Word of Tanks 5th Anniversary Extravaganza!

zombieyeti, on Aug 18 2015 - 20:05, said:   Let me begin by thanking you for years of fun tournaments!   TL;DR version If your findings are BADD 4th, DARKS 5th, xBBx 6th, then: BADD is 4th with an opponent strength, final rank of (1,2,3,5) DARKS is 5th with an opponent strength, final rank of (1,2,3,4) xBBx is 6th with an opponent strenght, final rank of (1,2,4,7) This was addressed in my 'spoiler' earlier post B.3.   Is the intent of Tiebreaker 3 to permit a team with an opponent strength final rank of (1,2,3,5) to place higher than a team with an opponent strength final rank of (1,2,3,4)? I'm going to guess that was neither your intent or the intent of Tiebreaker 3.   My argument (condensed): 4th DARKS final opponent strength is (1,2,3,5) 5th BADD final opponent strength is (1,2,3,6) 6th xBBx final opponent strength is (1,2,4,7)   The placement I'm arguing for fully satisfies the letter and spirit of Tiebreaker 3, as it looks at final opponent strength. As previously mentioned I examined all [13] possible permutations of the three teams and their final standings before finding this one perfect fit.   Summation: - There is no question that Tiebreaker 3 is a complex rule to implement and interpret as it fundamentally requires determining a final score of opponents when the final score is the issue at hand, kinda like saying 'The winner is the team who won'. I got around that conundrum by exploring all possible final scores [permutations of 4th, 5th and 6th place] and (purely coincidentally) there was only one that resulted in a 100% logical fit. Offhand, I'd say that this is a rare case. - As a result, I renew my appeal to place DARKS 4th, BADD 5th and xBBx 6th as that's the only way I exhaustively concluded that Tiebreaker 3 is resolved logically and truthfully. - I'm not requesting that any other teams lose gold, only that ours receives our award. - After all is said and done, I am satisfied with (but was not requesting) a final result of DARKS 4th place (BADD 4th place and/or xBBx 4th place) as long as DARKS receives the full 1000 gold award. - As always, love the tourneys and ty ty ty!   If you're still not convinced, the details of my argument are in the 'new' spoiler. I can't claim to have evaluated every possible permutation (that's an unknown unknown) but I can assure you I've assessed every permutation I could think of.   If I understand the Tiebreaking Rules correctly, the Tiebreaking rules are: a) Applied hierarchically and iteratively (If there is a n-tie by points, then apply Tiebreaker 1 and only Tiebreaker 1. If some tied teams remain, and the Tie is unresolved, apply Tiebreaker 2 and only tiebreaker 2, and if still unresolved, apply Tiebreaker 3 and only Tiebreaker 3). b) When exhausted with no resolution, the result is Tiebreaker 4 (Complete Draw).   IF it is your position that Tiebreaking 3. results in a draw for BADD and DARKS, then, as we are tied, shouldn't Tiebreaker 4 (complete draw) apply, with the result BADD and DARKS tie for 4th place? This was not the argument I was making, but I will be satisfied with that result.   IF instead, it is your position that Tiebreaking 3. results preliminary in a draw between BADD and DARKS, but then DARKS loses, because BADD lost to DARKS a) How does the fact that DARKS lost to BADD relevant to Tiebreaker 3? If so, why isn't the fact that DARKS beat xBBx, who beat BADD also considered in Tiebreaker 3? b) Even if we accept that DARKS loses to BADD, arbitrarily (for the sake of argument), then BADD holds 4th place with a final opponent strength of (1,2,3,6) and DARKS holds 5th place with a final opponent strength of (1,2,3,4)! That is, of course, an unacceptable result because Tiebreaker 3 clearly appeals to final opponent strength.   It doesn't seem then that Tiebreaker 3 is applied at all - and T3 does clearly say final rank.   Seems counter-intuitive, but this is the stuff I explored in the TL;DR spoiler in my original post.
Given the starting position:
- DARKS lost to 1,2,3 and BADD (unknown final rank), so DARKS lost to 1-3 and BADD - so DARKS prelim 'score' is 1,2,3,(4/5/6).
- BADD lost to 1,2,3 and xBBx (unknown final rank), so BADD lost to 1-3 and xBBx - BADD's prelim 'score' is 1,2,3,(4/5/6).
- xBBx lost to 1,2, DARKS (unknown final rank) and 7, so xBBx lost to 1,2,7 and DARKS. xBBx prelim score 1,2,(4/5/6),7.   Now, as dance210 collapses the many possible final outcomes, into one final state, we open Schrödinger's box, the cat is now alive, and ... With BADD in 4th place, DARKS in 5th and xBBx in 6th ... Opponent Strength: What was the final rank of any team(s) that have defeated the tied teams? - DARKS lost to 1,2,3 and BADD (which is 4th place) and ended up with a final Opponent Strength of (1,2,3,4) but is somehow in 5th place. - BADD lost to 1,2,3 and xBBx (which is 6th place) and ended up with a final Opponent Strength of (1,2,3,6) but is in 4th place (how???) - xBBx lost to 1,2, DARKS (which is 5th place) and 7, ending up with a final Opponent Strength of (1,2,7,5) and is in 6th place.   However (my argument), if DARKS is 4th, BADD is 5th and xBBx is 6th, then we open Schrödinger's box and ... 4th DARKS final opponent strength is (1,2,3,5) 5th BADD final opponent strength is (1,2,3,6) 6th xBBx final opponent strength is (1,2,4,7)   I think I explored all [13] possible combinations of DARKS, BADD and xBBx in 4th-6th place, but I can't be sure [transparently in my original 'spoiler']. If I did, then the only logical conclusion is DARKS 4th BADD 5th xBBx 6th. There is an alternate possibility that xBBx 4th, BADD 5th and DARKS 6th leads to a logically valid conclusion, but to be true a final opponent strength of (1,2,6,7) would need to be better (for placement) than a final opponent strength of (1,2,3,4).   Even if I didn't explore all possible combinations (for example, I missed one or more possible combinations) of teams and placements, DARKS 4th BADD 5th and xBBX 6th is still valid, and stacked up against one or more valid propositions, then a finding of 'tie' would be the rational outcome.   If additional detail is required, my original reply [open spoiler] assesses the [13] permutations I could come up with.