# Challenger de Yeongwol 2013 (dobles masculino)

Yeongwol Challenger Tennis :
| Campeones: | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić | Marin Draganja Mate Pavić |
| Subcampeones: | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin | Lee Hsin-han Peng Hsien-yin |
| Resultado: | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] | 6–4, 4–6, [10–7] |
| Modalidades |||||||||||
| Individuales | Masculino | | | |
| Dobles | Masculino | |

## Presentación previa

La pareja croata formada por Marin Draganja y Mate Pavić son los primeros campeones del torneo tras vencer en la final a los taiwaneses Lee Hsin-han / Peng Hsien-yin.

## Cabeza de serie

| Marin Draganja / Mate Pavić (Campeones) Rameez Junaid / Frank Moser (Cuartos de final) | Sanchai Ratiwatana / Sonchat Ratiwatana (Cuartos de final) Lee Hsin-han / Peng Hsien-yin (Final) |

## Cuadro

| Primera ronda | Primera ronda | Primera ronda | Primera ronda | Cuartos de final | Cuartos de final | Cuartos de final | Cuartos de final | Cuartos de final | Semifinales | Semifinales | Semifinales | Semifinales | Semifinales | Final | Final | Final | Final | Final |
| 1 | M Draganja M Pavić | 6 | 6 |
| WC | Chung Hyeon Y-K Lim | 2 | 1 |
| WC | Chung Hyeon Y-K Lim | 2 | 1 | 1 | M Draganja M Pavić | 4 | 6 | [11] |
| | | | | | R Gonzales A Sitak | 6 | 1 | [9] |
| | R Gonzales A Sitak | 6 | 6 | | R Gonzales A Sitak | 6 | 1 | [9] |
| PR | A Bury D Kosakowski | 3 | 2 | | | | | |
| PR | A Bury D Kosakowski | 3 | 2 | | | | | | 1 | M Draganja M Pavić | 6 | 7 | |
| | | | | | | | | | | M Bašić T Chen | 3 | 6$^4$ | |
| 3 | Sa Ratiwatana So Ratiwatana | 7 | 6 | | | | | | | M Bašić T Chen | 3 | 6$^4$ | |
| | T Daniel J McGee | 5 | 1 | | | | | | | | | | |
| | T Daniel J McGee | 5 | 1 | 3 | Sa Ratiwatana So Ratiwatana | 0 | 6$^5$ | | | | | | |
| | | | | | M Bašić T Chen | 6 | 7 | | | | | | |
| | M Bašić T Chen | 6 | 6 | | M Bašić T Chen | 6 | 7 | | | | | | |
| | Wu Du Zhang Ze | 1 | 3 | | | | | | | | | | |
| | Wu Du Zhang Ze | 1 | 3 | | | | | | | | | | | 1 | M Draganja M Pavić | 6 | 4 | [10] |
| | | | | | | | | | | | | | | 4 | H-h Lee H-y Peng | 4 | 6 | [7] |
| WC | J S Nam N Sang-Woo | 4 | 4 | | | | | | | | | | | 4 | H-h Lee H-y Peng | 4 | 6 | [7] |
| | B Klein T-h Yang | 6 | 6 | | | | | | | | | | |
| | B Klein T-h Yang | 6 | 6 | | B Klein T-h Yang | 3 | 4 | | | | | | |
| | | | | 4 | H-h Lee H-y Peng | 6 | 6 | | | | | | |
| | L-C Liang E López | 3 | 5 | 4 | H-h Lee H-y Peng | 6 | 6 | | | | | | |
| 4 | H-h Lee H-y Peng | 6 | 7 | | | | | | | | | | |
| 4 | H-h Lee H-y Peng | 6 | 7 | | | | | | 4 | H-h Lee H-y Peng | 6 | 3 | [10] |
| | | | | | | | | | WC | H-T Lee D Udomchoke | 2 | 6 | [6] |
| | M Reid J Ward | 4 | 1 | | | | | | WC | H-T Lee D Udomchoke | 2 | 6 | [6] |
| WC | H-T Lee D Udomchoke | 6 | 6 | | | | | |
| WC | H-T Lee D Udomchoke | 6 | 6 | WC | H-T Lee D Udomchoke | 3 | 7 | [10] |
| | | | | 2 | R Junaid F Moser | 6 | 5 | [3] |
| | Sergey Betov B Kavčič | 3 | 0 | 2 | R Junaid F Moser | 6 | 5 | [3] |
| 2 | R Junaid F Moser | 6 | 6 |