Well, I’m convinced! That I don’t want to limit prizes - either make them ineligible for the section altogether or else let them be eligible for all prizes. With all the different hypotheticals and explanations I have no idea what the correct way to deal with limits is.
That would make it easier to calculate prizes, but it would (a) discourage unrated players from entering, or (b) make it too easy for them to win undeserved prizes. Also, it isn’t just unrated players who get prize limits. At the bigger CCA tournaments there are also prize limits for players with provisional ratings or whose rating has been more than 30 points over the section maximum within the past year. The same argument applies to these players: CCA would like them to play but doesn’t want them to take too much prize money away from players with more reliable ratings.
Correct. If someone drops down, it means that they’ve won their max, and you’re just looking for the prize group that ends up providing it. You never break up and reconstruct prizes or create new prizes. Again, there’s no difference if there are just overall prizes; so it’s just a question of philosophy regarding “under” prizes. This method is consistent with top scorers maxing out their joint winnings.
Tom Doan, how would your algorithm have calculated prizes at the 2012 World Open, Under 1200 Section?
Prizes after reduction:
1st $4225
2nd $2112
3rd $1014
4th $845
5th $676
6th $591
7th $507
8th $422
9th $338
10th $253
1st U1100 $1014
2nd U1100 $507
7.5
Lawson (limit $1500, U1100)
Papageorgiou (limit $1500, U1100)
7.0
Hoyos (limit $2000)
Liang (U1100)
Ash (U1100)
6.5
Chen
Soares (U1100)
Cenezal (U1100)
Goldberg (U1100)
Moshkevich (U1100)
Ramachandran (limit $800)
6.0
Santiago
Vanterpool
Ana Smith
Vaklivia
Kothapalle
Seymour
Aviva Smith (U1100)
Fang (U1100)
Do (U1100)
7.5’s would pull in $4225 + $2112 which are too high. So analyze the 7.0 pack as
4225()
2112()
1014 (3rd place)
1014 (U1100)
845 (4th place)
That’s a total of $9210 for the top five. Lawson and Papageorgiou get their max of $1500 each out of that. The average of the remainder is > 2000 so Hoyos gets $2000 and the other two get $2105.
6.5 would split $676+$591+$507+$507+$422+$338 using normal methods since the share is below the $800 limit.
6.0 would split the $253 10th prize.
That’s how I calculated the prizes in 2012, but my reasoning was a little different. In my method, the 7.5 score group got $4225 for 1st, $2112 for 2nd for a total of $6337. $1500 each for Lawson and Papageorgiou, with $3337 dropping down. The 7.0 score group got $1014 for 3rd, $845 for 4th, $1014 for 1st U1100, plus $3337 from the 7.5 group, for a total of $6210. $2000 for Hoyos and $2105 each for Liang and Ash. I calculated the other prizes the same way you did.
I think the difference between your method and mine (which I hope is now equivalent to Bill Goichberg’s) is in a situation like the 2014 Boardwalk Open with a very big “under” prize.
The problem with your train of thought (in terms of implementing an algorithm for it) is that the 7.0 score group has four “prizes” (the three new ones pulled in plus the extra dropping down) which breaks the basic logic of prize distribution. In this case, you could make that still be 3 and 3 by adding part of the $3337 drop down to 3rd place to push that up to $4225 and the remaining $126 onto 4th place—which seems like a rather kludgy way to handle it, even if it’s mathematically equivalent. Consider what happens if the two top players were at 8.0 and Hoyos alone at 7.5. In my procedure, you would end up pushing everything down to 7.0 and would end up with exactly the same analysis. With yours, it looks like (if done sequentially), you would give the first two their $1500 each, float money down to raise 3rd place to $4225 and 4th to $971. Then give Hoyos his $2000 out of $4225 and add the remaining $2225 onto the already adjusted 4th place (making sure, of course, that that new prize doesn’t get above $4225 itself).
I don’t consider the $3337 to be a separate prize. It’s money that “goes to [the] next player(s) in line” as stated in the tournament announcement, subject to the restriction that no player can win more than he would have won if the players with prize limits weren’t in the tournament.
If I were writing a subroutine for this, I’d first calculate the prize each player would win if the players with prize limits weren’t included in the prize distribution:
Liang $3168.50
Ash $3168.50
Chen $1030.80
Soares $1030.80
Cenezal $1030.80
Goldberg $1030.80
Moshkevich $1030.80
Santiago $225.23
Vanterpool $225.23
Ana Smith $225.23
Vaklivia $225.23
Kothapalle $225.23
Seymour $225.23
Aviva Smith $225.23
Fang $225.23
Do $225.23
This becomes the prize limit for each of those players.
Now calculate the prizes.
8.0
Lawson and Papageorgiou win $4225 + $2112, with $1500 going to each player and $3337 dropping down.
7.5
Hoyos wins the 3rd prize of $1014 plus $3337 from the 8.0 group for a total of $4351, but can only win $2000 with $2351 dropping down.
7.0
Liang and Ash win $1014 (1st U1100) plus $845 (4th) plus $2351 from the 7.5 group for a total of $4210, or $2105 each, which is less than their limit so they each win that amount. The prize distribution would be the same as it was at the actual tournament.
In this example it turned out not to be necessary to calculate the prize each player would receive if the players with prize limits weren’t included in the prize distribution, but the general case you need to do that, at least as I see it. It would take a little longer to calculate the prizes but on a computer the delay wouldn’t be noticeable, and I think it’s more important to calculate the prizes correctly than to calculate them quickly.
Note that in my method the money which drops down goes to a score group, to be divided among the players in that score group, not to individual prizes like 3rd or 4th.