Anton84 wrote:
M30-34 has twice as many people, but gets the same # of slots? WTF?
First off, don't think of it as proportional anymore. There's 40 slots, 23 of which are given to each AG, leaving only 17 to be proportionally distributed. So it's really more of a flat distribution than a proportional one now.
But anyway, here we go. One time and one time only. This is how WTC does it.
40 slots to give.
- 23 AGs with a starter, so there go 23 slots
- 17 slots remaining to "proportionally" distribute amongst all 23 AGs. This results in a lot of 0.x, 1.x, 2.x, 3.x slot distributions to each AG. Lop off the fractional part, the .x, and allocate the 0,1,2,3 discrete slot to each AG. This results in 6 slots being given away with 11 yet to find a home.
- Uh oh. We still have 11 to go. What do we do? Round the fractional part (the .x)? No, that's not right. Rank the fractional part and give away the top 11 .x fractional ranks? No, that's not right either. But, yes, that is what WTC does.
It's not the correct way, but this is what it does. This essentially says that an AG earning 2.25 slots deserves another slot before an AG earning 5.24 slots does. Which is false. And is why the situation at IMWI occurred where M18-24's .x fractional component ranked high (and thus got another slot) when there were 40 slots but ranked low (and thus did not) when there are more.
The correct way to proportionally distribute discrete elements is to disregard the fractional component altogether. Skip step 3 altogether. A slot is not earned until an entire slot it is earned. So, 1.99 does not earn 2, but 2.00 does.
There's multiple mathematical ways to go about this. But the simplest to conceptualize is simply to run the algorithm with more slots until the actual total number (17) have been given away as of step 2 above. In this IMMD case, if WTC were to run its algorithm attempting to give away 48.8 slots then it would have properly distributed the 17 it actually has as of step 2 above.