Any Golfers Good at Math

[zeg]

Originally Posted by MEfree

Assuming you agree with Black_Death, Zeg, Saevel25, orchmund, The Rec C, ghalfair and others who see that the intention was to use a weighted average, what is the best way to convince the non-believers?

An "average" simply must have the property that if every value being averaged is the same, the average equals that value.  B's method doesn't preserve this, so it is flat out not any kind of average.

If that doesn't work, you could appeal to authority.  You've got a Caltech astrophysicist on your side, that must count for something.  :-)

[saevel25]

well most people just know normal average math, sum of totals divided by the number of items. But with statistics and weighting averages, theres a whole new dimension, the weights. Averaging out like normal assumes all the weights are 1

So for example you have 5 grades in a class, there equally weighted.

So lets say 100,100,100,50,50 each equally weighted at 1.0 that means you would have,

100x1 + 100x1 + 100x1 + 50x1 + 50x1 divided by 1+1+1+1+1 = 80

So looking at the original weighted equation, if there all equal its just like a normal average, 5 scores, added up divided by 5, since the weight is 100%, 1, then its just normal

Even if you say, lets weight them all 50%, 0.5

50 + 50 + 50 + 25 + 25 divided by 2.5 = 80

the math still works, doesn't matter there

But lest say we weight them at 100%,90%,80%,70%,60%

Thats, 100+90+80+35+30 divided by the sum of the weights, because you have to sum the number of events (number of tests) multiplied by the weight. So easier way to think about it is, each even is 1 or 0, it happened or it didn't. You weight the score, so you have to weight that event happening. If it happened that 1.0 x weight. You add those up, so your still adding in the number of events taken place, its just those events are weighted as well as the scores.

So thats 100+90+80+35+30 divided by 4 = 83.75, which makes since since the lower scores were weighted less, and had less effect on the overall grade.

[jshots]

Originally Posted by zeg

Black Death nailed it in 1.  When you're averaging weighted scores, you have to divide by the sum of the weights, not by the number of events.  It is dumbfoundingly obvious that this is the intention of the committee since they posted weights for the tournaments.  I don't know what to tell B except that he's wrong.

Maybe it would help slightly to multiply the weights by 10.  Then they're all > 1 and you can explain it as follows.  The score with weight=1 (original 0.1) counts as one score in the average.  The score with weight=10 (original 1) counts as ten scores.  Then what you're doing is dividing by the total number of scores, where you've counted some of the rounds more than once.  The weighted sum formula is unchanged by scaling the weights like this.

I want to know how many tournament commitees have a good understanding of weighted average.

[MEfree]

Originally Posted by jshots

I want to know how many tournament commitees have a good understanding of weighted average.

IMO, Not very many, even at the highest levels!

[sean_miller]

The only question here is whether or not the commitee agreed with B. It kind of sounds like they did, which is totally insane.

How could someone enter more events, win every event they entered, then not be named the player of the year?

[jshots]

If they knew how to calculate a weighted average when the rule was created it seems to me that it would be fine however if they thought what most people think then they probably screwed up royally. It makes absolutely no sense if you are using the wrong method to create the rule the way they did because the more tournaments a person played at the beginning of the season the worse their points will end up being.

[k14]

Couldn't you argue that since the maximum number of events is 8 then the sum of the top 8 year scores (weighting x score) should be averaged over 8 regardless of how many events you have played (not saying the weighted average is wrong just another way of looking at the situation/arguing for player A). If you do that then player A averages 60 and B 40.5. The logic the commitee is using is nonsensical and stupid. They are penalising player A for playing more tournaments. Are they saying that if someone only played 4 tournaments and finished behind A and B in 3 but won the final one that they would be the winner instead?

[TheGeekGolfer]

Originally Posted by MEfree

This past year's "Player of the Year" competition at My Golf Club was dominated by Player A who won all 9 events he entered, but Player B is contending that he won based on the following from the Tournament Committee:

Player of the Year is to be awarded to the player with the highest points average based on 10 tournaments held throughout the year with more weight given to tournaments later in the year.  A minimum of 4 and maximum of 8 results will be counted for each player.

I agree with most everyone here, player A is definitely 'Player of the Year'.  Using weighted averages is the correct thing to do, otherwise you get screwed for playing in early tournaments and winning those.  If the rules committee doesn't understand that they 1) don't understand math, and 2) didn't write their rules correctly and took these situations into account, then you should use their 'rules' against them.  No where in the 'rules' does it say you HAVE TO USE all or 8 results for each player.  It just says a 'minimum of 4 and maximum of 8 results' .  So, Player A should elect to just use the last 4 scores and 'average' those according to the 'rules' and then he has a total of 85 and wins over the Player B-dbag 's 'average' of 64.8.

[delav]

Originally Posted by TheGeekGolfer

I agree with most everyone here, player A is definitely 'Player of the Year'.  Using weighted averages is the correct thing to do, otherwise you get screwed for playing in early tournaments and winning those.  If the rules committee doesn't understand that they 1) don't understand math, and 2) didn't write their rules correctly and took these situations into account, then you should use their 'rules' against them.  No where in the 'rules' does it say you HAVE TO USE all or 8 results for each player.  It just says a 'minimum of 4 and maximum of 8 results'.  So, Player A should elect to just use the last 4 scores and 'average' those according to the 'rules' and then he has a total of 85 and wins over the Player B-dbag's 'average' of 64.8.

It appears that the tournament organizers didn't understand how or when to use weighted results when laying out the scoring rules.  In any event, I'm still at a loss for why player B would rationally petition against player A in this case.

[zeg]

Originally Posted by k14

Couldn't you argue that since the maximum number of events is 8 then the sum of the top 8 year scores (weighting x score) should be averaged over 8 regardless of how many events you have played (not saying the weighted average is wrong just another way of looking at the situation/arguing for player A). If you do that then player A averages 60 and B 40.5. The logic the commitee is using is nonsensical and stupid. They are penalising player A for playing more tournaments. Are they saying that if someone only played 4 tournaments and finished behind A and B in 3 but won the final one that they would be the winner instead?

In short, no, you can't do that and still call it an "average."  The number of "things" you average is what you divide by, so if you want to divide by 8, you can't use the weights.

Looking at this more closely, there are some problems with the ranking system even with a properly-computed average.  I'm not quite sure what the reason for weighting the later tournaments more is, but presumably the competition gets stiffer toward the end.  Thus, someone who played only in the early events and did well, then did not play in the later events at all would have a very good average.  Someone who showed up for the last event and had an ok finish would find his average dominated by that score, which doesn't seem right.  To make sense, all players should have to play in all or some of the heavily-weighted events.

So given what we're told, the only correct thing to do is the weighted average.  The other reasonable thing would be to rank based on the sum of weight*score with no division by anything.  That is very different, but would at least be sensible.  Whatever your method, if player A out-plays player B in every event and he still loses, your system is so obviously broken that Player B should be ashamed of himself for making this claim.

[MEfree]

Would anyone's opinion change if the calculation was based on a two year rolling period and used to determine seeding and eligibility for various events with the following weightings?

Week (ago)	Weight
1	1.00
2	1.00
3	1.00
4	1.00
5	1.00
6	1.00
7	1.00
8	1.00
9	1.00
10	1.00
11	1.00
12	1.00
13	1.00
14	0.9891
15	0.9783
16	0.9674
17	0.9565
18	0.9457
19	0.9348
20	0.9239
21	0.9130
22	0.9022
23	0.8913
24	0.8804
25	0.8696
26	0.8587
27	0.8478
28	0.8370
29	0.8261
30	0.8152
31	0.8043
32	0.7935
33	0.7826
34	0.7717
35	0.7609
36	0.7500
37	0.7391
38	0.7283
39	0.7174
40	0.7065
41	0.6957
42	0.6848
43	0.6739
44	0.6630
45	0.6522
46	0.6413
47	0.6304
48	0.6196
49	0.6087
50	0.5978
51	0.5870
52	0.5761
53	0.5652
54	0.5543
55	0.5435
56	0.5326
57	0.5217
58	0.5109
59	0.5000
60	0.4891
61	0.4783
62	0.4674
63	0.4565
64	0.4457
65	0.4348
66	0.4239
67	0.4130
68	0.4022
69	0.3913
70	0.3804
71	0.3696
72	0.3587
73	0.3478
74	0.3370
75	0.3261
76	0.3152
77	0.3043
78	0.2935
79	0.2826
80	0.2717
81	0.2609
82	0.2500
83	0.2391
84	0.2283
85	0.2174
86	0.2065
87	0.1957
88	0.1848
89	0.1739
90	0.1630
91	0.1522
92	0.1413
93	0.1304
94	0.1196
95	0.1087
96	0.0978
97	0.0870
98	0.0761
99	0.0652
100	0.0543
101	0.0435
102	0.0326
103	0.0217
104	0.0109

[MEfree]

Originally Posted by zeg

So given what we're told, the only correct thing to do is the weighted average.  The other reasonable thing would be to rank based on the sum of weight*score with no division by anything.  That is very different, but would at least be sensible.

That's actually what the committee in this example did prior to going to their current point "averaging" method.  I am not privy to all the details, but I think they simply did not understand how to compute weighted average when they switched over and are now too embarrassed to admit their mistake.

[TheGeekGolfer]

Originally Posted by MEfree

Would anyone's opinion change if the calculation was based on a two year rolling period and used to determine seeding and eligibility for various events with the following weightings?

Apples and Oranges - The OP's posting is to determine 'Player of the Year', the second is to determine who has been playing the best 'lately' to get into events or for world rankings.

[MEfree]

Originally Posted by TheGeekGolfer

Apples and Oranges - The OP's posting is to determine 'Player of the Year', the second is to determine who has been playing the best 'lately' to get into events or for world rankings.

I understand they are different things, but assuming that you are going to give recent events more weight than older results (which, IMO, makes more sense to determine who has been playing best lately, not who played the best for the whole year), doesn't it make sense to use a weighted average (as described in the first reply by Black_Death) in both cases?

[anotherday]

So what incentive would anyone have to play all of the events? Just play the minimum.

And what sense does it make to divide by Player B's 5 tournaments yet use all 8 of Player A's tournaments.

All math aside... Player A won all of the tournaments he was in. Isn't that enough said? Even if he had a lower handicap, there are systems in place to create a level playing field and determine a winner.

Go Player A you BAMF.. enjoy your championship.

[delav]

Originally Posted by anotherday

So what incentive would anyone have to play all of the events? Just play the minimum.

And what sense does it make to divide by Player B's 5 tournaments yet use all 8 of Player A's tournaments.

All math aside... Player A won all of the tournaments he was in. Isn't that enough said? Even if he had a lower handicap, there are systems in place to create a level playing field and determine a winner.

Go Player A you BAMF.. enjoy your championship.

The way the tournament series was structured, evaluating a winner by weighted average would answer this questions.  The whole "between 4 and 8" negates the point of a weighted average scoring system, and illustrates that things weren't really well thought out.

If the league was scored on weighted average alone, with increasing weights as the season progressed, players who played all events would stand a better chance of winning overall (0 times any weight is 0).  But, a player could miss a few early events and make up with strong performances later in the league when the weights were higher.

[B-Con]

I 100% agree with Black_Death and have been trying to explain this, but it is amazing how stubborn people can be in admitting their mistake when they don't fully comprehend correct math.  Here is an example of a reply I got:

Nowhere in the definitions do we use the term “weighted average” and even after more than one hour on the phone with you neither of us fully understand this in its pure mathematical sense.

It’s simply not practical to make a major change in the algorithm.

The points won are weighted then divided by events played, everyone can understand that!

Using this logic (and fuzzy math), Fairway_CY correctly predicts B as the "winner" with a 64.8 "average" but still agree A is the winner as weighted average is described even if the exact term is not used.

What on earth is the point in having weighted scoring if they're not even going to try to calculate a weighted average? They're not averaging anything, just adding one set of numbers and dividing by a fairly unrelated sum of other numbers. They probably wanted the tournaments to get more "important" as the year went on, but with that flawed system they actually [i]penalize[/i] players playing in the earlier tournaments. [quote name="MEfree" url="/forum/thread/46946/any-golfers-good-at-math#post_608429"]

Assuming you agree with Black_Death, Zeg, Saevel25, orchmund, The Rec C, ghalfair and others who see that the intention was to use a weighted average, what is the best way to convince the non-believers?

[/quote] The other method is not even an average. You can't just add and divide numbers arbitrarily and call it an average. Tell them that, in layman's terms, the units on the top don't cancel the units on the bottom. It's pretty much a non-sensible number. [quote name="zeg" url="/forum/thread/46946/any-golfers-good-at-math/18#post_608434"]

An "average" simply must have the property that if every value being averaged is the same, the average equals that value.  B's method doesn't preserve this, so it is flat out not any kind of average.

[/quote] That right there should be a huge red flag that something's wrong.

[mdl]

@zeg,

Obviously using weighted sum instead of weighted average introduces an advantage to the player who plays in more tournaments, with the advantage increasing when player A plays in more later tournaments than player B.  So this isn't really measuring average performance in tournaments which you participated in, with increasing weight given to later tournaments which presumably had more difficult fields, or at least are considered more important.

However, if you consider the same number of tournaments for each player (say best 4 for each player, instead of at least 4 but up to 8 if you participated in that many), then the weighted average and weighted sum should maintain rankings, no?

Assuming you want to allow members who don't have time to participate in all or almost all the tournaments to have a shot at player of the year, then obviously weighted average (the correct one, not weighted sum divided by unweighted tournament count, which is sort of a meaningless quantity) is the best measure.  I'm just wondering about the rankings question.