Playing-off 3 remaining players
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Does anyone have a fair and efficient way to play off three players? Assume
that there is supposed to be two prizes, $200 for first and $100 for
second.
This might come up with a tournament that started with 12 players, and
instead of byes it played down to 3 'finalists'.
Any ideas are appreciated.
Gregg C.
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Chuck Bower writes:
The old fashioned way is to offer the players a chance to buy the bye.
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David Montgomery writes:
You could have round-robins with decreasing match lengths until one of the
players wins both his matches.
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Jordan Lampe writes:
In Sumo they settle a 3 wrestler playoff with a "circle-fight". The three
players draw lots: A, B, and Wait.
First, A and B play. The winner plays Wait. If he wins, then the tournament
is over. If he loses, then Wait plays the loser of the first match. If Wait
wins, the tournament is over, but if Wait loses, then A and B play each
other again. And you keep going on and on until somebody wins two in a row.
Some math will show you this is somewhat unfair to the person who gets
picked as "Wait" (assuming all 3 players are equal strength). The way to
"fix" this is to redraw the lots if the first 3 matches end with all 3
players 1-1.
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Chuck Bower writes:
Total prize pool: $300. Expectation per player = $300/3 = $100.
So after the bracketing (but before any matches are played), each player's
expectation should remain $100.
The player who gets the bye (by random draw) thus must have an expectation
of $100 and since he's 50-50 to win, 0.5*first + 0.5*second = $100. ==>
first+second = $200. The remaining $100 goes to the two semifinalists.
Solution 1: Winner of semifinal match receives $100. Winner of final match
receives $200. This is still mathematically equitable. Note that the player
who gets the bye but fails to win a match gets nothing (and there seems to
be considerable justice in this :).
Solution 2: Everyone gets $50 before play begins; winner of semifinal match
gets (additional) $50; winner of final match gets (addditional) $100. Note
that the bye-player gets $50 + 0.5*($100) so his/her expectation is $100,
as it should be. By symmetry and conservation of $, the other two players'
expectations are also $100 each so the division is equitable.
Solution 3: Everyone gets $25 before play begins; semifinal match winner
gets (additional) $75; final match winner gets (additional) $150. (Again
you can check that each player's expectation is $100.)
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Maik Stiebler writes:
As much as I have embraced, Chuck's solutions, they still have the problem
of giving the player with the bye a much greater chance of "winning" the
tournament. That might be a real concern in either small stakes/amateurish
or highly prestigious tournaments.
A fair way of solving this would be handicapping the player who received
the bye such that his chance of winning the final is 1/3. This can be done
by giving the handicapped player a suitable score disadvantage.
Alternatively, a series of final matches could be played. Once the
handicapped player loses, he is out, if he wins, the handicap changes to
the other player. In the Nth match, the score disadvantage method could be
used.
The fair payout plan to go with that is to pay the same amount to both
losers, and more to the winner.
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