Robert-Y
Member
FM mo3 takes three hours which is a lot of time. I'm wondering if there's a way to determine the winner "sensibly" and save time. How likely is it to get two lucky FM solutions in a row? I haven't seen/heard of such a case before.
How about this format: Everyone does 2 solves and the winner is determined by finding the person with the best worst solve. I'm not sure if this has some name.
Example: Let's say these are the results for a final round of FMC:
1st: Shaniel Deppard: 28, 30
2nd: Yobert Rau: 29, 31
3rd: Mames Jolloy: 25, 40
In this case Shaniel is the winner, Yobert is second, Mames is third. Let's assume Mames got lucky with the 1st scramble. Even though she got a 25, it was not enough to let her rise above Shaniel and Yobert whose results are (arguably) more respectable, because of her worst solve. This seems sensible, right? In the case of a tie, I would suggest determining the winner by their best solve.
Currently I cannot see anything particularly wrong with this, but I haven't thought this through that much, admittedly.
Thoughts?
How about this format: Everyone does 2 solves and the winner is determined by finding the person with the best worst solve. I'm not sure if this has some name.
Example: Let's say these are the results for a final round of FMC:
1st: Shaniel Deppard: 28, 30
2nd: Yobert Rau: 29, 31
3rd: Mames Jolloy: 25, 40
In this case Shaniel is the winner, Yobert is second, Mames is third. Let's assume Mames got lucky with the 1st scramble. Even though she got a 25, it was not enough to let her rise above Shaniel and Yobert whose results are (arguably) more respectable, because of her worst solve. This seems sensible, right? In the case of a tie, I would suggest determining the winner by their best solve.
Currently I cannot see anything particularly wrong with this, but I haven't thought this through that much, admittedly.
Thoughts?