| How to average a traveller |
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Dear Martin, Thanks, Maury Reply Martin Sinot Dear Maury,I recently gave a detailed explanation about revoking. You can read it here. Unfortunately, it is not very easy to do it more simply, because it is a fairly large subject, as you can see from my answer there. It would help if you can give a specific situation you have questions about. The other question is about calculating the scores on a travelling score sheet. Assuming that we are talking about match point scoring, here is a method that works for any number of tables (although – admittedly – we would rather use a computer if we have a large number of tables).First recall that you get two points (US readers are accutomed to one point) for every score in your line that is worse than yours, one point (US: 0.5 point) for every score that is equal to yours, and zero points for every score that is better than yours. Then proceed as follows: - Order the scores from high to low. Also keep track of how many times a score occurs. Write the scores down with the frequency of occurrence next to each one. - Now add a third column, which will contain the awarded score. - Write -1 (minus one) in the score column BELOW the bottom line, and a zero in the frequency column next to it. - Then start scoring, beginning at the bottom, as follows: take the last score entered in the score column, add the frequency to the left AND the frequency in the row above that, and write the total in the score column. Then take this score, add the frequency next to it (yes, it is added twice!) plus the frequency one row higher, and write down the next score. Proceed to the top of the table. - Finally, enter the scores on the traveller (in the US: halving each one in order to get the matchpoints 1, 0.5, 0). - Remember that you need to do this for both NS and EW. However, once one column is completed, you can take a shortcut for the EW column, assuming that the director did not award adjusted scores: the sum of the NS and EW scores is equal to one top on the deal. A top on a deal is always equal to the number of pairs, rounded down to the nearest even number, minus two. So if there are 6 tables, hence 12 pairs, the top will be 12−2=10 (if there are only 11 pairs, the top is still 12−2=10). The EW score can then simply be found by subtracting the NS score from one top. An example to make things clearer: Suppose on a certain deal, NS have the following scores: −420, +50, −450, −420, −590, +50 (so EW have the complementary scores: +420, −50, +450, +420, +590, −50.Step 1: order the scores, keeping track of the frequency:
Step 2: add the score column, with −1 below the bottom line and a zero frequency:
Step 3: start scoring bottom up:
−450 gets 0+1+1=2 The table now shows:
−420 gets 2+1+2=5. By now I think you will see that +50 gets 5+2+2=9. So the table becomes:
Calculating the EW scores I will leave as an exercise to you. You can verify the results by subtracting the NS scores from the top (10); this should yield the same result. This is basic scoring, applicable if nothing special happens on a deal. Now suppose that one pair gets an artificial score (say, 50%). Let us say that the −590 in our example is replaced by this artificial score. That pair simply gets 50% of a top (10) which is 5 MP (2.5 in 1, 0.5, 0 scoring, US). For the other pairs, calculate the score without that pair:
But after this calculation, the scores need to be scaled up to the actual number of pairs. This is done by awarding one point (0.5 in 0.5, 0 scoring, US) for every comparison with the artificial score (there are other — better — ways, but they are best left to a computer). So our adjusted table becomes:
Finally you can insert the artificial score:
(In the US all scores will be halved). One final remark: the shortcut of calculating the EW scores from the NS scores is only valid if nothing special happens on the deal. If the director adjusts scores, then the NS and EW scores no longer balance. You need to calculate both lines separately then. Regards, Martin Sinot |