Just to give people some realistic expectations of the cuts at worlds here are some quick
calculations.
These numbers aren’t exact, they will depend on who wins the odd-man-out matchups when there are odd numbers in a record bracket.
The Worlds format for IA is 5 rounds of Swiss followed by a cut to the top 54, then 2 more rounds of Swiss followed by a cut to top 16. Your record after the first 5 rounds will stay, but the tiebreaker points are reset. Here are the percentiles for Swiss after 5 rounds:
round 5 |
||
record |
% |
percentile |
5-0 |
3.13% |
3.13% |
4-1 |
15.63% |
18.75% |
3-2 |
31.25% |
50.00% |
2-3 |
31.25% |
81.25% |
1-4 |
15.63% |
96.88% |
0-5 |
3.13% |
100.00% |
So if 100 people enter the tournament, you need a 3-2 record to guarantee the top 54 cut and approximately 4 people with 2-3 records will move on.
To figure it out based on a different number of participants, just take 54/(# of participants) to figure out what percentile you need to be in to move on. If 64 people enter, a 2-3 record should guarantee top 54.
Here are the percentiles after 7 rounds of Swiss:
round 7 |
||
record |
% |
percentile |
7-0 |
0.78% |
0.78% |
6-1 |
5.47% |
6.25% |
5-2 |
16.41% |
22.66% |
4-3 |
27.34% |
50.00% |
3-4 |
27.34% |
77.34% |
2-5 |
16.41% |
93.75% |
1-6 |
5.47% |
99.22% |
0-7 |
0.78% |
100.00% |
So if 100 people enter you need to have a 6-1 record to guarantee a spot in the top 16 and approximately 9 people with 5-2 records will make it.
To get the dice you need to be in the top 32.
If 100 people enter a 5-2 record will guarantee top 32 and approximately 9 people with 4-3 records will get the dice. If 64 people enter you need a 4-3 record for dice (1 or 2 people might get screwed) and everyone with a 5-2 record should make the top 16. 16/(# of entries) to figure out percentile.
If someone has a good estimate for the number of participants I can get you better numbers.
-Brett
Edited by brettpkelly