Should you bet in poker against Darth Vader?

Question

This is a theoretical question about poker-type games. I'm not going to specify the rules. You can consider No Limit Texas Hold'em or some simple theoretical model, where each player holds a number from the $(0,1)$ interval. We only consider playing heads-up, i.e., when two players play.

Playing against Darth Vader is quite tough because he can read your mind, and thus he knows your hand. You don't know Darth Vader's hand, but you have no hidden information.

Should you ever bet against the Dark Lord?

In games where cards are revealed at later stages, like Texas Hold'em, the answer can be yes, but even there it seems that you should bet/raise at most once after each time a new card is revealed. I'm interested in any sort of related results, which is why I haven't given precise definitions.

Update. After Douglas's examples, the following interesting question is left open.

If you bet/raise, should you always go all-in in a No Limit game?

Answer 1

Here is an answer to the updated question:

Suppose that there are two betting rounds. Darth Vader has three types of hands. Type 1 wins with probability 1. Type 2 is a draw that hits (becomes a winning hand) with probability $1/10$ between the betting rounds. You can't see whether type 2 hands hit. Vader has Type 1 $1/100$ of the time, and type 2 $99/100$ of the time.

You act first. The pot is $1$. The effective stack depth is $4$.

Option 1: Push all-in (bet $4$). Then Vader calls with type-1 hands and folds type-2 hands. You gain $1$ unit $99\%$ of the time but lose not just the pot but an extra $4$ units $1\%$ of the time, for a net gain of $95\%$ of the pot.

Option 2: Check to Vader. Vader can choose to check behind. Then Vader has a winning hand with probability $1/100 + 99/1000 = 109/1000$ on the river with an effective stack depth of $4$. The Nash equilibrium strategy is for you to check to Vader, who bets $4$ for value with all winning hands, and neutralizes your folds by bluffing $4$ times for every $5$ value bets. So, Vader bets $981/5000$ of the time with $10.9\%$ value bets and $8.72\%$ bluffs. This means you might as well fold all of the time, although you call $1/5$ of the time to neutralize Vader's bluffs. If you fold every time Vader bets, you win $80.38\%$ of the pot.

We have not considered all of Vader's possible actions. (Vader can get more by raising some of the time.) However, this is enough to say that checking is worse than pushing all-in.

Option 3: Bet $0.5$, then play to neutralize Vader's raises and later bets with type-2 hands. Before we do the calculations, as long as it is possible to neutralize drawing (type-2) hands, this is better hand-by-hand than pushing all in (option 1) because Vader might as well fold type-2 hands, and we might fold against type-1 hands, saving some of the remaining $3.5$ units.

Suppose when Vader raises to $x$ total, then bets $4-x$ all-in on the next street, you call the raise with probability $p(x)$ and you call the push with probability $q(x)$.

We can set the river calling frequency $q(x)$ so that these have the same equity, so that bluffing on the river has the same expected value as giving up. Vader risks $4-x$ to try to gain $2x+1$, so we call with probability $q(x) = \frac{2x+1}{x+5}$. That means if Vader raises with a draw and gets called, this costs Vader $x$ to get a situation worth $\frac{1}{10}(2x+1 + \frac{2x+1}{x+5}(4-x))$, a net cost of $\frac{10x^2+32x-9}{10x+50}$.

On the first round, Vader risks $\frac{10x^2+32x-9}{10x+50}$ to get a reward of $\frac{3}{2}$. We can neutralize his decision by calling with probability $p(x) =\frac{\text{reward}}{\text{risk+reward}} = \frac{15x+75}{10x^2+47x+66}$. This means Vader's payoff comes only from type 1 hands.

Suppose we call with probability $p(x) = \frac{15x+75}{10x^2+47x+66}$ on the first betting round and $q(x) = \frac{2x+1}{x+5}$ on the second round. Given a type-1 hand, Vader gets $\frac{3}{2}$ with probability $\frac{10x^2+32x-9}{10x^4+47x+66}$, $x+1$ with probability $\frac{60-15x}{10x^2+47x+66}$, and $5$ with probability $\frac{30x+15}{10x^2+47x+66}$. The average is $ \frac{486x+243}{20x^2+94x+132}$. If Vader knows your strategy, the best he can do is to set $x=\frac{3\sqrt{2}-1}{2}=1.62132$ and get $3.05944$ every time he has a type-1 hand, or $3.06\%$ of the pot. This is lower than the $5\%$ Vader gets when you push, and lower than the $18+\%$ Vader gets if you check, so betting less than all-in into Vader is better than checking or pushing, though another amount may be better than $0.5$.

Answer 2

Getting all-in while behind

It is not just when you are ahead that you might want to get all-in against someone who has an information advantage. Suppose the pot is $1$ and the effective stack depth is $1$, and there are two streets of NL-betting with the chance round between the betting rounds. Suppose Darth Vader's hands all have $60\%$ equity against yours, and you won't know which hands have hit. Given the information you will have in the second round, Darth Vader will be a sure winner $60\%$ of the time and a sure loser $40\%$ of the time. If you push, Darth Vader calls, and you pay $1$ for $1.2$, a net gain of $0.2$ compared with abandoning the hand immediately. If you check through the first round, then you have to defend against bluffs on the next round. The optimal strategy is for you to neutralize Darth's bluffs by calling half of the time, while he neutralizes your calls by bluffing once for every two times he bets for value, a total of $90\%$ bets. (If this were over $100\%$ then neutralizing bluffs would be wrong. You would just have to fold to any bet in the second round, including smaller bets.) Your equity is the same as if you fold all $90\%$, so you get $0.1$ if you check the first street through. Even though you are behind, it makes sense to get all-in because letting Darth Vader bet for value on the next street is so dangerous.

A slight variation of this is if Darth Vader makes a small bet with his whole range. Raising all-in as an underdog might be better than calling and suffering the disadvantage on the next street. For example, if the pot is $0.7$, and Darth Vader bets $0.15$ with $1$ behind (left to bet or call), then folding is worth $0$, calling is worth $-0.05$, and raising all-in is worth $0.05$.

This idea is relevant in practice. If you are considering defending the big blind against a short stack's open-raise in late position, you might call or reraise all-in. If you call, you have a positional disadvantage. You have to act first on each later street. If you push, you negate that disadvantage. So, there are times in practice when you are a slight underdog but you prefer reraising over calling.

By the way, this shows that many people misunderstand the Fundamental Theorem of Poker (not a theorem, and false in some settings though it is ok here), which says that whenever you get your opponent to act differently from how he would act with the cards face-up, you benefit. If the cards were face up, you would want to check the flop through and it would be Vader's mistake not to bet. However, here you have to weigh the FTOP-accounting mistakes in the first round against your anticipated FTOP-accounting mistakes in the second round. You expect to make huge FTOP-accounting mistakes by paying off value bets or folding to bluffs if you don't get all-in first.

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Bet/Raise/Reraise with no hidden information

It can be right for both players to bet/raise/reraise even if the cards are face-up, so in particular, it can be right to raise or reraise against a player reading your mind. Let's consider a limit betting structure with bets of a fixed size, $1$ unit, with two chance rounds. Suppose there is a round of betting, then after one chance round, with probability $p_i$ the players will both know that player $1$ will win with probability $q_i$, with $\sum_i p_i=1$, then there is another round of betting, and then the winner is determined (if no one has folded).

If the pot is $x$, and your chance to win is $q$ (decided after this round of betting), what is your equity? If $q<1/2$, your opponent bets. If $q>1/(x+2)$, then you call. Otherwise you fold. We'll use a baseline of calling. The ability to fold lets you benefit by $\max(0,1-q(x+2))$ compared to calling. If $q>1/2$, then you bet and your opponent either calls or folds. The ability for your opponent to fold costs you $\max(0,1-(1-q)(x+2))$.

Let's suppose $\sum_i p_i q_i = 1/2$. Each player has $50\%$ hot-and-cold equity. If no one folds, then each player will put in $1$ in the second round, either betting or calling (we can ignore $q_i=1/2$), so the players split the pot equally. The advantages come from finding profitable folds.

Let's suppose that the pot is initially $2$ and bets are fixed at $1$. (You do not choose the bet size.) Let $q_1 = 4/5$. Let $q_2=1/7$. Let $q_3=8/9$. If neither player bets, then neither player can call as an underdog in situations $1-3$ in the second betting round because you need $1/4$ equity to call when you get $3:1$ odds from a bet of $1$ into a pot of $2$. If each player puts in $1$ bet on the first betting round, then the pot will be $4$ and player $2$ has to call in situation $1$. If each player puts in $2$ in the first betting round, the pot will be $6$, so the first player has to call in situation $2$. If each player puts in $3$ in the first betting round, the pot will be $8$, so the second player has to call in situation $1$. So, if $3$ or more bets go in, the players don't fold in situations $1-3$ and they split the pot.

If only $2$ bets go in from each player in the first round, the pot is $6$, and the second player gains $p_3(1-(1/9)(8))=p_3/9$ from being able to fold. So, if the second player raises, the first player should $3$-bet to eliminate that advantage.

If only $1$ bet goes in from each player on the first round, the pot is $4$, and the second player gains $p_3(1-(1/9)(6)) = p_3/3$ from being able to fold in situation $3$. The first player gains $p_2(1-(1/7)(6))=p_2/7$ from being able to fold in situation $2$. If player $2$ raises, player $1$ will reraise and the players will split the pot, so the second player should raise if $p_2/7 - p_3/3 \gt 0$.

If the pot stays $2$, the second player gains $p_3(1-(1/9)(4))=5p_3/9$ from being able to fold in situation $3$ and $p_1(1-(1/5)(4)) = p_1/5$ from being able to fold in situation $1$. The first player gains $p_2(1-(1/7)(4))=3p_2/7$ from being able to fold in situation $2$. If $p_1/5-3p_2/7+5p_3/9 \gt 0$ then the first player should bet.

These inequalities inequalities are satisfied if $p_3=x, p_2=3x, p_1=4x$. That makes the average value of $q$ equal to $1423/2520 = 0.5647$ so we can set $x=1/16$ and make a fourth situation, $(p_4,q_4) = (\frac{1}{2},\frac{1097}{2520})$. In the fourth situation no one folds.

Given $\{(p_i,q_i)\}=\{(\frac{1}{4},\frac{4}{5}),(\frac{3}{16},\frac{1}{7}),(\frac{1}{16},\frac{8}{9}),(\frac{1}{2},\frac{1097}{2520})\}$, the net gains for the first player if each player puts in $0, 1, 2, 3+$ bets in the first betting round are $-\frac{11}{2520},\frac{1}{168},-\frac{1}{144},$ and $0,$ respectively. The correct action for the first player is to bet, and if raised, reraise. The best action for the second player is to check-raise. Raises after the $3$-bet are optional. So, the game-theoretically optimal action is bet/raise/$3$-bet.

This can be extended to create fixed-limit betting situations where the game-theoretically optimal action with no hidden information is for the players to put in any number of bets on one street, where any deviation from this has a positive cost.