this is a two player, zero-sum game.

the state is 2 vectors and two scalars (one vector and scalar per player).

call the vectors x and y and the scalars p and q respectively. x and y have some fixed dimension k. (and if it makes things easier, feel free to let k be infinite)

player 1 wins if p > 0 and q <= 0, and gets 1 reward (player 2 gets -1) and vice versa.

intuitively the idea is that x_0 (and also y_0) is the amount of resources that player 1 and 2 have

player 1 can spend x_0 to decrease q, since it wants to get q to 0

player 1 can also spend x_0 to increase x_1, which increases x_0 each time step -- or it can increase x_2, which increases x_1, which increases x_0

(the same goes for player 2)

formally:

at time t = 0, x_0(t) = y_0(t) = 1 and x_i(t) = y_i(t) = 0 for 1 <= i <= k

also at t = 0: p(t) = q(t) = 1

an action by either player is a positive k-dimensional vector.

call the action vectors a and b for player 1 and 2 respectively

then

x_i(t+1) = x_i(t) + x_{i+1}(t) + a_i(t) for 1 <= i <= k if we say x_{k+1} = 0

x_0(t+1) = x_0(t) + x_1(t) - sum(a(t))

p(t+1) = p(t) + min(a_0(t)-b_0(t), 0)

and all three of these rules hold similarly for player 2 (y). the only major difference is

q(t+1) = q(t) + min(b_0(t)-a_0(t), 0) . notice the swap of b and a

finally there is the constraint that a_i(t) >= 0 and sum(a(t)) < x_0(t) and the same goes for player 2

player 1 and 2 make their moves a(t) and b(t) simultaneously

so the question is what a winning strategy is, or if not that, if there is a nash equilibrium. actually i'd be happy with figuring out a merely "good" strategy

so far all i've determined is that at a_0(0) should be less than 1 --

otherwise player 2 if sets b_0(0) to be a small epsilon, player 2 will not immediately lose, while player 1 is stuck with x = 0 and cannot make any nonzero moves from then onwards

## properties of a certain game

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: properties of a certain game

An interesting game. Research first, conclusions towards the end.

The game is perfectly symmetric, I'll consider most things from the view of player 1 but of course it applies to both players. I'll call x_0 "attacking potential", x_1 "income" and a_0 "attacking strength".

Player 1 can force a win if p+x_0 > q + y_0 + 1 at any time.

sum(a_i(0))=1 loses against [b_i(0)=0 for all i and b_0(1)=1], which loses against a_0(0)=1, which loses against b_0(0)=epsilon. I guess the latter is useful independent of the rest. Assuming the opponent follows the epsilon-approach for attacking strength, the initial attacking power is not sufficient to win. I'll ignore the epsilon investments now, as I don't think they play a role in the following rounds.

You have to invest in x_1 or higher. Investing in x_1 at t=0 reduces your attacking potential at t=1, gets neutral at t=2 and only starts increasing it at t=3. The first serious attack can occur here, attacking before does't give any advantage. It also means you don't have to save anything at t=0, and at t=1 you don't invest anything.

Whatever you put in x_2 doesn't matter at t=3, so we get a first strategy to consider: "Attack at t=3", with a(0)=(0, 1, 0, ...), then do nothing, then use a(3)=(2,0,0,..).

The attack will be successful if y(3)<1. We can express this as y(3)=1+b_1(0)-b_3(0)-b_4(0)-... - sum(b(2)).

From t=1 on you know your opponent's strategy, so player 2 can stop investing. I'll drop all the (0) for readability. They survive the first attack if 0 < b_1-b_3-b_4 - ...

The game is not over yet, however. At t=4 player 1 will have an attacking potential of 1 again, so we get the additional constraint 1 < 2b_1+2b_2-b_4-b_5-... where 0,2,2,0 are coefficients of a row of Pascal's triangle minus 1 (the -1 is the investment).

t=5 leads to the inequality 2<3b_1+5b_2+3b_3 - b_5-b_6-... where 0,3,5,3,0 are the next row.

It is easy to find vectors that satisfy all these inequalities, and once you have an income of more than 1 you are done. Some investment in b_1 and b_2 does the job. I don't find nice general conditions (none of the inequalities is redundant). Something we can conclude from this, however: "Attack at t=3" loses against more balanced approaches, probably with decreasing b_i.

How to beat a more balanced approach? For every such approach we can make inequalities again. You win if you invest a little bit more into higher classes, but you lose if you do that too extreme.

If we ignore investments after t=0, this is a (multidimensional) rock-paper-scissors system. You lose against strategies that are much more aggressive, you win against strategies that are just a bit more aggressive, you lose against strategies that are a bit more long-term oriented, and you win against strategies that are much more long-term oriented. The Nash equilibrium is some randomized mixture of these strategies.

I wouldn't expect investments after t=0 to change this. If the difference in strategy between the players is too large, the game is basically done. If the strategy is extremely similar, you just have another rock-paper-scissors system. If there are just small differences, you get biased rock-paper-scissors, where the margins are smaller for one side.

How exactly to distribute the investments? No idea. Something like (eps,0.3+eps,0.4-2eps,0.15,0.075,...) is probably not a bad approach for an approach that survives "attack at t=3" and has a nice long-term growth.

The game is perfectly symmetric, I'll consider most things from the view of player 1 but of course it applies to both players. I'll call x_0 "attacking potential", x_1 "income" and a_0 "attacking strength".

Player 1 can force a win if p+x_0 > q + y_0 + 1 at any time.

sum(a_i(0))=1 loses against [b_i(0)=0 for all i and b_0(1)=1], which loses against a_0(0)=1, which loses against b_0(0)=epsilon. I guess the latter is useful independent of the rest. Assuming the opponent follows the epsilon-approach for attacking strength, the initial attacking power is not sufficient to win. I'll ignore the epsilon investments now, as I don't think they play a role in the following rounds.

You have to invest in x_1 or higher. Investing in x_1 at t=0 reduces your attacking potential at t=1, gets neutral at t=2 and only starts increasing it at t=3. The first serious attack can occur here, attacking before does't give any advantage. It also means you don't have to save anything at t=0, and at t=1 you don't invest anything.

Whatever you put in x_2 doesn't matter at t=3, so we get a first strategy to consider: "Attack at t=3", with a(0)=(0, 1, 0, ...), then do nothing, then use a(3)=(2,0,0,..).

The attack will be successful if y(3)<1. We can express this as y(3)=1+b_1(0)-b_3(0)-b_4(0)-... - sum(b(2)).

From t=1 on you know your opponent's strategy, so player 2 can stop investing. I'll drop all the (0) for readability. They survive the first attack if 0 < b_1-b_3-b_4 - ...

The game is not over yet, however. At t=4 player 1 will have an attacking potential of 1 again, so we get the additional constraint 1 < 2b_1+2b_2-b_4-b_5-... where 0,2,2,0 are coefficients of a row of Pascal's triangle minus 1 (the -1 is the investment).

t=5 leads to the inequality 2<3b_1+5b_2+3b_3 - b_5-b_6-... where 0,3,5,3,0 are the next row.

It is easy to find vectors that satisfy all these inequalities, and once you have an income of more than 1 you are done. Some investment in b_1 and b_2 does the job. I don't find nice general conditions (none of the inequalities is redundant). Something we can conclude from this, however: "Attack at t=3" loses against more balanced approaches, probably with decreasing b_i.

How to beat a more balanced approach? For every such approach we can make inequalities again. You win if you invest a little bit more into higher classes, but you lose if you do that too extreme.

If we ignore investments after t=0, this is a (multidimensional) rock-paper-scissors system. You lose against strategies that are much more aggressive, you win against strategies that are just a bit more aggressive, you lose against strategies that are a bit more long-term oriented, and you win against strategies that are much more long-term oriented. The Nash equilibrium is some randomized mixture of these strategies.

I wouldn't expect investments after t=0 to change this. If the difference in strategy between the players is too large, the game is basically done. If the strategy is extremely similar, you just have another rock-paper-scissors system. If there are just small differences, you get biased rock-paper-scissors, where the margins are smaller for one side.

How exactly to distribute the investments? No idea. Something like (eps,0.3+eps,0.4-2eps,0.15,0.075,...) is probably not a bad approach for an approach that survives "attack at t=3" and has a nice long-term growth.

### Re: properties of a certain game

thanks for the observations. i actually played a few test games against a friend yesterday and did notice the things you mention (though that may just be confirmation bias). hastily written code attached.

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