Wednesday 7 April 2010

Scoring the Matches

I've already told you about my event in Hamburg during Easter. Out of it I'm sharing the following problem with you: 
After the Round Robin in Alster Act 17 we ended up with a five way tie!
One sailor winning all his matches, five sailors winning four matches each and two sailor losing all (but one) their matches: The scoring looked like this:


The rules in the book have a solution for this tie.
As a umpire you will be frequently asked by the RC to help or check the scoring in a case like this. The first thing the RC needs are the four sailors who will be going on to the Semi finals. And then you must calculate the places 5 trough 8. Please leave your solution in a comment.

I will wait publishing comments for three days, so you need to find to solution yourself.
Oooh, there's one more data sheet you need to solve this: The pairing list of the Round Robin.
Good luck!


15 comments:

  1. I believe further information is needed to resolve this.
    I would like to know the order of races so I can apply C11.3.

    ReplyDelete
  2. Sorry Ignore my previous. I have just read all the words!! and found the pairing list.

    ReplyDelete
  3. My order
    Monin 1, Gurgel 2, Walser 3, Kemmling 4, den Boer 5, Meyer 6, Sehested 7, Nyberg 8.

    ReplyDelete
  4. 1 - Monnin
    2 - Gurgel
    3 - Kemmling
    4 - Meyer
    ----------------
    5 - Walser
    6 - den Boer
    7 - Sehested
    8 - Nyberg

    Don D.

    ReplyDelete
  5. First apply rule C11.1(a), calculate the scores among the five tied boats, excluding scores of those boats against boats not in the tie, thus:

    Gurgel, 1, 0, 0, 1, total 2
    Kemling, 0, 1, 1, 0, total 2
    den Boer, 1, 0, 1, 0, total 2
    Meyer, 1, 0, 0, 1, total 2
    Walser, 0, 1, 1, 0, total 2

    All five boats remain tied.

    Skip rule C11.1(b) because the tie is between more than two boats.

    Apply rule C11.1(c)

    All boats have 0 points against first boat Monnin.

    All boats are tied for second place.

    Shested is the next boat behind the tied boats, calculate points against Shested, thus

    Gurgel, 1,
    Kemling, 1,
    den Boer, 1,
    Meyer, 1,
    Walser, 1,

    All five boats remain tied.

    Skip rule C11.1(d) boats did not sail in different groups.

    Skip rule C11.1(e) there are no scores in a previous recent stage of the event.

    Apply rule C11.3, Assuming that a sail-off is not possible: Eliminate the score for the first race for each tied competitor

    Gurgel, 1st race F1M1 v Meyer, score 0,
    Kemling, 1st race F3M2 v Meyer, score 1
    den Boer, 1st race F2M1 v Gurgel, score 1
    Meyer, 1st race F1M1 v Gurgel, score 1
    Walser, 1st race F3M1 v den Boer, score 1

    Applying these adjustments to the boats rr score:

    Gurgel 4-0=4
    Kemling 4-1=3
    den Boer 4-1=3
    Meyer 4-1=3
    Walser 4-1=3

    Decides Gurgel in 2nd Place.

    Now start again at rule C11.1(a)

    Kemling, 1,1,0, total 2
    den Boer, 0,1,0, total 1
    Meyer, 0,0,1, total 1
    Walser, 1,1,0, total 2

    Now Apply rule 11.1(a) again between Kemling and Walser,

    Kemling, 0
    Walser, 1,

    Decides Walser in 3rd place, and Kemling in 4th place.

    Apply rule C11.1(a) beteen den Boer and Meyer,

    den Boer, 1
    Meyer, 0

    Decides den Boer in 5th place and Meyer in 6th place.

    Hope I got the principles and the arithmetic right.

    ReplyDelete
  6. Dear Sirs,



    In my opinion ranking will be this way;



    1- Monin

    2- Meyer

    3- Gurgel

    4- Walser

    5- Kemmling

    6- Den Boer

    7- Sehested

    8- Nyberg



    Thanks and best regards,



    Burak



    Tur-NU

    ReplyDelete
  7. First Gurgel(C11.3(a)) because he was the only looser in his first match.
    Walser and Kemmling are the next, with Walser winning against Kemmling. (C11.1(a) and C11.1(b))
    Then Den Boer and Meyer, with Den Boer winning against Meyer (C11.1(a) and C11.1(b))
    Uli

    ReplyDelete
  8. OoooPPPSS .. i just sow now the pairing list...
    Let me check if the result stands....

    ReplyDelete
  9. 1° Monin
    2° Gurgel
    3° Kemmling
    4° Walser
    5° Den Boer
    6° Meyer
    7° Sehested
    8° Nyberg

    ReplyDelete
  10. Jos,
    I've got a question about "eliminating the score for the first race for each tied competitor" (C 11.3 (a)).
    Does this refer to the first race of a competitor in that stage, regardless whether this race was against someone he is currently tied with?

    Thanks a lot

    Martin

    ReplyDelete
  11. First tie-breaker, according to C11.3(a), eliminating score for each 1st race for each tied competitor, results in Meyer being 2nd (after Monnin) and Gurgel 3rd.
    now to break the tie between the last three: applying C11.1(a), Walser will be 4th, Kemmling 5th, and den Boer 6th.
    OK, Jos?

    ReplyDelete
  12. @All
    This has been an interesting post - in so far that it shows that the rulebook is not easy in calculating the correct scores.

    Besides meticulously following the rules in C11 you also need to make sure you double check your arithmetic.

    Because none of the rules in C11.1 work between these five - and a sail-off was not possible - you need to go to rule C11.3
    Eliminating the the score for the first match of each tied competitor (regardless against whom they sailed) and then look at the points again.

    This results in Gurgel as the only one with two points; He's the second one to go through to the final stages. And leaves four. Again we look at these four without the others.

    By having two points, Kemmling and Walser go through and because Walser won his match against Kemmling he goes through as 3rd.
    Kemmling is the last who enters the final stages as 4th

    Den Boer beat Meyer; he becomes fifth, Meyer sixth. Finally of course Sehested and Nyberg.

    The one advice I can give:
    Please make sure you have at least two people independently checking this. It is very easy to make a mistake.

    ReplyDelete
  13. I previously posted as Anonymous (#11). my mistake was eliminating the 1st match between the TIED competitors and not REGARDLESS, as Jos pointed out.
    But that means eliminating, for Gurgel, the match with Meyer (1-0), leaving Gurgel with 4 pts, not 2? any comments?

    ReplyDelete
  14. @Yoram
    Gurgel has indeed 4 points if you count all. But all the tied competitors have 2 point from races against nrs 7&8, so I'm leaving those out for everybody.

    ReplyDelete
  15. This has been interesting and I almost had the same results as you came up with.

    I scored the 5 tied boats, but I did not eliminate the races that they sailed against the other boats they were not tied with when figuring out which was their first, second, third race under rule C11.2a. The rule says to break the tie by eliminating the first (then second, then third, etc.)race of the tied boats, but it does not say "between" the tied boats.

    I don't read the rule to say that you should not score those races, but you deleted the races they sailed with 7 & 8 because everyone beat them. However, you did not eliminate the races they sailed against Monnin, who beat everyone. Anyone who sailed early against Monnin thus has an advantage under this tie-breaking scenario.

    Don D.

    ReplyDelete

Related Posts Plugin for WordPress, Blogger...