In another thread, James Schuyler highlights various flaws in the standard methods of distributing place prizes and class prizes.
These flaws have also been pointed out in other threads, such as:
Distribute Prizes
http://main.uschess.org/forums/viewtopic.php?f=2&t=208
and
Prize Distribution Revisited
http://main.uschess.org/forums/viewtopic.php?f=2&t=6710
Various solutions have been proposed, making use of complicated proportional or fractional methods. Understandably, however, TDs don’t want to add 6/13 of one prize plus 4/7 of another when players are milling around asking for their money so they can begin the long drive home.
Before I go any further, let me make a fervent plea. Please do not hijack this thread with posts about the inherent unfairness of class prizes. I actually agree with that point of view (at least to some degree), but that’s not the purpose here. If you want to rail about class prizes in general, please do so in one of the following:
Splitting Money Prizes
http://main.uschess.org/forums/viewtopic.php?f=2&t=7365
or
Counterplay: Letters to the Editor
http://main.uschess.org/forums/viewtopic.php?f=24&t=7661
Suddenly, just the other day, I slapped my forehead. “I could have had a V-8.” An idea struck me which was so simple, and yet so logical, that I wondered why nobody had thought of it before. Stay tuned, and find out if you, too, should have had a V-8.
Ideally, a prize distribution method should:
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be continuous. A small change in the prize structure should result in only a small change in the actual prize won by any player.
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be monotone. An increase in one of the prizes should never result in a decrease in the actual prize won by any player.
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award class prizes to the players in that class.
-
be in agreement with the philosophy behind class prizes.
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be simple to apply in the heat of tournament aftermath.
Here’s my brainstorm:
A. First award the place prizes, without regard to the class or “under” prizes. In other words, figure out what the place prizes would be if there were no class prizes.
B. Next, award the class or “under” prizes to the players in those classes. If some players already receiving a prize in step A would now have their prize increased, get the increase from the class prizes, rather than tinkering with the place prizes already calculated. Observe the limit-one-prize-per-player rule, awarding any leftover amount to other player(s) in the class.
C. If there are class or “under” prizes in multiple classes, award prizes in the more inclusive categories before those in the less inclusive categories. For example, figure the under-2200 prizes before the under-2000 prizes.
How would this work in James Schuyler’s example?
1st $200
2nd $100
3rd $50
1st U2000 $60
p1 2200 5.0
p2 2200 5.0
p3 1900 5.0
p4 2200 4.5
p5 1900 4.0
A. First award the place prizes. This one is easy, because there is a 3-way tie for the 3 place prizes. p1,p2,p3 each get $116.67.
B. Next, award the under-2000 prize. p3 has already won more than $60, so his prize is not increased. The under-2000 prize goes to the next under-2000 player. p5 wins $60.
Here is one of the examples from the Distribute Prizes thread:
1st $150
2nd $101
1st U1800 $99
p1 2100 5.0
p2 1700 4.5
p3 1900 4.0
p4 1650 3.5
A. Award the place prizes to p1 ($150) and p2 ($101).
B. Award the under-1800 prize. p2 has already won more than $99, so this amount goes to the next under-1800, p4.
Here is a slightly different example from the same thread, an example originally designed as a companion case to expose the flaws in the standard prize distribution methods:
1st $150
2nd $99
1st U1800 $101
p1 2100 5.0
p2 1700 4.5
p3 1900 4.0
p4 1650 3.5
A. Award the place prizes to p1 ($150) and p2 ($99).
B. Award the under-1800 prize. p2 has already won $99, so he is entitled to an additional $2. (More would violate one-prize-per-player.) This $2 comes from the U1800 prize. p2 wins $101 total, with the remaining $99 (of the U1800 prize) going to the next under-1800, p4.
Finally, here is the example that started the Distribute Prizes thread:
1st $32
Class B (1600-1799) $20
Class C (1400-1599) $20
Class D/E (under 1400) $20
p1 1707 3.0
p2 1575 3.0
p3 1688 2.5
p4 1322 2.5
p5 1584 2.0
p6 1541 2.0
p7 1114 1.0
p8 830 0.0
A. Award the place prize to p1 and p2 ($16 each).
B. Next, award the class prizes, beginning with the most inclusive class. In this case, no class is inclusive of any other; in fact, the three classes are pairwise exclusive, so it shouldn’t matter which class is calculated first. Starting with class B, p1 gets his prize upped from $16 to $20 (using $4 of the $20 class B prize), and the remaining $16 goes to p3, the next B player in line.
C. Likewise, p2 gets his prize upped from $16 to $20, and the remaining $16 of the C prize goes to p5 and p6 ($8 each). p4 wins nothing at this point, as D players are not eligible for B and C prizes.
D. Finally, award the D prize to p4 ($20).
As far as I can tell, this algorithm solves the discontinuity and anti-monotonicity problems of other methods, while still respecting the limit-one-prize-per-player rule. It also awards class prizes to class players, and is in harmony with the basic philosophy of class prizes (which is that class players can earn more than higher-rated players for the same score).
Additionally, this method seems simpler and less arbitrary than the artificial “standard” ways. Division by N is required only when there is an N-way tie for the same prize. And I have a hunch that it will be possible, more often than at present, for TDs to calculate some of the prizes even before all the games finish.
One and all, please cast your examples at this algorithm, and let’s see how it works. You can even try to break it, if you want to.
Bill Smythe