Is there a procedure for resurrecting an old topic? The topic with the OP below has a wide ranging discussion about computing prize distributions with players with limits. I would like to add to that. (I’ve done some recent work on that).
A gets $100 (max)
B and C share $750 (the most they could share if A was not involved) and thus each get $375
That leaves $50 for the 4-1 players.
Long logic and calculations:
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A takes $100 from first and pushes $400 to second ($650)
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second becomes the new first but cannot be grown to more than the original first so it is dropped to $500 with $150 going to third ($300)
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third becomes the new second but cannot be grown to more than the original second so it is dropped to $250 with $50 going to fourth (chaning it from $0 to $50)
Short logic and calculations:
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Find the place that is the smallest amount that is at least the amount of the limit and take $100 from the $150 third place leaving $50 for fourth
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The untouched first and second remain
Note that the short calculation logic and the long calculation logic come to the same numbers (automatic based on the long logic).
Additional note:
When there are place prizes and class prizes that a player is eligible for, use the type of prize that is linked to the largest unused prize (place prizes if the largest remaining place prize is the same amount as the largest remaining class prize). That allows consistent calculations in all cases.
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Yikes!
I would have just added the $400 to 2nd & 3rd proportionally.
1st: $100 - $500 = ($400)
2nd: $250 + (5/8) x $400 = $500 — 5/8 is 250 / (250+150)
3rd: $150 + (3/8) x $400 = $300 — 3/8 is 150 / (250+150)
edit:
wow that old thread was lively. A good read for sure!
The logic I gave also holds for up to a nine-way tie for first (nine is where the limit matches what everybody else would have received anyway).
32B3 (you cannot make more from a tie than you would have if you were alone at that score) makes it obvious why a 2-way tie for first is not $100 for the limited player and $650 for the non-limited player. That rule was designed for cases where class players tied for lower-level overall prizes but it is also available for limited prizes.
There was an ADM some time ago to modify 32B3 to also say that a group of people cannot make more from a tie with others than they would have if those others were not involved in that tie (the ADM was referred to committee and disappeared). It was designed for times such as where two masters and two experts were tied with 3rd ($300), 4th ($200), 5th ($100) and top Expert ($500) were left. A 4-say split is $275, which is more than the $250 the two experts get splitting the expert prize. However that means that the two masters get more than they would have if they had split only 3rd and 4th, even though it is less than what one master would have gotten if there was nobody tied. That leaves a bit of vagueness where players (or even organizers/TDs) can argue about the intent of the rule. In my humble(?) opinion I feel the intent of the rule is clearly met by the logic I gave but it might be time for that ADM to be re-introduced to clarify whether or not I am correct about that intent.
PS I noticed I didn’t give the two masters, two experts split. The masters would split 3rd and 4th ($250 each) and the experts would split 5th and expert ($300 each).
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I moved this topic to the running tournament category and updated the title.
The reason we close old threads after two years is that things can change over time, it is probably just better to start a new thread, even with the same question.
32B3. Ties for more than one prize.
If winners of different prizes tie with each other, all the cash prizes involved shall be summed and divided equally among the tied winners unless any of the winners would receive more money by winning or dividing only a particular prize for which others in the tie are ineligible. No player may receive an amount greater from the division of those prizes than the largest prize for which he would be eligible if there were no tie. No more than one cash prize shall go into the pool for each winner.
I saw a tortured reading of 32B3 in a case like this that had the two masters “eligible” for 3th, 4th and 5th (despite the fact that they can only take two of those) so the “unless…” clause wouldn’t apply. The correct logic is to let the masters eat first (taking 3rd and 4th for $250 each) and then give the experts a split of $500+$100. (“Particular prize” is also singular when it’s certainly possible for more than one “under” prize to be involved in a larger prize group).
This was the rewording of 32B3 (a Bob Messenger redraft of an idea of mine which seems to have gone nowhere) which eliminates the use of the confusing word “eligible” and makes clear that prize calculations are done looking at groups, not individuals.
If winners of different prizes tie with each other, all the cash prizes involved shall be summed and divided equally among the tied winners unless any of the winners, or group of the winners, would receive more money by being held out of the tie group and winning or dividing only the prizes that would remain after distributing prizes to the other winners. No more than one cash prize shall go into the pool for each winner.
I would probably alter the final sentence to “No more than one cash prize shall go into any pooled prize for each player in that pool.” just to clarify that that applies not just to the score group as a whole, but to any of the subsets that are analyzed.
I consider the following to be axiomatic (basically for any top-to-bottom score-based distribution of a prize fund)
- A player or group of players cannot win less than they would if they had lower scores.
- A player or group of players cannot win more than they would if they had higher scores.
In the Master/Expert example, a four way split is $275 each, but if you add .1 to the scores of the Masters they would split 3rd and 4th for $250 each so that would violate #2. You get the same if you drop the experts .1 points (they would clearly get $300 each if they had the lower score so #1 would be violated). Note that this is not the rule but is the principle that governs how the subsets within a score group can be handled.
In the previous thread, there were several descriptions of methods consistent with the rules for dealing with prize distributions where some players are subject to limits. These generally seem to produce the same results in most (all?) cases but I’m not sure they all work for situations if there are multiple players with limits, and some (which involve spreading unused prize amounts to the smaller prizes) can be really complicated. The following describes a process which works in more general cases and doesn’t require creating or adjusting prizes.
Obviously this only applies if there are players who would naturally win more than their limit. What the process does is to “downfloat” players who win their limits to a score group where the marginal prize is no larger than the limit. (If you have only one player who hits a limit and that limit is spot-on one of the prizes, you can just assign that to that player. This works even when there is more than one limited player or the limit isn’t an exact match to a prize).
To apply to the case in the original post:
Player A: prize limited to $100; 5-0 Player B: not limited; 4 1/2 - 1/2 Player C: not limited; 4 1/2 - 1/2 Guaranteed prizes: 1st $500. 2nd $250, 3rd $150
5.0 Score group. A would naturally bring in $500. $500 is bigger than the $100 limit. Give out no money at this point. Float A down to 4.5 but mark him as getting $100.
4.5 Score group. A+B+C would naturally bring in $500+$250+$150. $150 is still bigger than $100. Float A out of 4.5. Redo the calculation with just B+C who split $500+$250.
4.0 Score group. A+D+E+F… would naturally bring in $150+$0+$0+… $0 < $100 so this is where A’s money will be distributed. A gets $100 of the $150 so everyone else in 4.0 splits the remaining $50.
As a more complicated example, suppose prizes are 4000-2000-1000-700-500-400-300-300-300-300. Unrateds are limited to $400. Provisionally rated are limited to $1000.
6.5 A (limit $1000)
6.0 B (limit $1000), C(limit $400), D (unlimited)
5.5 E (limit $1000), F, G, H (all unlimited)
Score 6.5 A naturally brings in $4000 > $1000. So mark as $1000 and float down. Distribute no prizes.
Score 6.0 A+B+C+D naturally bring is $4000+$2000+$1000+$700. $700 > C’s limit of $400 so mark C to receive $400 and float down. Redo without C. A+B+D naturally bring in $4000+$2000+$1000. A+B both have $1000 limit. The average of the marginal $2000 and $1000 is $1500>$1000 limit (you have to deal with everyone with a given limit together). Mark B to receive $1000 and float A+B down. Redo the calculation with just D who gets $4000.
Score 5.5 A+B+C+E+F+G+H bring in $2000+$1000+$700+$500+$400+$300+$300. The marginal $300 is now less than C’s $400 limit and $500+$400+$300+$300 is less than the total of limits for all four limited players so this is where all the limit players are handled. A and B get their $1000, C $400. E+F+G+H split the remaining $2800 so $700.00 each (E doesn’t hit his limit).
Note that if the unrated limit were $200 (rather than $400), you would float C down further until you finally hit a score group where have you run out of prizes (so the marginal prize is $0).
Doing it manually I’d say that A takes the $1000 prize (the smallest prize at least as large as the limit) and leaves 4000, 2000, 700, 500. 400. 300x4
Then B takes 1000 of the 2000 (the smallest prize of at least the limit), C takes 400 (the smallest prize of at least the limit) and D gets the 4000, leaving 1700 (the 1000 left over from 2000 added to the 700 that was the next prize - since the smallest prize at least as large as the limit was used the rollover is automatically no larger that the prize it rolled over from), 500, 300x4.[if C was limited to 200 then it would leave 1700, 500, 400, 300x3, and a newly created leftover of 100]
Then E-H split 1700+500+300+300=2800 four ways giving 700 each and leaving 300x2 (in your example you overlooked including E in the final split - note that the 1000 prize limit was irrelevant for that calculation). [if C was limited to 200 then 1700+500+400+300=2900 would have been split to 725 each leaving 300x2, 100]
It sounds like your logic would work just fine and your program would not have made the very human error of missing E. My shortcut is a simple way to manually reach the end result of your programming with less calculation (actually the same underlying calculations with the shortcut doing multiple steps at once). Shortcuts are useful when using good human judgement for choosing the prizes to allocate. Computer calculations are safe rote ways of doing it without needing human judgement that is good.
Yes, I corrected that to include E (which doesn’t really change much since E doesn’t hit his limit).
I know that was how some of the CCA TD’s described how they did it but I find that really hard to describe how the excess gets added to the other prizes. That’s where you get the people taking $3000 out of the $4000 first prize and adding it to the $2000 to get $5000 (that’s literally what “Balance of any limited prize goes to next eligible player(s) in line” means). You aren’t supposed to do that since that creates a prize bigger than the original maximum prize of $4000 so you instead take $2000 to make $4000 then the extra leftover $1000 to add to the $1000 to get $2000. (But if that prize had been $1800, you would take $800 to make $1800 and then rollover the extra $200 to augment the 4th prize). Yikes.
It looks complicated but in practice it is simplified by going to the smallest prize at least as large as the limit and adding any excess to the next prize. Since the next prize is less than the prize chosen the roll-over cannot exceed that prize. It is a way of automatically incorporating all upstream roll-overs to that point. It is mathematically equivalent to how you are programming it and easier to manually implement (as opposed to the also mathematically equivalent method of meticulously doing each and every roll-over step starting from the top).
Also if the prizes were 4000, 2000, 800 and the limit of sole first was 1000 then you would take 1000 from the 2000 and roll it into third place - leaving third place at 1800 (still not exceeding the 2000 the roll-over came from). I am guessing that if you run your program with a limited 6-0 and two 5.5-0.5 tied for second/third then it would have 1000 for first and the other two getting 4000+1800=5800 split to 2900 each.
Jeff typed. . .
“It was designed for times such as where two masters and two experts were tied with 3rd ($300), 4th ($200), 5th ($100) and top Expert ($500) were left. A 4-say split is $275, which is more than the $250 the two experts get splitting the expert prize. However that means that the two masters get more than they would have if they had split only 3rd and 4th, even though it is less than what one master would have gotten if there was nobody tied.”
Why can’t the 2 masters and the 2 experts each get $150[($300 + $200 + $100) ÷ 4] each while the the 2 experts additionally get $250(500 ÷ 2)? How can 2 masters win any top experts prize?
Ken, the two masters took 3rd and 4th (for $250 each) while the two experts took expert and 5th ($300 each). The $150 and $400 split gives each expert 1/2 of one prize and 3/4 of another (well three prizes of 1/4 each) for a total of 1.25 prizes while the rules limit a player to no more than one prize each. 32B1 explicitly prevents that 1.25 prize allocation.
It’s important to not think about what mix of prizes each player gets when you do a pool and average calculation. If the prizes are such that an even split is correct (change top Expert to $300), then you get ($300+$200+$100+$300)/4=$225 each. The masters share a prize pool including the expert prize so the experts can share in the place prizes. (The masters don’t “win” expert money. Their contribution to the pool is the place money).
A helpful way to look at this is that the $900 pool consists of $600 of orange (place) money and $300 of blue (expert) money. The $225 for the experts can be paid out as $150 each in blue money and $75 each of orange money. The $225 for the masters can be paid out as $225 each all in orange money.
I’ll go further and say that the $900 pool consists of $500 in orange (3rd/4th place) money and $100 in red (5th place) money and $300 of blue (expert) money. Any payout to the two masters is limited to the two greatest prizes the masters bring into the split. Thus $225 each consists of $225 orange money for the two masters and $25 orange, $50 red and $150 blue money to the two experts who also tied for 3rd. If expert is $500 then there is $500 orange, $100 red and $500 blue. A four-way split at $275 would have the masters each getting $250 orange (fine) plus $25 red (not fine, that money was not brought in by them) while the experts also get $250 blue and $25 red. Seeing that red money in the non-expert hands is a tip-off that the money brought in by the experts went to the wrong players.
Is there any way to word the TLA so that the 2 masters and 2 experts split the 3rd, 4th, and 5th monies ($600 ÷ 4) evenly and then the 2 experts split the ($500 ÷ 2) for top expert? Otherwise, having to split ($300 + $200 + $100 + $500) ÷ 4 = $275 each seems unnatural. Having to know how to properly pay winners of different calibers may be not natural knowledge, and I don’t want to pay each winner inacurately.
I agree that the two are mathematically equivalent if there are only place prizes. If there are under prizes (or any other non-place prizes), they may not be as your prize adjustment process will change prizes before they would naturally be used. That wouldn’t affect most CCA tournaments since the limits are usually in ratings-defined classes without under prizes.
There is probably a way by treating it as a major variation and putting it in the TLA and all publicity. Realize that if an expert comes in sole first then that expert gets all of first (masters are used to that risk) and all of top expert (severely annoying those experts who accepted that master-strength players would get the top money while they fight for expert money). Use class prizes rather than under prizes to (partially) safeguard against a severely underrated C player (such as an Illinois K-8 scholastic player who got a 1500 rating and then stopped rated chess while continuing to play IL High School Assn non-USCF chess with a very strong coach) could end up taking first place, top U2200, top U2000, top U1800 and top U1600.
Properly paying players in different rating classes is in Rule 32. Different people have different ideas about what is natural so the rule was written so that everybody followed the same logic.
PS I heard that the reason Rule 32B3 was written the way it was is due to a major tournament where an expert tied for top expert and 8th to 10th places with four or five masters and GMs. All of the money was pooled together and divided equally, leaving the expert with less money than if he had scored a half point less and thus been allotted all of first expert.
That is why I said earlier that you take the prize limited player and run through either the place prizes or the class prizes depending on the largest place or class prize still available for the player. If the top prize in the eligible class is $1000 and the top place prize is at least $1000 then run through the place prizes. If the top place prize is less than top prize in the eligible class then run through the class prizes.
This is relatively straightforward when using the manual method and shortcuts and should be programmable where the carried forward players are not only tagged with their prize limit but also the class (overall being a class) that they were initially assigned to for prize purposes.
This is the bottom part of my first post.
Additional note:
When there are place prizes and class prizes that a player is eligible for, use the type of prize that is linked to the largest unused prize (place prizes if the largest remaining place prize is the same amount as the largest remaining class prize). That allows consistent calculations in all cases.
32B1. One cash prize per player.
No winner shall receive more than one cash award. The award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others. Prizes such as biggest upset, best game, or brilliancy are standard exceptions from this rule. Any other special prizes should be announced and designated as such. A clear winner of more than one cash prize must be awarded the most valuable prize.
I would strongly urge you to learn how to do the prize allocations based upon the standard US Chess rules. You could make the under/class prizes “bonus” prizes but that’s contrary to practice. In a typical US Chess tournament, the prize money comes from entry fees provided by players at all skill levels (rather than coming from sponsors as is the case in many professional individual sports). The use of (not insubstantial) under prizes allows the players who are actually providing the prize fund to play for some of that. If you have a 40 player one-section tournament, instead of five place prizes (say $500-$300-$200-$100-$50) which would all be won by the strongest players, you can do the same prize fund at $400-$250-$100 U2200 $100 U2000 $90 U1800 $80 U1600 $70 U1400 $60. This is still fairly top heavy (the place prizes are about 2/3 of the prize fund) but spreads the money to more players at more skill levels. Bonus prizes (such as biggest upset) are usually for modest amounts compared with the other prizes (maybe $25-$50 in this case). Even though you could do the class prizes as bonuses (which you would have to announce in publicity and probably be very very specific about since it is so far out of US Chess standard), you would not want to do it in the case described above where the expert prize appeared to be on the level of maybe 2nd prize. $50 is one thing; $500 is completely different.
Note that this thread is about how to do prize calculations where there are players who have an upper bound on what they can win, which is much more complicated and (as you can see) not entirely worked out. Calculation of prizes by US Chess rules (at most one prize into the pool per player, give average unless some group within the score group can do better by letting the others go first) is actually not all that complicated. Just don’t try to interpret 32B3 too literally and don’t, repeat don’t, think that the masters are getting paid expert money. For instance with:
$400-$250-$100 U2200 $100 U2000 $90 U1800 $80 U1600 $70 U1400 $60
how would you do
5.0 Able(2050)
4.5 Baker(2307),Charley(2210),Dog(2030)
4.0 Easy(2250),Foxtrot(2105),George(1975),How(1875)
3.5 Item(2170),Jig(1935),King(1730),Live(1695),Mike(1530)
3.0 Nan(1775),Oboe(1599),Peter(1420),Queen(1355),other people