Results[]
Ulti's Analysis[]
| Division | Northwest |
|---|---|
| Match # | 30 |
| Match Date | Thursday, November 17, 2011 |
| Vote difference | 3429 |
| Oracle Expectations |
Siegfried vs. Nightmare - 55.15% 92 for - 15 against |
| GameFAQs Prediction |
Siegfried vs. Nightmare - 68.03% |
| Battle Contest Predictions |
Banked: 470 |
I'm seriously convinced the Soul Calibur guys only won this because DAT TOP OPTION, though obviously I'm biased. Ramza and Delita might seriously be the best rivalry in this entire bracket, and it's not just me being a big fan of FFT. I'm a big fan of history, and the relationship between those two reminds me of how human history works. Very few people know the true heroes of war, the soldiers, but kings and politicians are remembered forever even if most don't deserve it. Go look up the lyrics to War Pigs some time. Saying Ramza and Delita aren't rivals because they never fought is like saying LBJ and Kennedy weren't rivals; LBJ had a maniacal desire to become president and viewed Kennedy as an obstacle, and he wasn't exactly upset at Kennedy getting assassinated. There are tens of thousands of examples like this throughout history, and FFT touches upon that subject without really trying.
Well actually it does try, but I don't think Yatsumi Matsuno realized the depth he'd manage to explore that aspect of history when he was writing FFT's plot. The man is an absolute genius, but his tragic flaw is never being able to seal the deal, so to speak. He always resorts to "kill the big bad magical things" by the end of his plots, whereas if he would ever finish his tales of medieval political intrigue and conspiracy he would go down as one of gaming's best authors.
Fun Fact: Ramza and Delita were originally going to be the exact same character, but Square killed the idea of having two branching paths for the plot.
I have nothing more to really add here, except that Ramza will likely never win a contest match but he might have pulled it off were the seeding switched in this match.
Match Trends[]
