Dr Ian Plummer
Technical
Ties in Swiss Tournaments
Kevin Carter writes
It is usual when running a Swiss event to aim for n+2 rounds, where 2^{n} is
the next highest power of two above the number of participants. This is said
to optimise the probability of a single winner.
2^{3} = 8

2^{4} = 16

2^{5} = 32

2^{6} = 64

2^{7} = 128

2^{8} = 256

To test this I have written a computer simulation of a Swiss tournament for
up to 32 players.
It shows, for instance, that in a 20player competition run over 7 rounds
(n+2 where 2^{n}=32), there is a 62% chance of an outright top and
a 26% chance of a 'resolved tie'  that is two or more players on the same
number of wins but one has beaten the other(s). So, this gives an 88% probability
of a single winner. The other 12% are 'unresolved ties'  typically three players
equal on wins and each having beaten one of the other two.
In this example n+2 rounds are superior to n+1, at only 60%, and n+3, at 68%.
However, nine rounds (n+4) proves to be slightly better for a 20player event,
at an 89% probability of a single winner.
Running the simulation over several different player numbers and rounds, we
find that up to 24 players there is generally a double peak optimum, at about
90%, for n+2 and n+4. Above 24 players interestingly the double peak moves
to n+3 and n+5. For instance, for 32 players 10 rounds are optimum, at 92%,
just ahead of 8 rounds at 88%, with 7 rounds (n+2) down at only 69% and 9 rounds
(n+4) at 74%.
The above results all assume each player in every game has a 50% chance of
winning. In reality this will not be so. In a handicap tournament we would
generally expect a shallow normal distribution of abilities and in a level
event a skewed distribution.
The simulation was rerun for both of these distributions. Surprisingly the
results did not vary a great deal  an improvement of 24% being typical.
So, what should be done about the one in ten of all Swiss events that do not
yield an outright winner? I have seen counting net points and shooting at the
peg, but both are terrible solutions. Mathematically the fairest would seem
to be trying to ascertain which of the tied players has faced the strongest
opposition, for instance by totalling the number of wins by the opponents of
each.
Regards, Kevin Carter; 27.5.98
Author: Kevin
Carter
All rights reserved © 1998
