. In solving that game from Player A's perspective, we looked at Player B's decision against our range – whether he should call or fold. In general, if we are trying to play optimally, we will look at our own range in a given situation, and see how to make it unexploitable. That means we will try to balance our value hands with our bluffs and vice versa, as well as our bluff-catches with folds. By incorporating this sort of strategy into our game, we can make it difficult for our opponents to exploit us, but on the other hand we do not ourselves attempt to exploit our opponents. Since most of our opponents have weaknesses in their games, GTO concepts apply best to opponents who are very good or against whom we have no reads as well as in situations where we are unsure of the best play. On the other hand, we can use our understanding of these concepts to identify in our opponents which we can then exploit.

Back to Articles