Hannie, on Sep 6 2005, 11:34 AM, said:
Zar, on Sep 6 2005, 08:50 AM, said:
Goren 9 tricks 10 tricks 11 tricks
26 50% 50% 0%
27 22% 73% 5%
Sounds interesting Zar, a low standard deviation indeed seems to indicate a good evaluation method. I'm looking forward to see how your point count compares to methods like HCP, BUMRAP+321+fit, and Binky+fit.
I wonder if the table I'm quoting here is yet another joke, or really something you found when you were adding and dividing (just kidding). The actual numbers seem rather unlikely.
The numbers were to illustrate how the statistics would be calculated. The "chances" these are the real numbers are of course ZERO....Since some 26 point hand make 13 tricks, some make 6 or 7 tricks. This is jsut to highlight he approach, in simple terms.
As for average (mean) versus median question.
The average is easy, total the number of tricks, divide by the number of hands. Zar showed this as "Mean = (50*9 + 50*10) / 100= 9.50 tricks" What this means, hand played 100 times, won 9 tricks 50 times (add 50 nines) and 10 tricks 50 times (add 10 50 times), add those together, and divide the sum by 100 (the number of hands). IF it was played 101 times and the 101st time won 5 tricks, you would add 5 to the total and divide by 101.
For the Median, ZAR presented the equation as.... Median = 1/2 * [ 9 + 10] = 9.5 tricks, which was questioned. is well, 1/2 right and 1/2 wrong. For some applications, my understanding is the Medium (in unranked data), is the middle item when data is ranked from lowest to highest. Here, 9 is the lowest, 10 is the highest, so by this definition the Median is 9.5. Another definition is the value where exaclty half the data is higher than the current number, and half is lower the number. Here, just taking half the range (as ZAR did), could lead to a false number.
Lets imagine, for sake of arguement this data set...
9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 11
The Medium ZAR way would be 10 (1/2 * (11+9) = 10, but the value with exaclty half the number above, and have below would not be 10. (10 had one above, and 8 below). According to my guestimate (and recollection) the calculated Median for this set of numbers would be 9.00000, not some 9 point something like the mean (which is... 9.36)
It looks liek ZAR intends to use the Median is the less descriptive way which may distort the finding. I believe I would want to know the average number of tricks with the SD rather than the median. But he is giving us both, and he is showing the way he chooses to calculate it. Which is one acceptable way, but it give a different number than if you use the more traditional value.