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.



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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.



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Should you bet in poker against Darth Vader?
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 $0.5$.


