Any Golfers Good at Math

[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.

Here are the point results and tournament weights.  Who won Player of the Year?

Tournament	Weight	Player A	Player B
10	1	100	80
9	0.9	100	80
8	0.8	100	70
7	0.7	100	80
6	0.6	DNP	100
5	0.5	100	DNP
4	0.4	100	DNP
3	0.3	100	DNP
2	0.2	100	DNP
1	0.1	100	DNP

Edit (add): Would your answer change if this was used to compute seeding in the match play tournaments and to determine who gets into certain events and team competitions?

[Black_Death]

Player A wins "Player of the Year".

The Club is finding the Weighted Mean for each player and awarding the title to the player with the highest number.  In your example, Player A has a weighted mean of 100 (counting the 8 most highly weighted scores) and Player B has a weighted mean of 81 (only counting the 5 tournaments that he/she played in).

For the detailed solution see below:

Weighted mean can be calculated using the following formula:

where w represents the weight of the event and x represents the score for the event.

Player A has scores of {100,100,100,100,100,100,100,100} and weights of {1,0.9,0.8,0.7,0.5,0.4,0.3,0.2}.  Using these values we get 480/4.8 giving us a weighted mean of 100.

Player B has scores of {80,80,70,80,100} and weights of {1,0.9,0.8,0.7,0.6}.  Using these values we get 324/4 giving us a weighted mean of 81.

Player A wins.

For more information check out the wikipedia article on weighted mean at http://en.wikipedia.org/wiki/Weighted_mean .

[Fairway_CY]

Player A = 60.0

Player B = 64.8

Based on the committee rules... Player B is correct.  Based on everything else... Player B is a douche.

CY

[MiniBlueDragon]

I'll get my coat...

If Player B only has a maximum of 5 tournament entries then surely Player A's last 5 tournament entries weighted scores should be averaged? e.g.

PLAYER A

1 * 100 = 100
0.9 * 100 = 90
0.8 * 100 = 80
0.7 * 100 = 70
0.5 * 100 = 50
Average weighted score = 78

PLAYER B

1 * 80 = 80
0.9 * 80 = 72
0.8 * 70 = 56
0.7 * 80 = 56
0.5 * 100 = 60
Average weighted score = 64.8

[Whiskey84]

I've got Player A averaging 60 points, and Player B averaging 64 points. My math might be a bit rusty, but I'd imagine the committee should use a little common sense and award Player A as "Player of the year". 9/9 is damn impressive.

[Fairway_CY]

What you're missing, MBD, is that they're saying the average points.

Player B's 324 points divided by 5 events is 64.8 points, average.

Player A's 480 points divided by 8 events (the maximum) is 60.0 points, average.

Player B has a case for Player of the Year... but as has been said... common sense should prevail and Player A should be awarded Player of the Year.

CY

[saevel25]

Its player A, the weighted average formula is the right one.

[ochmude]

If you're trying to calculate average, you divide by the number of events (8 and 5 respectively).  You're trying to calculate weighted average, however.  For that you divide by the sum of the weights (4.8 and 4 respectively).  Black Death is correct, Player A's weighted average is 100 points for the 8 events played.  Player B's weighted average is 81 points for the 5 events played.

[MiniBlueDragon]

Originally Posted by ochmude

If you're trying to calculate average, you divide by the number of events (8 and 5 respectively).  You're trying to calculate weighted average, however.  For that you divide by the sum of the weights (4.8 and 4 respectively).  Black Death is correct, Player A's weighted average is 100 points for the 8 events played.  Player B's weighted average is 81 points for the 5 events played.

This makes no sense to me; why should one player have 8 scores averaged and one 5 scores averaged? To make it fair if one of the players has only entered into the last 5 tournaments then BOTH players should have their last 5 scores used.

The more tournaments entered into, the more the average is likely to be lowered by new scores. If that's the case the best thing to do is to not enter any tournaments at all until the final 4 as those are the most heavily weighted and will give the best scoring opportunity.

[The Recreational Golfer]

Yes, Black_Death is right. Since each tournament's score counts differently, you have to make an allowance for that. It's a weighted average.

The formula he presented says: the sum of the (weights times the scores) divided by the sum of the weights.

[ghalfaire]

Black_Death has it correct.  Player A is the winner.

[MEfree]

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.

[Nole77]

While it would appear unfair at first glance, you also have to take into consideration the other side of it.  The more numbers that are averaged, the less each individual score will affect the overall average.  It can hurt you or help you depending on what happened.  To me it looks like the rules are a little ambiguous.

In your example below, a person could do OK in the those last f events and end up with a very average score.  On the other hand, if that same person had competed in events earlier in the year and done better, those scores would help bolster his later scores and increase his overall score as well as help mitigate some of the damage form the later scores.  I suppose at the end of the day you could view it as a way to strategize.

Originally Posted by MiniBlueDragon

This makes no sense to me; why should one player have 8 scores averaged and one 5 scores averaged? To make it fair if one of the players has only entered into the last 5 tournaments then BOTH players should have their last 5 scores used.

The more tournaments entered into, the more the average is likely to be lowered by new scores. If that's the case the best thing to do is to not enter any tournaments at all until the final 4 as those are the most heavily weighted and will give the best scoring opportunity.

[Nole77]

That's just silly.  I agree with the Fairyway_CY, player B is a first class dbag.

Originally Posted by MEfree

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.

[ochmude]

Originally Posted by MiniBlueDragon

This makes no sense to me; why should one player have 8 scores averaged and one 5 scores averaged?

Because that's how I interpreted the guidelines in the first post.  It stated that, per the tournament committee, a minimum of 4 scores and a maximum of 8 scores can be used.  Player A posted 9 scores, so he can use 8 of those 9 based on the guidelines.  Player B posted 5 scores, so he is able to use all 5. If I interpreted those guidelines incorrectly, then that would affect my math.

[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.

[MEfree]

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.

You're spot on Zeg...if the weights were 10 to 1 instead of 1 to 0.1, then the weighted average would still be the same for both (with A the winner), but the "average" used by B (weighted points/# results) would now favor A instead of B (even though the relative weights of the events is the same with event 10 being worth double event 5 and ten times event 1).

So for those of you still siding with B, the weighted average formula used by Black_Death with Zeg's elaboration is the equivalent of saying

If X = 2Y then 10X = 20Y

while B's methodology would say that this is not true.

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?

[bottlerocks]

That is an embarrassingly bad failure of logic on player B's part.  Putting aside the fact that he does not comprehend the math, his reasoning implies that anyone who plays the minimum amount of tournaments at the end of the year is automatically going to have a advantage on those who play more and do better.  Should have been the first flag in his mind that he's making a doofus of himself.