1999 Challenge Bell – Doubles
Lori McNeil and Kimberly Po were the defending champions, but lost in the quarterfinals to Amy Frazier and Katie Schlukebir.
Doubles | |
---|---|
1999 Challenge Bell | |
Champions | |
Runners-up | |
Final score | 6–2, 6–3 |
Frazier and Schlukebir went on to win the title, defeating Cara Black and Debbie Graham 6–2, 6–3 in the final.
Seeds
Amanda Coetzer / Corina Morariu (Quarterfinals, withdrew) Lori McNeil / Kimberly Po (Quarterfinals) Sonya Jeyaseelan / Seda Noorlander (Quarterfinals, withdrew) Cara Black / Debbie Graham (Final)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | 6 | 6 | |||||||||||||||||||||||||
Q | 2 | 3 | 1 | ||||||||||||||||||||||||
6 | 5 | 6 | w/o | ||||||||||||||||||||||||
4 | 7 | 0 | 6 | 2 | |||||||||||||||||||||||
4 | 6 | 6 | 4 | 7 | 6 | ||||||||||||||||||||||
WC | 4 | 4 | 4 | 6 | 6 | ||||||||||||||||||||||
3 | 6 | 5 | 4 | 4 | |||||||||||||||||||||||
6 | 3 | 7 | 4 | 2 | 3 | ||||||||||||||||||||||
4 | 6 | 7 | 6 | 6 | |||||||||||||||||||||||
6 | 3 | 6 | w/o | ||||||||||||||||||||||||
3 | 4 | 3 | |||||||||||||||||||||||||
3 | 6 | 6 | 2 | 6 | |||||||||||||||||||||||
3 | 6 | 6 | 7 | ||||||||||||||||||||||||
6 | 7 | 6 | 6 | ||||||||||||||||||||||||
2 | 2 | 2 | 1 | 3 | |||||||||||||||||||||||
2 | 6 | 6 |
gollark: > [Edit] Worth to note is that Gradual was designed to be a strategy that outperforms Tit for Tat. It has similar properties in that it is willing to cooperate and retaliates against a defecting opponent. Unlike Tit for Tat, which only has a memory of the last round played, Gradual will remember the complete interaction and defect the number of times the opponent has defected so far. It will offer mutual cooperation afterwards again, though.
gollark: The *description* of "Gradual" is pretty understandable.
gollark: How exciting.
gollark: Its score is actually identical.
gollark: ```scheme(define actually-forgiving-grudge (lambda (x y) (let* ( (defection-count (length (filter (lambda (m) (= m 1)) x))) (lookback (+ 1 (inexact->exact (floor (expt 1.8 defection-count))))) (result (if (member '(1 0) (take lookback (zip x y))) 1 0)) ) result)))```I think this detects betrayals properly now.
References
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