1-5 against n_c

it was fun melee duel

me 3 : 5 n-c
me 2 : 5 perz
me 0 : 5 ferdia
me 5 : 0 imbaboun
me 5 : 0 swaan
me 5 : 2 mikeyjd
me 5 : 0 T1mmy_83
me 2 : 5 seffer
me 2 : 5 xriox
me 5 : 0 sliky
me 5 : 0 cadagan
me 1 : 5 beay

12 duels done, got 32 points ( not 29 perz! ;p )
5 duels left. ( maximus_le_roi - blizz , joukesantman - fury , carloanalo - nec , heftylld - fury , minifoot - hammer )

total 29 points in my calculations.
come to perz/123 and return my insight plz.

This post was edited by Perserker[PvP] on Dec 6 2010 06:42pm

Now I with you completely agree! I thought 2:5 give 2 points.

done



This post was edited by snatch on Dec 6 2010 06:51pm

5:4 ferdi = 3
2:5 snatch = 1
5:3 Perz = 3
5:0 timmy = 4
1:5 steffer = 1
5:1 sliky = 3
5:3 n-c = 3
5:1 cad = 3



thanks

steffer's point re: sliky; is in relation to his total score. (3 points vs 4). So getting that one extra point might be a factor if your doing the KO stage the same way as me @ 49.

Whats your plan for the KO stage perz ?

This post was edited by ferdia on Dec 7 2010 02:17am

i was thinking that letting only 8 players continue might be too few since we are 18 members in stage 1. do you have any suggestions?

18 members, then a last 16 proceed to straight KO. easy eveyone got a chance to duel each other, only 2 players don't advance. you will need to enforce strict time frames for each KO round, i.e. make sure everyone completes their duels asap.

sounds good to me and it will be the best way to complete it before christmas while allowing ppl to make the KO stage. Use what i did @ 49 tours with last 16 (re: based on score) i wouldnt advise another, smaller group stage, straight KO means you can be lucky / unlucky - It increases the probabilities of more varied builds to advance further. [ i.e. an element of chance, where otherwise so many builds would not get as far ].

This post was edited by ferdia on Dec 7 2010 03:40am

so the person with the most points duels the one with the least points etc?
i was also thinking about putting matchups together totally random.