Author: Stephen A. Boak
Date: 16:55:55 12/05/99
Go up one level in this thread
On December 05, 1999 at 18:50:59, Imran Hendley wrote:
>On December 05, 1999 at 18:45:28, Imran Hendley wrote:
>
>>On December 05, 1999 at 17:54:50, pollie lackey wrote:
>>
>>>if i hold a 6 round tourney, and 64 players enter, using all possible
>>>possibilities of win draw and losses, what is the most number of players that
>>>can finish with 4 1/2 points or more,,,, and
>>>
>>>in a 5 round with 32 what is the most number of players that can finish with
>>>31/2 or more points
>>
>>(number of games) / (minimum number of points) = (number of players that can
>>finish with minimum number of points or more)
>
>(number of games in the tournament) / (minimum number of points) should be
>rounded down to the nearest integer.
>
>eg: 384 games in tournemnt, 4.5 minimum number of points
>
> 384 / 4.5 = 85 and 1/3 therefore 85 players can finish with 4.5 or more
>(three players can finish with 5.0)
Your math is incorrect. I don't know the exact answer, without doing some more
figuring, but it isn't that simple.
I don't think there is an exact formula that can be utilized to calculate the
answer to the indicated question. The method you *seem* to be trying to use
might establish an *upper bound* to the real answer, but would not be the exact
limit requested--it would be too high.
In a 6-round tournament with 64 participants there will be 192 total games
played, if each player plays all the rounds.
Assume the result of every game is only Win or Loss (no Draws), since this is
easily calculated manually as an example case. Then there will be exactly 1
finisher with 6 points, plus 6 finishers with 5 points. This will be seven
players with *at least* 5 points.
The difficulty in calculating an exact answer to the original question lies in
the huge number of combinations and permutations of ways to achieve a certain
final score--among many contestants. These possibilities grow exponentially as
the total participants and rounds increases.
The exact number of drawn games versus games with decisive results is not known
in any particular tournament, let alone under the special conditions requiring a
determination of the maximum quantity of players that can score at least x.x pts
in a tournament. Yet the number of draws is a large component for the
calculation requested.
I can use a bell curve (gaussian distribution curve) from probability theory to
calculate (approximate) the 'most likely' number of players at, or above, or
below a certain point score. In a large tournament, the quantity of players
with final scores from 0 to max (max = total number or rounds) will approximate
a bell curve when graphed, and those quantities may be closely estimated using
the bell curve.
The odds of deviating 'greatly' from the 'most likely' amount are very high in
general. The odds of actually encountering in real life the
'max number of players that can finish with x points or more'
in a large tournament with 6 rounds or more, where x/(total rds) is between 0.6
and 0.8, would be astronomical, in my opinion. I'd like to see this calculation
also, as a matter of curiousity, for the exact scenario--64 players, 6
rounds--to see if my conjecture holds up even under such limited circumstances.
Certain major tournaments in the United States advertise a 'consolation' prize
fund, say $10,000 U.S. or more, to be divided among all players with at least
4.5 pts or larger, in perhaps a 7 or 8 round tournament. It is interesting to
forecast, using probability theory, what the typical payout per person will be
for a share of the consolation prize fund. Normally it is not a lot of money,
per person.
Interesting math question, not one with an easy answer IMO. Many problems in
math have no exact formula for calculating the answer--they require a knowledge
of the subject, knowledge of all the possibilities, and perhaps lots of manual
techniques to derive an exact answer.
--Steve Boak
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