This concerns the distribution of prizes when some players are prize-limited and there is more than one “flight” of prizes. The correct method for dealing with players who are limited when there is only one prize schedule I think has been figured out (there are several mathematically equivalent ways to do that). Unfortunately, it’s not so clear when there are different eligibility sets. Take the following:
Prize Fund
Overall
1000
500
300
200
100
U1400
600
300
Results
5.5 A (U1400), B (subject to limit)
5.0 C (U1400), D (subject to limit), E, F
4.5 G (U1400),H
Without limits, A and B split 1000+600, C,D,E and F split 500+300+300+200 and G and H split 100.
With a 600 limit
A and B can still split 1000+600 with A getting 1000 and B 600. The rest is as before.
With a 500 limit
There are two distinct ways to handle this depending upon the source of the 1500 that A and B will split:
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They can split 1st and 2nd (A = 1000, B=500) which means that C gets the 600 U1400 prize and D, E and F split the 300+200+100, and G gets the 300 2nd place U1400 money, or
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They can continued to “split” the 1000+600, with the extra 100 being added to the 2nd overall money. C, D, E and F would split the (now) 600(augmented 2nd place)+300+300+200 and G and H would split 100.
To me, (b) is more principled (as much as possible you do the normal calculations, and flow unused money down). However, suppose now that the limit is 0. (As in the CCA on-line tournaments where people who did not do the extra monitoring were ineligible for prizes). The “obvious” way to handle that (basically ignore the ineligible players) is (a). A gets 1000, B nothing. C gets 600, D nothing, E and F split 500+300. G gets 300, H 200, with 100 left over for the next score group.
(b) applied to that situation would have A gets 1000. Extra 600 flows to 2nd place (take 500 to make it 1000—can’t go above the 1000) and 3rd (from 300 to 400). C, D, E and F would bring in 300 (U1400, 2nd), 1000 (now 2nd), 400 (now 3rd) and 200 (4th). D can’t take any, so C, E and F split 1700 (567 each), leaving 200. 5th place gets upped to 200 (can’t go above 200), and “6th” to 100. G and H will split 200+100.
Note that in (b), six players “take” seven prizes, which seems to be a fundamental error. However, it really is eight players taking six prizes, two just get calculated at zero. If the limit were $1 instead of $0, then you would have effectively the same calculation.
Where there is just one set of prizes, (a) and (b) are basically the same—whether you float players down until they hit the right spot given their limit or flow unused money down end up being mathematically equivalent. They aren’t, and in fact, can be quite different, when there are multiple flights of money.