- ----- Original Message -----
**From:**Michael Tamada**Sent:**Saturday, May 24, 2003 6:20 AM**Subject:**RE: [APBR_analysis] Re: Defensing the Mavs<snip>Pretty much the same results can be obtained by using a simpler technique: really, the only combinations with large numbers of observations are 2 FTA and 4 FTA. Bowen's FT% appears to rise from 52.4% to 57.1% over that range, implying that each FTA raises his FT% by 2.35 percentage points.If we take the regression equation literally, then each additional FTA per game raises Bowen's FT% by 2.3 percentage points -- implying that the marginal FTA has a 4.6 high percentage point probability than the previous one. And given his estimate y-intercept of .460, if Bowen ever attempts 11 FTs in a game, he'll have over a 100% probability of making his 12th FT! ;)Serves me right for saying something without taking two minutes to check it out for myself. As penance, I'll re-run the Bowen FT analysis from a slightly different angle. (BTW, if anyone wants to perform their own analysis, I've posted a csv of Bowen's stats from all his games to date at http://members.rogers.com/strudel/deleteable/BruceBowenCareer.csv)Instead of looking at Bowen's FT%, I looked at the binomial probabilities for each of his games. That is, I calculated the probability of Bowen making x FTs in n attempts, given his career FT% of 54%. The compliment of that probability (1 - p) is what I'll call his Q-score, and is (I think) a more reasonable measure of the difficulty of various FT shooting performances.The table below shows how FTA affect the Q-score.FTM-A FT% Q-score0-1 .00 .00

1-1 1.00 .54

2-2 1.00 .79

1-3 .33 .16

2-3 .67 .56

3-3 1.00 .90

1-5 .20 .05

4-4 1.00 .96

5-5 1.00 .98

1-8 .13 .01

7-8 .88 .98

8-8 1.00 1.00FT percentages for performances of 1-1, 2-2, 3-3, 4-4, and 8-8 are all 100% -- yet the difficulty level for the latter is surely higher than the previous ones. Now see how the Q-score reflects the increasing difficulty: .54, .79, .90, .96, 1.00.Since the Q-score is the compliment of the binomial probability of Bowen hitting x number of FTs in n attempts by chance (based on his FT%), we can think of the Q-score as the opposite -- his "effort" value or something. Since there is a .10 chance of Bowen hitting 3 of 3 from the line through chance alone, he gets a .90 Q-score for going so far above his expected number. Likewise, going 8-8 from the line is so unlikely for a .54 FT shooter, that he would get a 1.00 Q-score for such a performance. OTOH a 1-8 performance is also unlikely, and Bowen would receive a .01 Q-score. (Aside: Shouldn't that be a negative number, for decreasing his expected FT%? I don't know how to conceptualize the math here -- maybe someone has a better idea?)Bowen, CareerFTA Games FT% Q-Score

1 16 .38 .80

2 106 .52 .63

3 12 .47 .67

4 43 .57 .50

5 5 .64 .41

6 10 .57 .47

7 4 .43 .68

8 3 .67 .22

9 1 .67 .18

10 1 .50 .52

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total 201 avg .54 .60There is a high negative correlation between Bowen's Q-score and the number of FT attempts he has in a game -- the r^2 is .46. You'll notice that Bowen's Q-score drops dramatically at FTAs and more (except for the four game blip at 7 FTA), meaning that he has fewer games in which he exceeds his binomial probability for those games. The difference is significant at the p < .05 level.(The more I think about it, the more I don't like the fact that the Q-score is not negative for underachieving performances -- it rewards players for exceeding their probability, but doesn't punish them for, er, subceeding this probability. [What exactly is the antonym of "exceed"?]. However, I really don't know anything about probability, and am unable to think confidently about the matter. Can anyone with some math background give me some advice?)Getting back to FT% proper, another way to look at it is to divide Bowen's games into "good" free throw shooting games, "expected" games, and "bad" games. Any game in which Bowen made more than 2/3 of his FT attempts was labelled "good," any game lower than 1/3 "bad," and the middle third "expected."Fta Good Expected Bad

1 6 0 0

2 30 51 0

3 0 5 7

4 5 31 4

5 1 3 1

6 2 6 2

7 0 1 3

8 0 3 0

9 0 1 0

10 0 1 0

After combining pairs of rows, the relative frequencies are here:Good Exp Bad1-2 41% 59% 0%

3-4 10% 69% 21%

5-6 20% 60% 20%

7-8 0% 57% 43%

9-10 0% 100% 0%A chi-square test reveals that distribution to be highly significant, at the p < .001 level.My thoughts are to conclude that Bowen has fewer good FT shooting performances in games in which he shoots higher than usual numbers of free throws.ed - ----- Original Message -----
**From:**igorkupfer@...**Sent:**Sunday, May 25, 2003 2:22 PM**Subject:**Re: [APBR_analysis] Bowen's FT shooting (was: Defensing the Mavs)<snip>Getting back to FT% proper, another way to look at it is to divide Bowen's games into "good" free throw shooting games, "expected" games, and "bad" games. Any game in which Bowen made more than 2/3 of his FT attempts was labelled "good," any game lower than 1/3 "bad," and the middle third "expected."DeanO brought to my attention that the table I originally presented contained incorrect information. The following table is accurate.FTA Good Exp Bad Total1 10 6 0 | 16

2 76 0 30 | 106

3 7 5 0 | 12

4 7 19 17 | 43

5 1 3 1 | 5

6 2 4 4 | 10

7 3 1 0 | 4

8 0 0 3 | 3

9 0 0 1 | 1

10 0 1 0 | 1

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106 39 56 201

After combining pairs of rows, the relative frequencies are here:FTA Good Exp Bad1-2 70% 5% 25%

3-4 25% 44% 31%

5-6 20% 47% 33%

7-8 43% 14% 43%

9-10 0% 50% 50%The distribution is significant at the p < .001 level.ed - -----Original Message-----
**From:**igorkupfer@... [mailto:igorkupfer@...]**Sent:**Sunday, May 25, 2003 11:23 AM**To:**APBR_analysis@yahoogroups.com**Subject:**Re: [APBR_analysis] Bowen's FT shooting (was: Defensing the Mavs)

Instead of looking at Bowen's FT%, I looked at the binomial probabilities for each of his games. That is, I calculated the probability of Bowen making x FTs in n attempts, given his career FT% of 54%. The compliment of that probability (1 - p) is what I'll call his Q-score, and is (I think) a more reasonable measure of the difficulty of various FT shooting performances.*** I don't understand why we'd want to look at Q-scores rather than FT%, since FT% under most circumstances that I can think of is the statistic which really matters. But if we agree to examine Q-scores ...FT percentages for performances of 1-1, 2-2, 3-3, 4-4, and 8-8 are all 100% -- yet the difficulty level for the latter is surely higher than the previous ones. Now see how the Q-score reflects the increasing difficulty: .54, .79, .90, .96, 1.00.*** Okay, high Q-scores indicate the achievement of a relatively unlikely event, for a .54 FT shooter. But then there's something very fishy about the table below:Bowen, CareerFTA Games FT% Q-Score

1 16 .38 .80

2 106 .52 .63

3 12 .47 .67

4 43 .57 .50

5 5 .64 .41

6 10 .57 .47

7 4 .43 .68

8 3 .67 .22

9 1 .67 .18

10 1 .50 .52

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total 201 avg .54 .60*** Bowen's highest Q-scores occur in games where he has 1 FTA. (BTW, what do the Q-scores in this table represent: his mean Q-score over the games in which he had that number of FTAs?) Implying that Bowen was having unusually good FT shooting performances in those single FTA games. But this table says that his FT% was 38% in those games, his worst FT% of any category!Similarly at the other end of the table the numbers don't make sense: Bowen clearly shot 6-9 in his 9 FTA game, and 5-10 in his 10 FTA game. Even without going through the binomial calculation, it must surely be the case that his 6-9 game represents a better, less likely, and therefore higher Q-score game than his 5-10 game: he got more FTM in fewer FTA. But the table claims that his 6-9 game comes out to a lousy .18 Q-score, whereas his 5-10 game rates at .52. This doesn't make sense, if the Q-scores are supposed to be measuring how good or difficult or unlikely his FT performance was.So I'm skeptical of the data in these tables.I don't like the good-expected-bad tables, corrected or otherwise. It is almost always a mistake to take detailed numeric data, such as FT% or Q-score, and transform it into less informative, less detailed categorical data such as good-expected-bad. Information is getting wasted.It's like the difference in the amount of information if we measure how many points Tim Duncan scores, vs. putting his scoring into categories (20 or above is "lots", below 20 is "few"). What we've done is taken a highly informative variable, points scored, and transformed it into a less informative one: lots of points or few points.--MKT