Transfers after 1D-X
				
						#1
						
								 
							  
								
						
					
				
				Posted 2013-August-04, 11:34
transfers over 1♣ double make a lot of sense and is easy to come up with a system that is nice
1♦ double a little more difficult. I think your gonna lose natural redouble. Anyone got a method?
if it makes a difference 1♦ is unbalanced
				
						#2
						
								 
							  
								
						
					
				
				Posted 2013-August-04, 13:43
				
						#3
						
								 
							  
								
						
					
				
				Posted 2013-August-04, 15:02
				
						#4
						
								 
							  
								
						
					
				
				Posted 2013-August-04, 18:00
				
						#5
						
								 
							  
								
						
					
				
				Posted 2013-August-04, 20:12
				
						#6
						
								 
							  
								
						
					
				
				Posted 2013-August-05, 01:44
So:
RDbl = 4+♥
1♥ = 4+♠
1♠ = transfer to NT
1NT = transfer ♣, pretty much any strength
2♣ = good ♦ raise
2♦ = weak ♦ raise
				
						#8
						
								 
							  
								
						
					
				
				Posted 2013-August-07, 12:55
				
						#9
						
								 
							  
								
						
					
				
				Posted 2013-August-07, 13:38
 fromageGB, on 2013-August-07, 11:52, said:
fromageGB, on 2013-August-07, 11:52, said:
In general I think you should follow the same style that you play over 1♣-(X)-transfer.
				
						#10
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 03:11
				
						#11
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 09:38
Playing T-Walsh it is easy to play a similar system after:
1x-(1D/H)-?
(with DBL is transfer to 1 over 1).
or after:
1C/D/H-(DBL)-
(with RDBL is transfer to 1 over 1).
=> I would recommend these transfers when playing T-walsh.
				
						#12
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 09:54
 kgr, on 2013-August-08, 09:38, said:
kgr, on 2013-August-08, 09:38, said:
1x-(1D/H)-?
(with RDBL is transfer to 1 over 1).
Its my understanding that under ACBL GCC conventional meanings may be given to redoubles and responses
(including free bids)
so you can use 1♦=♥ and 1♥=♠ etc. as these are free bids
				
						#13
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 10:13
 kgr, on 2013-August-08, 09:38, said:
kgr, on 2013-August-08, 09:38, said:
Playing T-Walsh it is easy to play a similar system after:
1x-(1D/H)-?
(with
 steve2005, on 2013-August-08, 09:54, said:
steve2005, on 2013-August-08, 09:54, said:
(including free bids)
so you can use 1♦=♥ and 1♥=♠ etc. as these are free bids
I obviously did mean DBL iso RDBL
(and all other 1-level bids are transfer, with 1♠ transfer to 1NT)
				
						#14
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 11:19
 steve2005, on 2013-August-08, 09:54, said:
steve2005, on 2013-August-08, 09:54, said:
(including free bids) thereto
Assuming you are referring to "Competitive" section, item 2, you missed a word.
1♣ (1♦) 1♥=♠ is not GCC-legal (unless 1♣ was strong and forcing, not the case under discussion here, or 1♦ was conventional, unlikely).
				
						#15
						
								 
							  
								
						
					
				
				Posted 2013-August-08, 11:44
Woe betide you if you find the one pair in the country that plays penalty doubles. Having said that, I might try that one day just for the director calls :-)
				
						#16
						
								 
							  
								
						
					
				
				Posted 2013-August-18, 17:49
 steve2005, on 2013-August-04, 11:34, said:
steve2005, on 2013-August-04, 11:34, said:
You can also try my Systems Index for this sort of thing. I just added this thread.
-- Bertrand Russell

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