mathguy
mathguy
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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
Switch
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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
Ayecarumba
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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
AcesAndEights
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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
dwheatley
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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
ThatDonGuy
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July 18th, 2012 at 10:57:45 AM permalink
My first instinct was, "Ask a high school/college tennis coach - they probably have any number of stories about this."
SOOPOO
SOOPOO
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July 18th, 2012 at 3:10:49 PM permalink
See dwheatley and acesandeights answers. If you knew the actual winning percentage of your best versus ALL of the opponenets, and the winning percentage of your second versus ALL of the opponents, etc... then you could answer this question...
charliepatrick
charliepatrick
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July 20th, 2012 at 6:01:54 PM permalink
I tried six players who's ranks were 99.999999 80 60 40 20 0.000001 so their probability of winning was proportional to their rank (hence use of 0.000001 to prevent division by zero). I haven't played with all perms but (100vs20, 80vs100, 60vs80, 40vs80, 20vs40, 0vs0) gives 3.06. Trying the same with 75,65...25 also gives an advantage, so it seems best strategy to win the first with maximum chance and slightly reduce the odds of winning the others. This seems counter intuitive.
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