The guidelines for a scholastic swiss tournament state that all players in a section who score at least x points in a y-round event will receive a trophy. There is a limit of one half-point bye. And of course, if there’s an odd number of entries, there could be a full point bye awarded to someone every round.
Does anyone have a formula to calculate in advance the maximum number of trophies an organizer would need available to award, once the number of entries is known?
Thanks!
Grant Neilley
I’m not sure a cut-and-dried formula is possible.
Suppose you have a 4 round event and your criteria is that everyone with 1.5 or higher gets a trophy.
The worst case scenario would be if every game was a draw, in which case everybody is tied at 2 points and gets a trophy.
On the other hand, if your cutoff was 2.5 points to get a trophy, then nobody gets one.
But since most games are not drawn, there should be a distribution around the midpoint (ie, 2 points in a 4 point event). My math is a bit rusty here, wouldn’t this be a binomial distribution?
I can probably pull some data from the ratings system showing the typical spread of
points, which will give you some kind of min/max range on trophies based on typical events.
Can you give me some idea what size of event you are looking at? 20 player events are likely to be ‘lumpier’ than 100 player events, for example.
I think limiting it to mostly scholastic events might be necessary because there tend to be fewer draws by lower rated players.
Among players rated 400 to 1100 about 8% of the games are drawn. For players rated 1200 or higher, the percentage of draws increases in a very linear way as their rating goes up, 1500 players draw about 13% of the time, 2000 players draw about 19% of the time, 2300 players draw about 22% of the time, and 2600 players draw about 31% of the time.
Oddly enough, when I look at players rated below 400 the percentage of draws goes up to about 11%. Could this be due to stalemates?
Bill Smythe’s model for plus score tournaments might also suit your needs. Bill has written at length about those events in another thread some time ago.
Tim
OK, here are two tables covering events since 2006 with from 3 to 7 rounds showing the percentage of players in each final score group. (I’ve taken out events where the number of players is close to the number of rounds, because events like quads could show up as either 3 or 4 round events depending on how the TD submitted it.)
Keep in mind this represents data aggregated over a fairly large number of events, so a single event could deviate a bit in one direction or the other.
This first table is only for events with no players rated above 1000 in them, so it should include mostly scholastic events:
[code]rounds count 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7
3 337 10.1 5.7 30.0 11.8 27.3 5.2 9.9
4 1279 4.6 3.1 17.9 9.9 28.9 8.8 19.1 2.9 4.9
5 1148 2.8 1.9 9.3 8.7 21.4 11.9 22.3 6.9 10.5 1.7 2.6
6 67 2.1 1.9 6.7 6.0 15.4 11.8 20.0 9.1 14.9 4.0 5.9 0.8 1.4
7 15 0.3 1.0 4.0 5.1 9.2 9.3 17.5 10.6 17.7 7.4 10.8 2.6 3.2 0.8 0.4[/code]
This next table is only for events with no players with an established rating under 1200, so it should include mostly adult events:
[code]rounds count 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7
3 1027 10.5 6.6 27.8 14.5 24.9 8.8 6.9
4 1430 5.2 4.7 16.3 14.7 24.0 14.1 13.2 5.2 2.7
5 848 4.3 2.9 9.5 10.3 18.5 16.1 17.4 10.4 7.0 2.7 0.9
6 213 3.5 2.3 6.7 6.9 12.7 13.4 16.7 14.4 11.6 6.6 3.8 1.0 0.4
7 70 4.5 2.4 6.2 4.3 8.9 9.3 14.9 12.1 14.7 8.5 7.6 3.6 2.4 0.5 0.1
[/code]
The statistics appear to be a bit bottom-weighted, especially for the adult events, I assume that’s because of dropouts and re-entries
The tables are interesting, thanks.
I figured some of the math whizes here might like the challenge to (or already did) develop a formula, but maybe there are too many variables for that?
The specific scenario I was trying to figure is a 6-round event for scores of 4 points or more, with 50-100 players per section. I was doing some trial and error stabs at different scenarios to maximize the number of qualifying scores, but my head started hurting. 
For example, I thought if everyone drew the first 4 games, everyone would stand at 2 points. If half the field won game 5, then half of those won game 6, 25% of the players would finish with exactly 4 points. I thought this would maximize those who qualify, yet in a real life tournament I looked at, about 30% finished with 4 points plus. So something’s wrong with my logic. Is there a scenario that would work?
Ouch, my head’s hurting again.
Grant Neilley
If what you’re looking for is a guarantee that you won’t have more people qualify for trophies than the number you order without having too many extra, that’s probably not possible.
However, from the first table you can see that on average 27% of the field will score 4 or more in a 6 round scholastic event.
Thus with 70 players you should order at least 19 trophies, a few more would give you some cushion.
Here’s another math quiz:
What’s purple and commutative?
What’s green and very very far away?
What’s yellow and equivalent to the Axiom of Choice?
Bill Smythe
Here’s another results table. This one includes events since 2004 with no players higher than 1499 and where no players had more than one unplayed game. (That should largely eliminate the dropout/re-enter factor while allowing for an odd number of players and the occasional half point bye.)
[code]rounds count 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7
3 1055 9.1 3.8 27.6 10.3 32.6 5.5 11.1
4 2994 3.7 2.5 16.1 8.1 30.4 9.0 21.4 3.4 5.3
5 2123 1.5 1.5 7.9 6.4 22.5 10.5 26.0 6.7 12.2 2.0 2.7
6 89 0.7 1.1 4.5 4.6 14.2 9.2 23.0 10.6 18.6 3.8 6.8 1.1 1.9
7 9 0.0 0.4 1.8 2.7 9.7 7.5 20.3 9.1 20.8 7.1 12.7 3.2 3.7 0.7 0.2[/code]
In this table, 32.2% of the players in a 6 round event scored 4 or more points, so the number of trophies predicted for a 70 player event would be 23.
Here’s the equivalent table for events which had no established players under 1000 in them:
[code]rounds count 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
3 1611 7.8 4.9 26.7 13.4 29.7 8.7 8.7
4 1308 3.5 2.4 14.2 10.5 28.1 13.2 18.4 5.6 4.0
5 278 1.7 1.3 6.9 6.6 19.4 13.9 23.9 10.7 9.9 3.2 2.4
6 12 1.7 0.3 5.2 2.4 15.2 12.5 17.6 16.3 16.3 5.2 4.5 2.1 0.7 [/code]
If anything this last one is a bit top heavy.
These are tough! Perhaps #3 is a reference to Zorn’s Lemon?
Jim
and #1 would be an Abelian Grape.
Jim
Since you asked for the maximum number (i.e., the worst case), statistics won’t help you. Alas, I’m too tired to iron out the last wrinkle - but I’ll give you an approximation that should be very, very good.
It seems to me that the limiting factor is Swiss pairings. We can derive an expression for the worst case assuming incompent pairings…but those are unduly pessimistic.
As already noticed, if the target score (your x) is y/2 or lower, then ALL of the participants might qualify (all games drawn).
As x gets bigger, you can use the formula:
Players / (2^(2x-y))
As an example, suppose you have 50 players in a 5 round event. Then this formula gives:
Qualifying score ................. #players who might qualify
5.0 ................................... 1.5625
4.5 ................................... 3.125
4.0 ................................... 6.25
3.5 ................................. 12.5
3.0 ................................. 25
2.5 ................................. 50
Round up, to allow for the odd perfect score being paired down.
These estimates are probably very good at the top end, and increasingly pessimistic as you approach an even score.
As you can see, you don’t need a fancy formula to compute this. Just set the #of possible qualifiers to be equal to the entire field at an even score, and then divide by 2 for each decrement of 0.5…until you get to 1.
A “plausible scenario” for each score is that you play a perfect 1/0 Swiss (all decisive games) for some number of rounds, and then all games are drawn (in the top score group).
And, here is where we see that the above is only an approximation. While the top score group is drawing out to the end of the event, players from a lower score group CAN catch up.
So…the “pessism” I noted above is fortunate; perhaps it counterbalances the second-order effect of players moving up after an initial loss. Given time, we could probably get a closed form for this - but I think that the table above is adequate for tournament planning.
Good luck!
This thread demonstrates two things to me: 1. Its much easier to simply use tie breakers and award trophies to places rather than point totals; and 2. Math is certainly not my forte (it is good to be reminded of that once in awhile).
There are various theories about how many awards to give out at scholastic events. Some people think everyone should go home with SOMETHING.
The Lincoln Chess Foundation used to have ribbons with a chess piece printed on them that went all the way down to 25th place. If someone finished below that, we had ‘participant’ and ‘honorable mention’ ribbons too.
We once discussed printing up ribbons with 1 point, 1 1/2 points, 2 points, etc, but the trophy shop we were getting the ribbons from had closed and the only source we could find wanted more than twice as much per ribbon.
The ‘Participant’ ribbon always sounded kind of dorky to me, but one of our board members was into equestrian events, they tend to find ways to give everyone 1 or more ribbons at each event. We have a friend whose son is in junior figure skating, they have a trophy case full of ribbons too. (Now that I think of it, my boy scout troop’s flag had about a 3 inch thick stack of ribbons won at various jamborees, too.)
Other ideas we discussed:
Keeping track of cumulative wins in our events and giving out ribbons at 5 wins, 10 wins, 15 wins, etc. (This was many years before MSA.)
The idea I still wish we had tried was giving out chess pieces, white pawns one time, black pawns the next time, etc, so that after several events someone had a complete tournament quality set. (We could get decent sets and boards for under $8 in quantities of 100 or more.)
One helpful tip: little brass tags are cheaper than entire trophies - and you can pry them off and apply new ones in seconds.
another: if you run a series of events, you can use the same style of trophies and “roll over” excess trophies from one event to the next.
But, yes. Tiebreaks are unfair - but they work.
Also, in my experience, kids learn RealFast that “participation” awards are “loser awards”. By giving out a limited number of very nice trophies (instead of a box full of trash) you demonstrate that excellence is rewarded. Parents take a long time to learn this - but the kids get it right away.
Excellent. For #2, think projective geometry (and fruit, of course).
Bill Smythe
That sounds like a cool idea, except most kids already have a chess set. And To accumulate the entire 32 pieces they will have have to play in 32 tournaments, which in most kids we’re talking about 2 to 3 years.
But having the chess piece on a Key chain is more plausible. I’ve seem many kids proudly adorned their backpacks with the chess key chains.
We run a series of tournaments during the school year. There are trophies for top 5 in each section. Tie breaks are used. Anyone scoring 3 points who isn’t in the Top 5 gets a trophy. We have plates and generic trophies for those extra place trophies. 2.5 scores get a medal. In the K-1 section all non trophy winners receive medals.
If you look at a cross table from one of our larger events last year you’ll see that our 3 points scorers generally make up around 10-20% of the section.
If you have a 5-round 48-player tournament then you could have 33 players at 3-2.
After round one you could have 24 1-0 and 24 0-1 scores.
The 24 1-0 could draw out to reach 3-2.
The 24 0-1 would all play decisive games. The 12 1-1 could play decisive games, giving you 6 2-1 (drawing out to reach 3-2) and 6 1-2. The 12 0-2 could give you 6 0-3 and 6 1-2.
The 12 1-2 could play decisive games to bring in another 3 3-2 scores.
The 33 3-2 scores consist of 24 players who started 1-0, 6 players who started 2-1 and 3 players who started 1-2.
At 4-1, with no draws in the first three rounds, you could have 6 players who started 3-0. The 18 2-1 players can give another 5 that can reach 4-1. Thus there could be 11 4-1 players in a 48 player 5-round tournament.
Accelerated pairings with a round two where the lower half round one winners beat the upper half round one losers could increase this to 12 or 13 (a 96 player tournament could have 26 4-1s, but in a 48 player tournament there are cross-score-group pairings that muddy the waters).
Back when we were considering the serialized chess set participation prizes, the sets most young players were bringing were those really cheap boxed sets. I suspect that’s still the case here, though the only event I run this days supplies sets and boards, but I still see mostly junk in the skittles area.
Also, the idea was that at event 1 they’d get 8 white pawns, at event 2 they’d
get the black pawns, etc, until they had a set and a board. We figured this would add less than $2 to the entry fees we needed to charge. Our data also suggested that most players who made it to 5 events kept on playing. (That still appears to be the case when I look at data from MSA.)
Ken Sloan wrote:
Thank you, that’s exactly what I needed!!
Grant Neilley