2013 Torneo Internacional AGT – Doubles

Doubles
	2013 Torneo Internacional AGT
Champions	Chris Guccione ; Matt Reid
Runners-up	Purav Raja ; Divij Sharan
Final score	6–3, 7–5

John Peers and John-Patrick Smith were the defending champions but decided not to participate together.
Peers played alongside Jamie Murray, Smith partnered up with Samuel Groth. Both pairs lost to Marcelo Demoliner and Franko Škugor in the first and second round, respectively.
Chris Guccione and Matt Reid defeated Purav Raja and Divij Sharan 6–3, 7–5 in the final to win the title.

Seeds

Nicholas Monroe / Simon Stadler (Semifinals)
Jamie Murray / John Peers (First Round)
Sanchai Ratiwatana / Sonchat Ratiwatana (First Round)
Purav Raja / Divij Sharan (Final)

Draw

Key

Draw

First Round					Quarterfinals				Semifinals				Final

1	N Monroe S Stadler	6	6
WC	Mauricio Astorga M Sánchez	2	1		1	N Monroe S Stadler	6	6
	M Draganja M Pavić	6⁶	6³			A Martin J Millman	3	3
	A Martin J Millman	7⁸	7⁷						1	N Monroe S Stadler	4	2
4	P Raja D Sharan	6	6						4	P Raja D Sharan	6	6
	H-h Lee H-y Peng	3	3		4	P Raja D Sharan	6	6
WC	JM Elizondo N Massú	3	6	[6]		J Cerretani A Menéndez-Maceiras	4	0
	J Cerretani A Menéndez-Maceiras	6	3	[10]									4	P Raja D Sharan	3	5
	A Krajicek B Reynolds	6	6											C Guccione M Reid	6	7
WC	Alfredo Moreno Alejandro Moreno Figueroa	3	2			A Krajicek B Reynolds	4	4
	C Guccione M Reid	6	3	[10]		C Guccione M Reid	6	6
3	Sa Ratiwatana So Ratiwatana	1	6	[2]						C Guccione M Reid	6	6
	S Groth J-P Smith	6	6							M Demoliner F Škugor	3	4
	J Kerr D Young	2	4			S Groth J-P Smith	4	4
	M Demoliner F Škugor	7⁷	7⁷			M Demoliner F Škugor	6	6
2	J Murray J Peers	6⁴	6⁵

gollark: ... yes, I forgot that.

gollark: Which I think gives you 2/15.

gollark: It would be quite annoying on larger things, but if you had, say, a 3-sided die, a 4-sided one, and a 5-sided one, and wanted to have 2 of them show a 1, then the possibilities are just 1, 1, anything and anything, 1, 1 (order is 3-sided, 4-sided, 5-sided).So you can work out the probability of each case (1/3 * 1*4 * 1 and 1 * 1/4 * 1/5) and add them.

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?

References

Main Draw

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