Assigned Adjusted Scores (WB 4.1.1.2) and Underplayed Boards
Am I correct in my assumption that if a TD adjusts a score on a board as discussed in section 4.1.1.2 of the White Book, but the Board is underplayed as discussed in sections 4.2.3 and 4.2.4 of the White Book, then then for the purpose of calculating MPs and XIMPs, the points awarded to the result are adjusted according to Neuberg and the corresponding XIMP formula?
Comments
It sounds right, but I'm less than absolutely certain what situation you are asking about. Could you give an example?
I hope this example is not too complex.
Consider a T9 R12 1W Pairs Event (3/4 Howell Movement) with a Missing Pair in Round 1 at Table 5, EW.
11 of the 12 Rounds were played, with Round 11 being skipped; this leads to some Boards being played 7 times and others 8 times.
This is in fact the Event that was played at Goring & Streatley on 20th May 2019. You can see it on Bridgewebs. It was scored using MPs.
Board 1 was played 7 times, some other Boards 8 times. Board 1's results therefore need adjustment up to 8.
If the TD was to make an Assigned Adjustment to the NS7-EW3 Result on Board 1, making the score, say, +400 instead of +460 (for whatever reason), would the MPs assigned to the this result need to be adjusted by the Neuberg Adjustment (because the board was underplayed)?
By the way, I tried to upload some .jpg files to illustrate the above situation, but they all failed to upload - is there a problem with this facility or am I doing it incorrectly?
For these purposes, an assigned adjusted score is exactly the same as the original score it replaced. If the board needed to be Neuberged because it had fewer results than some others, as in your Board 1, then this applies whether or not any of the results were assigned.
If there is an assigned adjusted score (a single score) this replaces the score obtained at the table and the conversion to match points and any further adjustment for missing scores is done with those results.
if there is an assigned adjusted score which is a weighted combination of scores then you can think of the process is two ways (which give the same answer)
1. Calculate the match points for the group of scores, including the weighted scores, and then apply neuberg to those match points to get fhe final match points.
2. Scale all the frequencies for results including the weighting and then calculate the match points; if there is one result missing, and six table results and an assigned score of 60% X and 40% Y, then the scaled frequency for the six results is 8/7 and the scaled frequencies for X is 0.6x8/7, and Y is 0.4x8/7
Thank you Robin, I'm treating an Assigned Split score (WB 4.1.1.3) in the same way as a Weighted and Split Score with NS getting 100% of one score and EW getting 100% of the other score.
I think this is case 1 above.
I wish to return to this topic having seen a result published on Bridgewebs for a BBO Tournament in which an Artificial score was assigned to one Result on one Board. All the Boards were played 11 times, except Board 2, in which 1 result was adjusted and assigned an Artificial score.
The Board (2) in question had the following scores published on Bridgewebs:-
NS EW Bid By Tks +Sc -Sc + -
1 11 4S W +3 0 510 15 3
3 13 4S W +3 0 510 15 3
4 19 4S W +3 0 510 15 3
1 20 4S W +3 0 510 15 3
8 21 n/a 50% 50% 10 10 Artificial Adjusted Score
2 22 3NT W +4 0 520 9 9
5 18 3NT W +4 0 520 9 9
6 16 6S W +1 0 1010 3 15
10 15 6S W +1 0 1010 3 15
9 12 6NT W +1 0 1020 2 16
7 14 7S W 0 1510 0 18
Am I correct in assuming that the MPs awarded above are incorrect because the 10 'non Artificial' scores have not been adjusted up according to the Neuberg formula and that the correct MPs should be:-
NS EW Bid By Tks +Sc -Sc + -
1 11 4S W +3 0 510 16.6 3.4
3 13 4S W +3 0 510 16.6 3.4
4 19 4S W +3 0 510 16.6 3.4
1 20 4S W +3 0 510 16.6 3.4
8 21 n/a 50% 50% 10 10 Artificial Adjusted Score
2 22 3NT W +4 0 520 10 10
5 18 3NT W +4 0 520 10 10
6 16 6S W +1 0 1010 5.6 14.4
10 15 6S W +1 0 1010 5.6 14.4
9 12 6NT W +1 0 1010 2.3 17.7
7 14 7S W 0 1510 0.1 19.1
While Neuberg scoring is an option in BBO2XML and might not have been selected, the first matrix looks wrong because the table scores do not add up to 20, except for the average. With simple (non-Neuberg) matchpoint scoring I think they should be
16 4
16 4
16 4
10 10
9 11
9 11
5 15
5 15
2 18
0 20
Your adjusted ones seem correct, though I've just looked at them by eye and not made any careful checks.
You can tell whether the Neuberg option was selected in BBOtoXML by looking at the software version number under the ranking list on BridgeWebs. If there is a lowercase "n" after the version number, e.g. 1.3.10/w1.6.5n, then that means that Neuberg was selected.
If Neuberg is not selected then BBOtoXML matches the BBO results. This is intentional. If there is a 50/50 score on the traveller it appears that BBO does not add extra matchpoints to the other rows on the traveller, i.e. as far as the other entries are concerned the matchpoints are allocated as if one fewer table played the board. This is the reason the matchpoints for the other rows do not add up to 20. Note that BBO does not show matchpoints on the traveller, but if this assumption is made in the BBOtoXML code then the percentage results calculated by BBOtoXML for the travellers match the percentages displayed by BBO on the travellers.
As EBU expect Neuberg to be used for computerised scoring, it should be selected in BBOtoXML for virtual tournaments for EBU affiliated clubs.
I should qualify this statement by saying that, when Neuberg is not selected, BBOtoXML only uses the BBO style of non-Neuberg scoring that I described above if the event was run on BBO and has not been merged with another event. If it is a merged file, or is derived from an imported XML file (which might have been played on a different online platform or face to face) then the normal form of non-Neuberg matchpoint scoring, as in the post by @gordonrainsford, is used.