2013 Dunlop World Challenge – Men's Doubles
Philipp Oswald and Mate Pavić were the defending champions, but Oswald chose not to compete. Mate Pavić partnered with Marin Draganja, but they lost in the semifinals.
Men's Doubles | |
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2013 Dunlop World Challenge | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 4–6, 6–3, [10–4] |
Chase Buchanan and Blaž Rola won the title, defeating Marcus Daniell and Artem Sitak in the final, 4–6, 6–3, [10–4].
Seeds
Marin Draganja / Mate Pavić (Semifinals) Sanchai Ratiwatana / Sonchat Ratiwatana (First round) James Cerretani / Adil Shamasdin (Quarterfinals) Pierre-Hugues Herbert / Frank Moser (withdrew)
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
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 77 | 6 | ||||||||||||||||||||||||
![]() ![]() | 64 | 2 | 1 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
![]() ![]() | 4 | 710 | [12] | ![]() ![]() | 4 | 2 | |||||||||||||||||||||
![]() ![]() | 6 | 68 | [10] | 1 | ![]() ![]() | 61 | 4 | ||||||||||||||||||||
Alt | ![]() ![]() | 5 | 3 | ![]() ![]() | 77 | 6 | |||||||||||||||||||||
WC | ![]() ![]() | 7 | 6 | WC | ![]() ![]() | 5 | 2 | ||||||||||||||||||||
![]() ![]() | 4 | 78 | [8] | ![]() ![]() | 7 | 6 | |||||||||||||||||||||
![]() ![]() | 6 | 66 | [10] | ![]() ![]() | 4 | 6 | [10] | ||||||||||||||||||||
![]() ![]() | 4 | 66 | ![]() ![]() | 6 | 3 | [4] | |||||||||||||||||||||
![]() ![]() | 6 | 78 | ![]() ![]() | 4 | 6 | [10] | |||||||||||||||||||||
![]() ![]() | 6 | 4 | [5] | 3 | ![]() ![]() | 6 | 3 | [4] | |||||||||||||||||||
3 | ![]() ![]() | 3 | 6 | [10] | ![]() ![]() | 1 | 4 | ||||||||||||||||||||
![]() ![]() | 4 | 2 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
WC | ![]() ![]() | 6 | 6 | WC | ![]() ![]() | 65 | 1 | ||||||||||||||||||||
![]() ![]() | 7 | 6 | ![]() ![]() | 77 | 6 | ||||||||||||||||||||||
2 | ![]() ![]() | 5 | 3 |
gollark: Enumerate all the different possibilities where you have X dice showing 3, work out the probability of each, then add them?
gollark: Just multiply the probabilities for getting side X on each die together?
gollark: You also are probably not running Haskell with its giant runtime on a microcontroller doing those things.
gollark: My friend likes Haskell but also spends time reading incomprehensible papers on logic and type theory and such.
gollark: Possibly. People *allegedly* use them for real world applications occasionally.
References
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