oh well then the answer is statistically simple: never attack as its below 50% chance to win. Always wait for your opponent to attack, if he does not do so, stack to 8 and +6 and proceed to attack, you gain everything if attack works, and lose nothing, you essentially just give up your 1st mover advantage, allowing them to stack to 8 and 6 and then attack.

Attacking at the 2v2 is a bad decision, as a the % of winning is lower then your possible 8v8, and losing the 2v2 would result in giving them the first possible 8v8 attack.

Also if you lose first attack and then stack to 8 and attack, your % to win is still not above 50%, you have a 44% and 47% each time.

This problem is kinda tricksy wicksy because not caring ot do the math I think I may be wrong. Player one would have 2 chances to win, where player 2 would have to defend twice in a row to win. I still think player 2s best option is to never attack, as his single attack % will always be higher to hope for a defend.

And sry for above posts, I misread the problem, although I think the problem i stated is interesting, and I do think that my solution to it is correct.

Dude, read the thread.

Sooner or later, someone will pga this game. Suppose that the payoff for the game is $100. What bribe would the first player accept to give up the first turn? I suspect it would be the difference in expected payoff in attacking in a 3 v 3 versus a 2 v 2. Likewise, further bribes would be accepted for giving up the 3 v 3, 4 v 4, etc. until it is no longer profitable to accept any more bribes. What would the the total amount of bribes received before the attack occurs?

Hmm, that situation would be zero sum. So it's never profitable both for the receiver and the donor of the bribe.

If I go first, I will flag for $64.11 or more.

Wiggum and The Brain: While I do agree that Wiggum's explanation that you should 2v2 and then build to the 8v8 to be *correct*, assuming that both player were playing optimally, you will never reach that 8v8 situation.

Player 1: 2v2 (Everyone agrees)
Player 2: 2v2 (He is looking at the exact same situation as player 1 previously was. Why does he have any incentive to play something different if we all agree that the initial 2v2 is the best outcome in that situation?)
Player 1: 2v2 (Same as his first turn, why would he then choose to build)

While getting the 2v2 then 8v8/8v8+6 may have a higher percentage, it can never be realistically achieved. I'm pretty sure that's the rollback equilibrium, I made a game tree.

Wiggin, sorry I spelled your name wrong twice in my response.

Yes of course 5/5 flying. I just described the other situations for the arguments sake.

Shivan, that's what was calculated in the infinite sum (the 64.1%). Wiggin was correct in saying that you have to attack 2vs2 because building to 8vs8 is not optimal :).

Skrumgaer, how about this: I assume the bribing player is not optimal (as he is trying to bribe, what a nublet! :P) and will build to 8+6. Of course, I'm going to attack 8+5 vs 8+6 and have a 55% chance of winning, so I would accept a bribe of about $10 or more to go 2nd? :P

Wiggin, your strategy for first player is correct, but your calculated odds of winning are wrong, since you are not assuming the optimal response from second player.
The optimal response is ALSO to attack 2 vs 2, by symmetry.
If x is chance to win a 2 vs 2, the chance for first player to win is then x times the sum from i=0 to infinity of (1-x)^{2i}. The sum is equal to 1/(1-a), where a=(1-x)^2. If x=.444, this gives chances of winnings at approx 0.642.