[jogs]

Trinidad, on 2014-March-26, 11:36, said:

How would they calculate the datum when there is an even number of tables (e.g. 10x 6♠ making (1430) and 10x 6♠-1 (-100))?

Seems very complicated to me...

Rik

From wikipedia.

Quote

In statistics and probability theory, the median is the numerical value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. The median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one (e.g., the median of {3, 3, 5, 9, 11} is 5). If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values

The median in your example is 665.

[aguahombre]

I guess that also works out to be fair, comparatively, even if the median of odd tables were a unique score (like +870...much closer to the 980 group than to the 480 group) in the given case.

[jogs]

When there is a large dataset, sometimes median is better than average for skew distributions. Like average income can be skewed by the addition of one billionaire. Median income better represents the financial state of the masses.

[aguahombre]

jogs, on 2014-March-27, 16:33, said:

When there is a large dataset, sometimes median is better than average for skew distributions. Like average income can be skewed by the addition of one billionaire. Median income better represents the financial state of the masses.

The billionaire and the pauper are eliminated before the datum (mean) is calculated, so you would have to have two whales for that to take effect.

[Fluffy]

aguahombre, on 2014-March-27, 16:48, said:

The billionaire and the pauper are eliminated before the datum (mean) is calculated, so you would have to have two whales for that to take effect.

I never understood this rule, what you should is take away x% highest and lowest, like 10% for example, if you have 15 results, you totally eliminate the upper, and remove only 'half' of the second for example. There is probably an even better formula that weights all results putting the most weight around the median.

[aguahombre]

The two methods I KNOW are prone to skewing are raw averaging without eliminating extremes at all, and whatever they came up with for on-line IMP pairs. Those dividing lines between scores and IMPs were put where they are for a reason. Fractional IMPs seem just plain wrong.

[gwnn]

aguahombre, on 2014-March-27, 17:52, said:

The two methods I KNOW are prone to skewing are raw averaging without eliminating extremes at all, and whatever they came up with for on-line IMP pairs. Those dividing lines between scores and IMPs were put where they are for a reason. Fractional IMPs seem just plain wrong.

Yes, they were put where they are for a reason, namely to compare reasonable bridge scores with other reasonable bridge scores. They are not put there to compare +420 to +295 or -50 to +231.5. So to me it makes more sense to compare real bridge scores and then average the IMP's than to compare a real bridge score with a average (usually impossible) bridge score.

[helene_t]

jogs, on 2014-March-27, 16:33, said:

When there is a large dataset, sometimes median is better than average for skew distributions.

Which is "better" begs the question: "better" for which purpose?

Anyway, raw bridge scores are generally not skewed. Using a robust statistic like the median can sometimes lead to absurd results. Suppose there are 19 tables. 10 NS pairs score +1430, 9 score -100. The median is +1430. So if you score +1430 you get 0 IMPs. Serves you right for making the slam on a randomly chosen two-way finesse, maybe. The problem is, however, as Rik shows, that NS can only lose on this board. At least in Rik's example, the difference between the average IMPs for NS and EW would not be more than a couple of IMPs. Here the difference will be more than 15 IMPs.

I think the notion that one should exclude extremes is flawed. If you are seriously concerned that there is a single very weak pair that produces nonsense results and which shouldn't be used for comparison, you might want to play Swiss, or do some Swiss-like postprocessing of the data like removing all boards involving pairs that scored less than -2 IMPs/board before recalculating the datum and butler scores.

I am not seriously suggesting this, though, since there is a better and simpler suggestion: X-IMPs:
- Produces zero average for both NS and EW
- Has similar tactical implications as a team match
- Evens out the discreteness of the IMP-scale
- Is reasonably robust because the single 7NTxx-13 gets IMPd before averaging so the impact is reduced.

Of course you can also just play matchpoints if you really want to reduce the impact of outliers.

[jogs]

helene_t, on 2014-April-15, 05:41, said:

Which is "better" begs the question: "better" for which purpose?

There is rarely a large dataset in bridge. The median is better for some economic measures. Median income paints a more accurate picture than average income.