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Old 07-26-2005, 01:40 AM
Hedge Henderson Hedge Henderson is offline
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Join Date: Jul 2003
Location: Tejas
Posts: 64
Default Re: Formula for a player of the year award?

In the home game I host, we play ring games most nights. We do have 6-8 tournament nights a year, though, ranging from 4-player "mini-tourneys" (you might play 5 or 6 in one night) to full two-table tournaments.

For tourney rankings, I use the natural log of the player's finish vs the number of entrants. I think I got that idea from someone here.

For a 8-player tournament, it works out to a score of ~2.2 for first, 1.5 for second, 1.1 for third, and so on.

For a 10-player tournament, first gets 2.4, second 1.7, third 1.3, etc.

Once I have everyone's raw score, I average those together, de-logify it, and multiply by 100 to get a number (percentage, if you like). I have it all set up in an Excel spreadsheet, so it's a lot less work than it sounds.

Our best players usually have scores of 60-75, the worst have scores of 25-40. A score of 50 means you nearly always finish in the middle of the pack.

We set an arbitrary minimum for number of tournaments played so the guy who just showed up once or twice and happened to get lucky doesn't win a title.

If your tournaments have many tables and radically different numbers of players each time, the above method might require some adjustment. It works well for us, though, since I can't really seat more than 16.



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In short:

- more points for beating more players
- more points for bigger buyins
- way bigger difference in points between first and second than, say, fifth and sixth.


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Why not just use the cash to keep score, then?


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...playing heads-up is so much more complex than a ring game, the winner should get a decent cut of the cake.


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In a perfect world, it would be. Unfortunately, in most home tourneys, since they're limited to just a few hours, how the top two or three places fall usually involves a lot of luck.
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