What do these numbers mean, technically ? For say, Offense luck : 0.40 . OK offense had it better than expected, but what does it quantify exactly ? This is still unclear for me.
Thanks for the link, Fizzer, although I must still miss something. When you mouse over the list of moves in a game, offense and defense luck appear as numbers around +/- 0. Maybe I need a very tiny example to get what these very numbers mean, and when exactly they are taken into the calculus. I do maths, I'll cope with it eventually :o)
Ahh, sorry, I thought you were talking about the offense and defense kill rates.
The offense luck is the difference between how many armies the attacking armies killed and how many they would have killed assuming an average roll. For example, 100 will kill on average 60, so if they kill 61 the luck is +1.
Defense luck is the same thing for the defense roll, but from the defender's perspective. For example, 100 defending would, on average, kill 70, so if they kill 68 the defense luck is -2.
I have been curious about these too. So are these numbers unaffected by the ratio of attackers to defenders? The attack and defense are two simultaneous dice rolls in other words?
This raises a statistical question: in long games the cumulative luck should generally be converging on zero, right? Regression to the mean and all that.
Strangely, no !
See [here](http://en.wikipedia.org/wiki/Random_walk) and do some search about brownian movement : the expected difference should converge as the square root of the number of turns * the standard deviation (which is not zero !).
Addendum : the mean is 0, of course ; the standard deviation is what I told you (I must re-do the maths but should be it). I hope the distinction is clear for you.
Actually, it does not converge to 0 as far as I know, offensive luck will converge to a value slightly smaller than 0. I think defensive luck will converge to 0, although I'm not 100% sure of this.
The reason: 2vs2 and 1vs1 attacks. In a 1vs1 the offense will not kill 0.6 as expected, but only 0.18.
If you ignore that, what Mian says is completely correct, except that it is not the square root of the number of turns but the number of armies involved. This is a binomial with a success probability of p (0.6, resp. 0.7) for every army, so the standard deviation will be square root of n*p*(1-p) with n the number of attacking/defending armies.
