Pairs, XIMPs and Weighted & Split Scoring
As an example take a Board that has been played 4 times in a Pairs competition that is being scored by the XIMP method. The scores for the 4 Results are:-
-110, +50, +170, +120
and the XIMPs for these results (I believe) would be:
| Result | NS XIMPs | EW XIMPs |
|---|---|---|
| -110 | -5.667 | +5.667 |
| +50 | -0.333 | +0.333 |
| +170 | +4.000 | -4.000 |
| +120 | +2.000 | -2.000 |
Now if there was a 5th table and (for some reason) its Result was Weighted & Split as follows:
| Result | NS Probability | EW Probability |
|---|---|---|
| -110 | 50% | |
| +50 | 50% | |
| +170 | 50% | |
| +120 | 50% |
I think (but am not sure) that the XIMPs for the other 4 results would become:
| Result | NS XIMPs | EW XIMPs |
|---|---|---|
| -110 | -4.750 | +5.875 |
| +50 | +0.250 | +0.875 |
| +170 | +4.250 | -3.250 |
| +120 | +2.500 | -1.250 |
But what would the XIMP for the Weighted & Split Result be and why?
Can anyone help me with this, or point me at a Technical document that defines how to score Weighted & Split scores in a Pairs event scored using the XIMP method ?
Comments
Hi Tony,
Your initial X-IMP table is correct, as are the X-IMPs for the split/weighted scenario.
The X-IMPs for Table 5 are surprisingly simple; You just take the relevant results from table 2 and weight them. So for NS, it's 0.5 x -4.75+0.5 x 0.25=-2.25 IMPs, and for EW it's 0.5 x -3.25+0.5 x -1.25=-2.25 IMPs. Table 2 has already taken into account the results for all 5 tables which is why it's as simple as taking the weighted scores.
4.1.1.4 and 4.1.1.5 in the White Book (https://www.ebu.co.uk/documents/laws-and-ethics/white-book/white-book-2019.pdf) give a few examples.
Great, thanks, that's what I thought (but didn't like to show my ignorance!)