You should attack approximately 1% of the time.

Let p1 be the probability that player 1 will attack. Let p2 be the probability that player 2 will counterattack. Then player 1's expected payoff would be p1q1 + (1-p1)(1-p2q2) where q1 and q2 are the probabilities of winning the particular attacks. Take the derivative of this and set to zero and you have

q1 = (1 - p2)q2

Likewise for player 2 (remember, both are optimizing) we will have

q2 = (1- p1)q3 where q3 is the probability that player 1 will win the counter-counter attack.

Now let both players have the same p. (An approximation). Then,

q1 = (1 - p)^2q3

or

p = 1 - sqrt(q1/q3)

Now for a 2 v 2, q1 = .44, and for a 3 v 3, q3 = .45, so the value of p is

p = 1 - sqrt (.44/.45)

which is approximately 1 - .99

so player 1 should attack 1 percent of the time and not attack 99 percent of the time.

I goofed with a parenthesis. The solution is the smaller root of a quadratic equation and is approximately 19%.

lol kdice.

Skrum, what are you talking about... given a specific situation, there must be a best move, either attack or don't. It can't be best to attack 19% of the time in that situation.

after reading your post, wiggin, i thought, hmmm, theres the right answer! but it still bugged me (cus i hate admitting defeat, im such a cocky sob).

but in the shower this morning i started thinking again about it. and i realized the assumption your 70.32 answer is based off of: that you are playing a conservabot. sure, if you assume that the bot will only attack if he has 8+6, then the best plan is to 2v2 and 8v8 him asap and you have a 70.32 chance of winning.

however, you cant assume anything about your opponent, that he will be dumb. i mean, if you were rolling against me, wiggin, and you lost your initial 2v2... why would i just sit and stack to 8+6? because then i would hold the advantage (according to you, a 70.32 advantage). so i should 2v2 and 8v8 you asap. but wait, as soon as i lose my 2v2, you would hold the 70.32 advantage, and no way would you forfeit that and wait for 8+6.

see what im saying? you need to visualize playing against a mirrorbot, not a conservabot. this bot is as genius as you.

so, in the end, you have a small range of outcomes: 2v2 ad infinitum for a slightly lower chance of winning, or 8+6v8+6 ad infinitum for a slightly higher chance of winning.

...

right?

Yes, I see what you are saying.

Let us instead assume I play a wiggin1-bot.
The wiggin1-bot 2v2s if possible, and if it's behind it 8v8s as soon as it can.

Again now the best move is to 2v2. Stacking to 8+6 is not an option, since it will 8v8 me before that. The bastard.

(Actually if you are playing 2nd against an optimal opponent, except he doesn't 2v2, then it's best to wait until 8+3 before 8v8ing him. Maybe he will lose the 8v5.)

assuming that both players are playing optimally, they would, of course, be mirror images of each other.

another thought: if someone knew that you were going to 8v8 them and have a better opportunity to 8v8, why wouldn't they 7v7? then the whole chain breaks down and you are back the 2v2 which i think is pareto optimal.

wiggin1:

Consider bluffing in poker. If you bluff all the time, you lose too many chips. If you never bluff, the other bettors will drop out if you bet high on a good hand. So you have to bluff a percentage of the time, in a way that is not predictable by the other players.

Yes, because poker is a game of incomplete information. Unlike kdice.