Forum
skrum on aggressiveness (attack/defend ratio)
skrumgaer wrote
at 4:49 PM, Wednesday January 23, 2008 EST
I have looked at the attack/defend ratios of the top 25 and they tend to run 55:45, 54:46, or 53:47. Since they tend to play each other a lot, one might ask: how can they all have attack/defend ratios greater than 50:50? After all, at a particular table, the overall attack:defend ratio has to be 50:50 for a particular game, so some players will have a ratio greater than that and some will have a ratio less than that.
The answer is in who attacks whom. To make the math simple, consider all attacks a player makes where attacker and defender have the same number of dice (8 v 8, 7 v 7, etc.) Consider 8 v 8's only for example. For all a player's 8 v 8 encounters, the expected value E of territories gained would be A.Pa - D(1-Pa) where A is the number of attacks, D is the number of defends, and Pa is the probabiliity of winning an attack. If you factor out the Pa, the expected gain is Pa(A+D) - D. The expected gain per encounter (e) would be obtained by dividing the above by (A+D), to get e = Pa - D/(A+D). Let the attack defend rato be expressd as a:d. Then the above formula is e = Pa - d or e = Pa - (1-a). For e to be positive, we have Pa > 1-a or a > 1 - Pa. Now, for an 8 v 8, the probability of winning an attack is about 45%, so your attack/defend ratio should be better than 55:45 for all your 8 v 8's. How can you do this? By attacking someone who can't attack back. In other words, beat up on the little guys. Now, did I really need a lot of math to come up with this idea? Similar analyses can be made for all your 7 v 7's, 7 v 8's, 8 v 6's, and so on. |