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Small Sub-fields

Section of the White Book state that where there is a small Sub-field of less than 3 Boards or 1/3 of the total number of results (whichever is the smaller - I guess), a variation of the Neuberg formula should be applied:-

  MPs = ((M-(A-1)+12)/10) x(E-1)

in compensation for the pairs involved for not getting a proper comparison.

I want to confirm, or otherwise, that this applies when the size of the sub-field is just 1.

As an example let us assume that a Board has been played 8 and just one of the results is Fouled. Does this 1 Foulded Result get 7 MPs, or 7x1.2, ie 8.4 MPs?


  • They get 60% of a top. This is just straight out of Law 12C2. The small sub-field formula is a combination of this with Neuberg, based on the fact that with very few comparisons it is not uncommon for pairs to get a very unfair result with just a strict application of Neuberg.

  • Thanks for your reply, Gordon, however I am still confused.

    The White Book has this formula in it for dealing with Small Sub-fields:-

    MPs = ((M-(A-1)+12)/10) x(E-1)

    If I apply this formula to the case where E=8 (8 results), A=1 (just 1 result in the small sub-group), and M=1 (the MPs for this result before applying the Neuberg adjustment) the result is 9.1MPs. This is considerably greater than 60% of a Top (in this case 8.4MPs).

    However if there is a typo in the White Book and the formula should be written (with the "+" replaced by a "x") as:-

    MPs = ((M-(A-1)x12)/10)x(E-1)

    then the result is 8.4 MPs, which is 60% of a top.

    Could you comment on this please?

  • The White Books says, of the small sub-fields formula:

    The formula is used if A = 2 or A = 3, and E ≥ 3 A

    If A =1 then Law 12C2 applies.

  • edited November 2020

    Thanks, so applying your answers above to a situation where a Board which has 9 results, giving it a Max MPs of 16:-

    If 1 result is fouled, the MPs awarded to the fouled result are:-

    If 2 results are fouled, the MPs awarded to the fouled results are:-
    10.4/8.8 for the Top/Bottom
    9.6 if the fouled results are tied

    If 3 results are fouled, the MPs awarded to the fouled results are:-
    11.2/8 for the Top/Bottom
    10.4/8.8 for a Top=/Bottom=
    9.6 for a Middle (or all fouled results tied)

    With MPs for NS and EW summing to 19.2 which is 120% of 16, the Max MPs.

    Is this correct?

  • One result = 60%, two different results 65%/55%, three different results 70%/60%/50%; if results are equal then the match points are shared.

    In response to earlier comments on the forum, the draft 2020 White Book has

    8.87.3 Law 87B: Pairs or individual scoring
    If a score on a board cannot be compared with any other score, it is scored as if ‘no result can be obtained’, Law 12C2.
    Where scores can be compared with other scores, apply above Neuberg’s formula (§ or the Small sub-fields formula (§

  • edited November 2020

    @TonyHedge said:
    With MPs for NS and EW summing to 19.2 which is 120% of 16, the Max MPs.

    Is this correct?

    This is as expected by the White Book: Small sub-fields
    Note The small sub-field ‘formula’ awards scores that sum to 120% (AVE+ + AVE+) –

  • From your answers I deduce that my MPs above are correct.

    In the case of 1 fouled result: 9.6 MPs is 60% of 16 MPs (the maximum)

    In the case of 2 fouled results, there are 2 scenarios:-
    (a) A Top and a Bottom: 10.4 MPs (65%) and 8.8 MPs (55%) respectively
    (b) 2 Equal Results: 9.6 MPs (60%) for both

    In the case of 3 fouled results, there are 3 scenarios:-
    (a) Top, Middle and Bottom: 11.2 (70%), 9.6 (60%) and 8 (50%) respectively
    (b) 2 Equal Tops and a Bottom: 10.4 (65%) for the 2 equal Tops and 8 (50%) for the Bottom
    (c) 3 Equal Results: 9.6 (60%) for all 3

  • I have a situation which applies better (North American scoring). 8 tables, Mitchell movement, skip after round 4, one board was fouled after 4 teams had played it. So 3 teams got an average, and 4 teams got a real score, and then we applied the Neuberg (which works fine if only one average). 3 averages causing problems (see below). Seems to me that, since board was only played 4 times, the average should be 2 (not 3) for the 3 teams who didn't play it. But if I give them an average of 2, my summation check (last row) fails. So my question is, does 3 averages out of 7 boards qualify for the Small sub-fields of the EBU White Book, and should I use the formula given there?
    Pairings Contract By Result Scores Matchpoints
    N-S E-W N-S E-W N-S E-W
    1 2 Av 3.00 3.00
    2 4 Av 3.00 3.00
    3 6 Av 3.00 3.00
    4 1 3 NT N +3 490 3.88 2.13
    5 3 3 NT N +3 490 3.88 2.13
    6 5 3 NT N +3 490 3.88 2.13
    7 7 3 NT N +1 430 0.38 5.63
    ✓ ✓

  • The small subfield formula does not apply to averages, ave+/ave/ave- equate to 60%/50%/40% of the top based on the maximum times any board is played.

    The remanining four scores are too many for the EBU small subfield formula to apply

  • @BobZeeTD said:
    (North American scoring).

    For those not familiar with North American Match Point scoring, the Match Points are simply half what we award in the UK (as you may have surmised from the traveller). Hence we get the expression "Point-a board" for some teams events.

    one board was fouled after 4 teams had played it. So 3 teams got an average, and 4 teams got a real score

    Are you saying that the board was not played at all at the last three tables? If so, then the pairs unable to play the board should have had Av+ as they were unable to play the board through no fault of their own.

    I would presume, though, that they did play the board in the mis-boarded state, and that means that they are a sub-field of three tables, for which we apply the small sub-field formula for three tables. At least, that's what we do in EBU-land. If your use of North American Match Pointing means you are in ACBL-land, I don't know what regulations apply there.

    Barrie Partridge - CTD for Bridge Club Live

  • Thanks all.
    No, the board was not played at 3 tables, because it was an unrecoverable fouled board (16 cards in one hand, 10 in the next!). So only 4 pairs played the board. I use my homemade scoring program which is good & accurate but nothing fancy; Mitchell movements for all turnouts, some with skip, some with relay chair, some played with straight Mitchell. I use the Neuberg adjustment for unplayed board, which works fine if one team could not play the board. My problem is that, with 3 teams unable to play the board (and 4 who did), it seems like the maximum available for the board should be 3, representing the 3 other teams team "4" played against. So if the max is 3 (& min is 0), the average should be 1.5 for the 3 teams who did not play the board. I'm sort of OK with that, but for the other 4 teams who did play the board, notice in my example, the total of the N-S & E-W MPs is 6 (as if everybody played it), while the total for the 3 teams who didn't play the board is 4 (as if only 4 teams played it, which is what happened). Neuberg makes my head hurt! Seems like I need an adjusted Neuberg formula when number of teams who couldn't play a board is >1.

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