Game 2: The preflop raise/shove/fold game
I will present here another toy game. This is a slight modification of a game described in Will Tipton’s excellent book “Expert Heads Up No Limit Hold’em” (Volume 1).
- 2 Players in the SB and BB paying 0.5 bb and 1 bb blinds, respectively.
- Both players have initial stack sizes of S bb (before posting the blinds).
- The deck is a standard 52 card deck and both players are dealt two cards from the deck at random.
- The betting structure is fixed as follows:
- The SB can either fold or raise to R bb; here, R is some given number and can’t be controlled by the SB for the purposes of this game.
- If the SB raises to R, then the BB can either fold or raise all-in.
- If the BB raises all-in, then the SB can either fold or call. If the SB calls, flop, turn, and river are dealt without further betting rounds, and the hand goes to showdown.
The betting structure can be visualized in a decision tree as follows:
My aim in this post is to derive the SB’s optimal strategy in terms of maximizing the expected value of their stack after the end of the hand, given the strategy of the BB. In other words, I want to find the SB’s maximally exploitative strategy given the strategy of the BB. In later posts, I will try to describe an approximate Nash equilibrium for this game.
Note: In this post, I will follow Will Tipton’s usage of the term EV as the expected value of a player’s stack after the end of the hand. This usage of the term EV is somewhat nonstandard but quite convenient for computations.
I first need to describe the BB’s strategy. The BB’s only decision point is in Node C, where they can either fold or go all-in. I will assume here that the BB shoves at Node C with the following hands, which contain 526 combos or 39.67% of all combos:
The basic idea is now to work backwards through the tree and compute the SB’s EV at each node:
- At nodes where the hand has ended because one of the players has folded (Nodes B, D, and F), the SB’s stack is deterministic and given in the tree diagram above.
- At nodes where both players are all-in (Node G), the pot is 2S and the SB will win
2S*EQ on average, where EQ denotes the equity of the SB’s specific hand against the BB’s shoving range. Thus, the SB’s EV in Node G is 2S*EQ.
- At nodes where the SB is to act (Nodes A and E), the SB seeks to maximize their EV and hence chooses the action which has the largest EV. Thus, the SB’s EV in those nodes will be the maximum of the SB’s EVs in all child nodes.
- At nodes where the BB is to act (Node C), the SB’s EV is the weighted average of the EVs in all child nodes, weighted by the frequencies with which the BB takes different actions.
Let’s start with the leaf nodes:
Node B:
If the SB open-folds, their stack will be S-0.5 (giving up the small blind).
Node D:
If the SB raises and the BB folds, the SB’s stack will be S+1 (winning the big blind).
Node F:
If the SB raises, the BB shoves, and the SB in turn folds, the SB’s stack will be S-R.
Node G:
If both players are all-in, the SB’s EV will be 2S*EQ, as explained above.
Having the SB’s EV determined in all leaf nodes, we can now work backwards through the remaining nodes:
Node E:
The SB chooses to call if 2S * EQ is larger than S-R or, equivalently, if EQ > (S-R)/(2S). For example, if the initial raise is to R=2.5 bb, then the minimum equity required to call as a function of the initial stack size S looks as follows:
Let’s assume in the following that the initial raise is to R=2.5 and that the initial stacks are S=10. Then the minimum equity needed to make the SB prefer calling over folding at Node E is 37.5%. To find the hands which have at least 37.5% equity against the BB’s shoving range, we can look at the equity matrix against that shoving range:
It turns out that the following hands all have at least 37.5% equity against the BB’s shoving range:
(This range contains 718 combos or 54.15% of all combos.)
The SB’s EV in Node E is thus
-
S-R = 7.5 for all hands that fold.
-
2S * EQ for all hands that call, where EQ is the equity of the SB’s specific hand and can be read off from the equity matrix above.
We can also write this as
EV(Node E) = max(S-R, 2S*EQ).
Node C:
This is a decision node for the BB. Thus, the SB’s EV is simply the weighted average of the child node EVs, weighted by the frequencies with which the BB takes the two actions. Recall that we assume that the BB is shoving p=39.67% of the time in Node C. Thus, the SB’s EV in Node C is
EV(Node C) = (1-p)*EV(Node D) + p*EV(Node E) = (1-p)*(S+1) + p*max(S-R, 2S*EQ).
Node A:
Finally, the SB needs to decide which hands to open-raise with and which hands to open-fold. To maximize their EV, the SB open-raises any hand for which EV(Node C) > EV(Node B) = S-0.5 and open-folds all other hands. For the specific values of R=2.5 and S=10, it turns out that EV(Node C) > EV(Node B) always, so that the SB’s optimal strategy in Node A is to open-raise 100% of their hands. Indeed, the following graph shows the SB’s EVs in Nodes B, C, and E as a function of the SB’s equity against the BB’s shoving range:
We see that EV(Node C) lies above EV(Node B) no matter what the SB’s EQ is. Thus, it’s optimal to open-raise 100% of hands and
EV(Node A) = EV(Node C) = (1-p)*(S+1) + p*max(S-R, 2S*EQ).
Summary
For the given parameters R=2.5 and S=10 and the given BB strategy, the SB’s optimal strategy can be summarized as follows:
- The SB open-raises 100% of hands.
- Against a BB shove, the SB calls with all hands that have at least 37.5% equity against the BB shoving range.
Outlook
These strategies do not form a Nash equilibrium. The BB can now in turn try to employ a maximally exploitative strategy against this SB strategy. This will likely include an increase of the BB shoving frequency. Then, the SB can again find another maximally exploitative strategy against this new BB strategy and so on… until some equilibrium or approximate equilibrium is reached and no player can unilaterally improve their EV anymore by changing their strategy.
