Here's how White (my new name for Player 1) wins in the $u=v=1$ case. The idea is of course for White to force the knight to an edge, where it can then be summarily captured. WLOG let's force the knight to the lower edge (in my coordinate system, that'll be the edge given by $n=1$). It's enough to show that whenever the knight stands at a point $(a,b)$, we can force a situation where its next move will either allow it to be captured, or will place it on a square whose $n$ coordinate is $<b$; that $n$ coordinate can't decrease forever, so White wins.
So suppose the knight starts at $(a,b)$, and the queen starts at $(c,d)$. Here's a painfully explicit strategy which works uniformly.
In case $c\ne a\pm 1$, White plays the queen to $(c,b+2)$ (the fact that $c\ne a\pm 1$ guarantees Black can't capture here). Now four of Black's moves head towards the bottom, so those are no problem. The moves to $(a-1,b+2)$ and $(a+1,b+2)$ are open to capture, so those are out. Thus Black must move to $(a\pm 2,b+1)$. But now White plays to $(a\pm 2,b+2)$, and Black's only safe moves are to squares with an $n$ coordinate of $b-1$, and the knight has been pushed down, establishing all that we need.
In case $c=a\pm 1$, White moves to $(c,b+4)$. Black's only safe non-retreat is then to $(a\pm 2,b+1)$. White plays to $(a\pm 2,b+4)$. Black has four safe moves; two place it on the $b-1$ row, and we are done. The other two place it back on the $b$ row, but such that we are now back in the previous case, so done.
I'm actually not sure how to be fully explicit in the $u=1,v=2$ case, so I'll just explain it conceptually. Roughly, if the two knights are far enough apart, it's like they aren't even part of the same game, leaving us with successive instances of the winning $u=v=1$ case. (I know very little about combinatorial game theory, but this idea of breaking up games into component sub-games is common. You can see the idea, for example, in this paper by Noam Elkies on pawn endgames, and I feel like this is a big part of thinking in Go, where various regions are their own little battles.) The second knight is irrelevant, the queen picks off one, then the other. On the other hand, if the knights are close enough, the queen will be able to attack both at once, and moreover can arrange that with either side to move, leading to a capture of one knight and reduction to the $u=v=1$ case.
Finally, if $u=1,v=3$, here's an initial position of the knights which White can't win. Place them at $(a,b)$, $(a+2,b+1)$ and $(a+4,b+2)$. They all protect each other here, and no matter where White places the queen, one of the "outer" knights has a safe square to move to, and can then just move back on the next move.
Generally, if $v=u+1$, White will usually be able to at least force repeated even trades of a single queen for a single knight, reducing to the winning $u=1,v=2$ case; but with huge numbers of pieces I don't know how to solidly argue this. For arbitrary $u,v$, the space of possibilities is such that I have absolutely no idea what can be said in general.
$0-1$ poker games have been studied by Borel, Von Neumann and Morgenstern, and others. The most detailed treatment I know is by Bill Chen and Jerod Ankenman, who wrote a series of posts in rec.gambling.poker in 2003, "The [0,1] game: Part 1-14." Some of this was included in their 2006 book, The Mathematics of Poker.
The game you mention is "[0,1] Game #9" on pages 198-203.
Before going into the solution, note that it is a common misconception that mixed strategies are ubiquitous. In fact, for games with finitely many options, mixed strategies are typically required for finitely many boundary cases. In a game with finitely many states like rock-paper-scissors that may be everything, but in a game like this with no atoms in the distribution of cards, mixed strategies are not required at all. Because of the hidden information, when you bet, your opponent still will not know whether you are betting for value (with a strong hand which is ahead on average when called) or with a bluffing hand (which is behind your entire calling range). At the Nash equilibrium, you may be indifferent between betting or checking with a range of hands, when checking will never win and betting as a bluff gives the same value as giving up. So, instead of choosing a function which gives you a probability of betting when dealt that card, you might as well partition the possibilities into sets for each action.
The following excerpts show the Nash equilibrium:
X's strategy consists of the following thresholds:
- $x_0$ between check-folding and bluffing.
- $x_1^*$ between check-folding and check-calling.
- $x_1$ between value betting and check-calling.
Y's strategy likewise consists of three thresholds:
- $y_0$ between checking and bluffing if X checks.
- $y_1^*$ between calling X's bet and folding to X's bet (if X bets)
- $y_1$ between value-betting and checking if $X$ checks.
...
$0 \lt x_1 \lt y_1 \lt x_1^*, y_1^* \lt y_0 \lt x_0 \lt 1$
[This follows the convention that lower cards are stronger, and corrects a typo.]
...
Solution:
$y_1 = (1-\alpha)/(1+\alpha)$
$ x_1 = (1-\alpha)^2/(1+\alpha)$
$y_0 = (1+\alpha^2)/(1+\alpha)$
$x_0 = 1 - \alpha(1-\alpha)^2/(1+\alpha)$
$y_1^* = 1 - \alpha$
$x_1^*= 1-\alpha$
where $\alpha$ is the probability with which you fold a bluff-catcher with a bet of that size relative to the pot to make someone indifferent to bluffing. Since you have a bet of $1$ into a pot of size $2a$, $\alpha = 2a / (1+2a)$.
Best Answer
Note that the distinction between the objectives of survival and killing the opponent can be important. Suppose some $p_i(n) = 0$ (for both 1 and 2) while $p_i(n-1) > 1/2$. The player who steps forward first is very likely to be killed, so to maximize your probability of survival your best strategy at distance $n$ is always to shoot. The opponent, also wanting to survive, will also shoot, and the game will go on forever without anybody getting hurt.
But if your objective is to kill the opponent, this strategy is clearly sub-optimal: it would be better to step forward and have a positive probability of killing the opponent. But that's not optimal either: there's no need to step forward right away, you could wait a while in the hope that the opponent steps forward first. Waiting $k+1$ turns before stepping forward dominates waiting $k$ turns, so there is no optimal strategy.
To avoid such problems, let's assume $p_i(d) > 0$ for all $d$. This will ensure that the probability of both players surviving indefinitely is 0. Then an optimal strategy can be found using dynamic programming. Let $V_i(d)$ be player $i$'s probability of winning under optimal strategies, starting with distance $d$ and $i$'s turn to shoot. Then $V_i(0) = 1$, otherwise $V_i(d) = \max(1 - V_{3-i}(d-1), W_i(d))$, where $W_i(d)=\min\left(\frac{p_i(d)}{p_1(d) + p_2(d) - p_1(d) p_2(d)}, p_i(d) +(1 - p_i(d)) V_i(d-1))\right)$. It is optimal to shoot if $W_i(d) > 1 - V_{3-i}(d-1)$, to step forward if $\lt$, and both are equally good if $=$.