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View Full Version : How should a "player of the year" be calculated?


johnnycakes
09-19-2004, 11:39 PM
Paul Phillips says [ QUOTE ]
All current points systems are better at ranking who plays the most tournaments than they are at ranking success because there is no penalty for failure. Even some hypothetical perfect ranking system would be of minimal value unless it was applied over a much longer period than a single year.

The obvious way to rank tournament players is in how many buyins of profit they average per tournament played. Doing this would require knowing which tournaments everyone enters, including the ones where they don't cash. I'm not sure that will ever happen, and I don't think I'd want it to anyway.

The nature of the game means you'll never come up with an unarguable ranking list. Everyone wants to resist this but hey, gambling's gambling. Tom Weideman has written a lot about this subject on rgp, everyone should read his posts.


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So, what do you (the reader of this post, not just Paul) think would be the optimal way to calculate a tournament player of the year?

I hate NASCAR, but I imagine they are in a similar situation. Anyone know how they do it?

Here is the formula that PokerSchoolOnline uses to calculate their rankings. There is no buy-in requirement or a mininum amount of tournaments played, so those would have to be defined if a tournament player of the year ranking calculation was to be modeled after something like this:

[ QUOTE ]
Rankings Formula

The Natural "Logarithm Rankings" system gives no advantage to simply playing more often -- but, it gives more weight to good results, and less weight to bad results, such that one exceptionally bad result does not 'kill' your ranking for the whole week, and one exceptionally good result will not give you a safe enough lead that you can then just 'sit on'...

Calculation details:

Where "X" is finishing place out of "N" players, each player's individual tournament result is valued at:

ln ( (N + 1) / X )

ln = "natural" base for logarithms -- a universal number known as "e" = 2.718282... ( second in fame only to "pi" = 3.141593...)

-- and "natural logarithms" are logarithms "to the base e" -- that is,

numbers expressed as powers of "e"... The base "e" is key in many mathematical applications.

...averaged over his number of plays...for example, with 300-player tournaments, a player finishing 1st, 300th, 300th would rate 1.9046, just slightly better than a player finishing 45th every time, 1.9004...

And then these 'natural logarithms' are converted back into "percentiles".

Converting logarithms back to percentiles

To convert the natural log score back to percentile, you need to use this formula:

(1 - exp (-L) ) * 100

The variable L is the average of all the natural log score. The 'exp' means the inverse of natural log.

For example, these are log scores from 3 different tournaments: 1.5, 2.0, and 1.0, so the average is 1.5.

Then plug that number into the above formula:

= (1 - exp ( -1.5 ) )* 100
= (1 - 0.22313016 ) * 100
= 77.686984


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What do you think?

Johnny

Nottom
09-20-2004, 01:38 AM
[ QUOTE ]
I hate NASCAR, but I imagine they are in a similar situation. Anyone know how they do it?


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Not much of a NACAR fan myself, but being from SWVA I can't help but be somewhat familiar with it. I know they changed to rules this year but I'm still pretty sure the Winston/Nextel Cup rankings would have the same basic problems that the CardPlayer rankings have: more events tend to lead to a higher ranking. Of course in NASCAR everyone that has a shot to win is in every event anyway so it doesn't really matter.