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, but you won't know which hands have hit. If you push, Darth Vader calls, and you pay $1$ for $1.2$, a net gain of $0.2$. 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 folds 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.) 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, 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, you might call or reraise all-in. If you call, you have a positional disadvantage. 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 the anticipated FTOP-accounting mistakes in the second round, and 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.
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. 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 $q$, what is your equity? If $q<1/2$ your opponent bets, and you either call a bet or fold to one bet. 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 $\max(0,1-(1-q)(x+2))$ compared to if they were compelled to call.
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 $1$ on the second round, either betting or calling (we can ignore $q_i=1/2$), so the players split the pot equally. The advantage of a player is if the player can find profitable folds.
Let's suppose that the pot is $2$, and bets are a fixed $1$ unit. 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 on the second betting round because you need $1/4$ equity to call when you get $3:1$ odds. If each player puts in $1$ bet on the first betting round, then the odds will be $5:1$ and player $2$ has to call in situation $1$. If each player puts in $2$ on the first betting round, the odds will be $7:1$ on a call in the second betting round, so the first player has to call in situation $2$. If each player puts in $3$ on the first betting round, the odds will be $9:1$ on a call in the second round, so the second player has to call in situation $2$. So, if $3$ or more bets go in, the players don't fold 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 player $2$ 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 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 and any deviation from this has a positive cost.