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Formula for calculating average cross-IMPs

#1 User is offline   Maponos 

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Posted 2012-October-26, 23:59

For cross-IMP pairs play, it appears that BBO's software calculates a pair's score on a board by dividing their total cross-IMPs by the number of tables against which their score has been compared (one less than the number of tables at which the board has been played). While this seems like the right thing to do, it turns out that the correct result is obtained by dividing total cross-IMPs by the total number of tables (including that pair's own table).

To see why, consider the case of only two tables. Suppose that one table makes 3NT for +400, and the other goes down one for -50. A difference of 450 is worth 10 IMPs, so one table will score +10 cross-IMPs and the other -10. Therefore the pair who made will score 20 IMPs more than the pair who went down, which is twice as many as they are supposed to score. Clearly we must divide by 2 to obtain the correct difference. It is easy to verify that this result generalizes to more than two tables.

Of course, on BBO each board is usually played a large number of times, so it make very little difference whether you divide by n or n-1. But the correct formula is just as easy to calculate as the incorrect one, so why not get it right?
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#2 User is offline   jbaptistec 

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Posted 2012-October-27, 12:39

I think this example might be better.

Suppose we run a tournament with 2 sections : one section with only two tables, one section with 10 tables.
On some deal, pairs A1 (in section 1) and A2 (in section 2) score 3NT when all the rest of their respective fields went down.

Should both pairs score 10 cross-IMPs ? Or should A1 score 5 IMPs and A2 10 IMPs (EDIT : actually, only 9), since A2's result was compared to more tables ?
Tired of red/black ♠♥♦♣.
For 4 suits, why not 4 colors ?
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#3 User is offline   barmar 

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Posted 2012-October-27, 20:39

View PostMaponos, on 2012-October-26, 23:59, said:

To see why, consider the case of only two tables. Suppose that one table makes 3NT for +400, and the other goes down one for -50. A difference of 450 is worth 10 IMPs, so one table will score +10 cross-IMPs and the other -10. Therefore the pair who made will score 20 IMPs more than the pair who went down, which is twice as many as they are supposed to score. Clearly we must divide by 2 to obtain the correct difference. It is easy to verify that this result generalizes to more than two tables.

IMP Pairs is like a series of team games, where each of the other tables is your teammates. A two-table IMP Pairs game is actually just like a regular team game. And in a normal team game, one team would get +10, and the other would get -10, which corresponds exactly to the result when we divide by the number of comparisons.

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