A maths puzzle

[Fnaar]

Miss Goodbody claims 69 is the answer to any problem.

 

[Tin Pot]

The points are derived solely from the rankings in the seven individual events, using the weighting I quoted, and the maximum/minimum values of 1000 and zero.

I'm concerned that there appears to be ambiguity in my description of the problem, but I'm having trouble seeing where it is. Help welcomed.

...1000 points for the first event, first place?

Or 1000 points for first overall after seven events?

 

[DaveReading]

...1000 points for the first event, first place?

Or 1000 points for first overall after seven events?

1000 points in total after seven events (assuming first place in each).

 

[twentysix by twentyfive]

Without spending huge amounts of time with pen and paper it seems to me that the question is the Puzzle.

Often times if the question isn't properly formed the answer comes out rather bizarre if at all.

That's the best I can do

 

[Tin Pot]

1000 points in total after seven events (assuming first place in each).

Okay so it's ranking.

You don't know the final position, and therefore score, of one person until you know the position of all the others.

The maths is similar, but must be applied to all the contestants so that you can determine the final rank.

a) The sum of this gives you a competition score that can then be ranked against the others

1 *0.5
8 *0.5
1 *0.5
46 *0.5
3 *1.5
13 *1.5
34 *0.6

Personally, I'd do it in SQL, it's a relational problem. Excel is all hassle.

With b) being 10th, 15th, 15th, 42nd, 23rd, 7th and 36th you can tell the competition score will be lower than a), but you don't know how many places it will be lower, so it can't be ranked yet, and therefore can't be scored 0-1000

 

[DaveReading]

With b) being 10th, 15th, 15th, 42nd, 23rd, 7th and 36th you can tell the competition score will be lower than a), but you don't know how many places it will be lower, so it can't be ranked yet, and therefore can't be scored 0-1000

Unless I'm mistaken, you can determine absolute scores for any team based purely on their placings in the 7 events, the weighting, and the (arbitrary) score range of 0-1000. The overall ranking is neither here nor there.

There are only 350 (50 x 7) data values, so it's perfectly doable in Excel.

 

[Pro Tour Punditry]

And there was me trying to do it in my head whilst cycling home from work.

I have reverted to McMaths to solve it

 

[DaveReading]

OK, I'm beginning to think that my mistake may have been over-simplifying the problem.

So if it helps, here's how the originator (a £3 bn UK company that everyone has heard of) puts it:

"The overall score can be a number between 0 and 1000. The final score for each team is calculated by adding up combinations of a team’s ranking position for each individual event and the weighting set for the given event. This means that in order to get a ‘perfect’ overall score of 1000 a team would have to be ranked #1 in all events across the board; conversely to score a 0 a team would have to be ranked #50 in all events in the competition. The amount of points deducted from the ‘perfect’ overall score depends on the weighting of individual events; for example ranking #5 in Event 5 will mean more lost points than ranking #5 in Event 2 due to higher weighting associated with Event 5 (150%) than the weighting associated with Event 2 (50%)."

(to be read in conjunction with the weighting values in my original post)

 

[Tin Pot]

OK, I'm beginning to think that my mistake may have been over-simplifying the problem.

So if it helps, here's how the originator (a £3 bn UK company that everyone has heard of) puts it:

"The overall score can be a number between 0 and 1000. The final score for each team is calculated by adding up combinations of a team’s ranking position for each individual event and the weighting set for the given event. This means that in order to get a ‘perfect’ overall score of 1000 a team would have to be ranked #1 in all events across the board; conversely to score a 0 a team would have to be ranked #50 in all events in the competition. The amount of points deducted from the ‘perfect’ overall score depends on the weighting of individual events; for example ranking #5 in Event 5 will mean more lost points than ranking #5 in Event 2 due to higher weighting associated with Event 5 (150%) than the weighting associated with Event 2 (50%)."

(to be read in conjunction with the weighting values in my original post)

Okay, so you need to determine an intermediary score for each event.

To get a 1000, you need 1st in all seven events. This implies that 1000/7 is the average score for getting first place in a single event, then use the event multiplier. So 1st place in the first event scores 1000/7*0.5.

50th place scores 0.

So now each place above 50th scores (1000/7*0.5)/50 per place.

Do the same for each event, changing the multiplier.

 

[Tin Pot]

My only concern is the weighting - it's not balancing out. Are you sure of your weighting values?

 

[byegad]

I consulted Baldric, and the answer is definitely:-

Some Beans!

 

[DaveReading]

My only concern is the weighting - it's not balancing out. Are you sure of your weighting values?

I'm not sure what you mean by "not balancing out". The weightings are purely arbitrary values, if you divided them all by 10, for example, they would still be in the same ratio to each other.

If it makes the maths easier, assume that there are no ties in any of the events, i.e. the ranking positions 1 to 50 are all filled with no instances of, for example "5th equal".

It's easy then to work out that the mean team score must be 500 (25,000 points available in total), and that the two teams in the question will be in the upper half of the overall results table.

 

[jefmcg]

http://mediacentre.heathrow.com/pressrelease/details/81/Corporate-operational-24/8581

 

[DaveReading]

Correct, I didn't want to get readers too excited by mentioning emissions.

 

[Tin Pot]

I'm not sure what you mean by "not balancing out". The weightings are purely arbitrary values, if you divided them all by 10, for example, they would still be in the same ratio to each other.

If it makes the maths easier, assume that there are no ties in any of the events, i.e. the ranking positions 1 to 50 are all filled with no instances of, for example "5th equal".

It's easy then to work out that the mean team score must be 500 (25,000 points available in total), and that the two teams in the question will be in the upper half of the overall results table.

Weighting is only ever done in relation to other values, if it is ever entirely arbitrary it becomes meaningless.

In this case it is noise values. If the event is more noisy than the other six it has a multiplier applied - if all are equally noisy then no multiplier would be applied.

So to get your weighting multiplier you need to know the cda values for each event, average them, then determine each as a percentage of that average. Then apply that to the event.