mathguy
• Posts: 1
Joined: Jul 17, 2012
July 17th, 2012 at 10:42:50 AM permalink
A group of 2N people want to start a competition with two teams. The teams
are supposed to play each other by first having each team rank its players
and then have each player compete against his correspondingly ranked equal
on the other team. For example if each team has N players, the 1st ranked
(strongest) players on each team would play each other. The 2nd ranked
players would play each other, etc. Therefore for a given pair of teams,
there would be N games played and each team could win up to N points.

If you assume the teams are all drawn from the same distribution of
players, then the above rule would mean each team could expect to win N/2
games. But what if one team cheated? If one team knew the rankings of the
other team, but played its higher ranked players against the lower ranked
players on the other team, could it expect to win more than N/2 games? The
answer is "yes", and the problem is to find the optimal “cheating”
strategy.
Switch
• Posts: 934
Joined: Apr 29, 2010
July 17th, 2012 at 11:13:53 AM permalink
My first instinct would be to place my weakest player against their strongest player and then to rank the rest of my players in order of strength so that my strongest would play their 2nd strongest, my 2nd would play their 3rd etc etc.
Ayecarumba
• Posts: 6763
Joined: Nov 17, 2009
July 17th, 2012 at 11:21:41 AM permalink
Quote: Switch

My first instinct would be to place my weakest player against their strongest player and then to rank the rest of my players in order of strength so that my strongest would play their 2nd strongest, my 2nd would play their 3rd etc etc.

I agree with Switch. If your ranking of both is accurate, Switch's strategy will produce only one loss. Getting away with it, if the difference in stregnth between the top and bottom players is easily observed, is a different story.
Simplicity is the ultimate sophistication - Leonardo da Vinci
AcesAndEights
• Posts: 4300
Joined: Jan 5, 2012
July 17th, 2012 at 11:23:35 AM permalink
Quote: Ayecarumba

I agree with Switch. If your ranking of both is accurate, Switch's strategy will produce only one loss. Getting away with it, if the difference in stregnth between the top and bottom players is easily observed, is a different story.

That's assuming that you are guaranteed to win if you are stronger than your opponent. Is there some amount of random chance in this, like most sporting events or casino games? If so you aren't guaranteed to lose only once, but your expectation would definitely be greater than N/2 wins.

Needs moar parameters.
"So drink gamble eat f***, because one day you will be dust." -ontariodealer
dwheatley
• Posts: 1246
Joined: Nov 16, 2009
July 18th, 2012 at 10:44:03 AM permalink
The answer definitely depends on the distribution of outcomes between any two un-evenly ranked players.

MOAR information.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
ThatDonGuy
• Posts: 6341
Joined: Jun 22, 2011
July 18th, 2012 at 10:57:45 AM permalink