How to handle score corrections when only a small portion of the field plays the board? (Neuberg)

00488172

During a recent online pairs event, BBO crashed and most of the field were unable to complete a board, through no fault of their own. The system crash, and subsequent delays for players getting reconnected, meant that one particular board was only completed at 3 tables (out of 9). The other 6 tables were all awarded 50/50 automatically by BBO, and inspection shows there was insufficient play at those tables to predict a valid outcome.

Normally it would be correct to apply a neuberg correction on the valid results. However applying this correction (through the BBO to XML converter) the results for the 3 tables that completed the board are still pretty extreme, with one NS pair getting 14 MP's (and E/W 2MP's) , another NS given 8 MP's (E/W also 8) and another N/S given 2 MP's (E/W 14).

This still feels unfair to me - without the neuberg correction the MP awards would be 16/0 8/8 and 0/16, so neuberg has hardly smoothed the MP awards for 4 pairs. On top of this, even the tables that did manage a result were distracted by delays and disconnections and messages fromn the TD, so those players might argue that the system problems affected the play.

This leads me to ask:
1. Is the neuberg correction valid when the actual number of results is small?
2. At what point is it better to simply award 50/50 for all players? (Is this legal?)
2. How would you recommend scoring this event?

Mark Humphris - Pinner Bridge Club

Robin_BarkerTD

After the section on Neuberg, the White Book has this, which I hope answers the question:

4.2.3.3 Small sub-fields
If the size of the group is at most three and is at most a third of the total number of results Neuberg’s formula is not used. Instead a group of two results is scored with a top as 65% and a bottom of 55%, and a group of three results is scored with a top of 70% and a bottom of 50%; with intermediate and tied results scored as for ordinary match-pointing.
The formula is used if A = 2 or A = 3, and E ≥ 3 A
Percentage = 60% + (M – (A – 1)) x 5%
Match points = ((M – (A – 1) + 12)/10) x (E – 1)
Note The small sub-field ‘formula’ awards scores that sum to 120% (AVE+ + AVE+) – this is as compensation to the pairs involved, for not getting a proper comparison.

00488172

Thanks Robin, very helpful