This is an informative post. If we take the the results of the seeding match to be the same as the necklace match (risky, but most accurate prediction), then we have this equation.Miku wrote:Kanade's Necklace Score might be
21.71% * 57=12.3747
while second place's Necklace Score might be
15.28% * 99=15.1272
.2171x < .1528y, where less than means Kanade loses
0 < 0.1528y - 0.2171x
0 < 0.1528y - 0.1528x - 0.0643x
0 < 0.1528 ( y - x ) - 0.0643x
0.0643x < 0.1528d
d > 0.4208x, where d = difference in SDO and x = Kanade's SDO
This means in order for Kanade to lose the necklace, the difference between the leader's SDO and Kanade's SDO has to be greater than 0.4208 times Kanade's SDO. Let's take Diamond of 2010, where Mikoto got 60 SDO and Shana got 99 SDO. Now let's assume Kanade has 60 SDO.
d > 0.4208 * 60 = 25.2
So if the difference between Kanade's 60 SDO and the top SDO getter is greater than 25.2, then Kanade loses. 60 + 26 = 86. Now let's analyze what this means.
1) Four characters gets 7 wins.
2) Kanade has to be the one with 60 SDO (lowest).
3) The one with the highest SDO must have 86 SDO or higher.
4) The one with 86 SDO or higher has to get 2nd in the necklace match.
Using (1) and (2), the chance Kanade has the lowest SDO is 1/4. Using (4), the chance one of the remaining 3 with the highest SDO get 2nd in the necklace match is (3 C 1)(1/3)(1/6) = 1/6. This probability of Kanade losing is (1/4)(1/6) = 4.17%. According to my definition of lock being 95% or more, these calculations would suffice making Kanade a lock.
But we're not done yet. We haven't used the part where the one with the highest SDO must get 86 SDO or more (3). This only happened once in 2010, so if we include this as a rough estimation (1/7), then the probability will be (1/4)(1/6)(1/7) = 0.00595%, which is the chance Kanade will lose. This means the chance Kanade will win the necklace is 1 - (1/5)(1/4)(1/7) = 99.4%. This is now a sure lock.