MinorKid Posted September 20, 2012 Report Share Posted September 20, 2012 Case 1:[hv=pc=n&s=shkqjt32d2cakqj32&d=n&v=b&b=13&a=7d7s]133|200[/hv] 7♦: "I am making 7♦ !" Case 2:[hv=pc=n&s=shkqjt32d2cakqj32&d=n&v=b&b=13&a=1d7s]133|200[/hv] 1♦: 11~15 HCP, 2+♦ These cases are not duplicates, of course there could be different hands among the three players. 1 Quote Link to comment Share on other sites More sharing options...
RunemPard Posted September 20, 2012 Report Share Posted September 20, 2012 7♦ by N is a preempt? This must be that so called goulash with agreements about this before hand. In that case I will go with the obvious 8♦. Quote Link to comment Share on other sites More sharing options...
RunemPard Posted September 20, 2012 Report Share Posted September 20, 2012 On a more serious note...I pass. Although double will most likely take the hand down...but I am paranoid any time I play in these tournies. Partner probably sitting on a ♠ void and the other opp has the other ♠ and void in our other suit... :) Evilness at it's best. Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted September 20, 2012 Report Share Posted September 20, 2012 There's a case for partner having 12 diamonds and A♥ and RHO having 13 spades, but I'm not sure I'm prepared to take the risk of RHO only having 11 or 12 of them :) Quote Link to comment Share on other sites More sharing options...
MinorKid Posted September 20, 2012 Author Report Share Posted September 20, 2012 I trust my partner for making 7♦. If he has AKQJxxxxxxxx and a small card, he will bid 6♦. But WTF is 7♠?? There's a case for partner having 12 diamonds and A♥ and RHO having 13 spades, but I'm not sure I'm prepared to take the risk of RHO only having 11 or 12 of them :) Partner could be 12 ♦ with the ♠A 1 Quote Link to comment Share on other sites More sharing options...
CSGibson Posted September 21, 2012 Report Share Posted September 21, 2012 obvious 7N bids, as long as its from my side (god, I hope lefty has a spade void...and no aces...) Quote Link to comment Share on other sites More sharing options...
gwnn Posted September 21, 2012 Report Share Posted September 21, 2012 double (negative) 2 Quote Link to comment Share on other sites More sharing options...
MinorKid Posted September 22, 2012 Author Report Share Posted September 22, 2012 How about case 2 ?Are you sure that West holds ♠ void this time? 1 Quote Link to comment Share on other sites More sharing options...
rogerclee Posted September 22, 2012 Report Share Posted September 22, 2012 How can you think LHO is likely at all to have a spade void? RHO's 7S bid doesn't even guarantee holding 10 spades, let alone 13. 2 Quote Link to comment Share on other sites More sharing options...
CSGibson Posted September 22, 2012 Report Share Posted September 22, 2012 To clarify, I was joking. At the table I would make Gwnn's negative double. Quote Link to comment Share on other sites More sharing options...
gszes Posted September 22, 2012 Report Share Posted September 22, 2012 Case 1:[hv=pc=n&s=shkqjt32d2cakqj32&d=n&v=b&b=13&a=7d7s]133|200[/hv] 7♦: "I am making 7♦ !" Case 2:[hv=pc=n&s=shkqjt32d2cakqj32&d=n&v=b&b=13&a=1d7s]133|200[/hv] 1♦: 11~15 HCP, 2+♦ These cases are not duplicates, of course there could be different hands among the three players. case 1 PASS not out of fear-- but because you want to convey to p a willingness to try 7nif they have the spade A. This means any other hand just x 7s even with zero. It is far tooimportant to leave us a chance of making 7n to use x in any other fashion even if thesituation is going to be rare. Note that our hand can be far weaker overall to make this pass (ie) void Axxxxx x Axxxxx case 2 a case can be made for treating this hand the same as hand 1. The number of hands(after a 1d bid) that would meet the requirements for a try for 7n can probably be counted using no more than your fingers and toes. This extremely low % means it is not mathematically worthusing this way and IMO pass here would merely mean too weak to x 7s and X is a plain oldpanalty x (this is what i would choose whatever top club your partnership leads looking for count). a direct bid of 7n in either case is just a bid designed to get rid of your current partner. 1 Quote Link to comment Share on other sites More sharing options...
Antrax Posted September 22, 2012 Report Share Posted September 22, 2012 gszes, do you think it's often people bid 7♠ missing the ace of spades? I mean, this is goulash so any discussion of probability is silly to begin with, but ignoring that. Quote Link to comment Share on other sites More sharing options...
MrAce Posted September 25, 2012 Report Share Posted September 25, 2012 To clarify, I was joking. At the table I would make Gwnn's negative double. You scared me for a second !! I was about to call 911 and send people to your house :lol: Quote Link to comment Share on other sites More sharing options...
gwnn Posted September 25, 2012 Report Share Posted September 25, 2012 Next week on Awkward. : 1NT-p-7NT-p;p-x-xx-??pass or pull?? Quote Link to comment Share on other sites More sharing options...
MinorKid Posted September 27, 2012 Author Report Share Posted September 27, 2012 case 1 PASS not out of fear-- but because you want to convey to p a willingness to try 7nif they have the spade A. This means any other hand just x 7s even with zero. It is far tooimportant to leave us a chance of making 7n to use x in any other fashion even if thesituation is going to be rare. Note that our hand can be far weaker overall to make this pass (ie) void Axxxxx x Axxxxx case 2 a case can be made for treating this hand the same as hand 1. The number of hands(after a 1d bid) that would meet the requirements for a try for 7n can probably be counted using no more than your fingers and toes. This extremely low % means it is not mathematically worthusing this way and IMO pass here would merely mean too weak to x 7s and X is a plain oldpanalty x (this is what i would choose whatever top club your partnership leads looking for count). a direct bid of 7n in either case is just a bid designed to get rid of your current partner. Case 1.Partner has ♦AKQJxxxxxxxx ♠A OR ♦AKQJxxxxxxxx ♥AI guess most pairs would like to defend 7♠ or 7♠X . 7NT will make when he has the first hand or RHO holds all thirteen diamonds, which is not a big chance. However, if you have (void Axxxxx x Axxxxx) instead than 7NT will make for sure because your partner must have the ♠A . Case 2 is a bit different from case 1.This time you judge your RHO, given that 7♠ is going to make.You can calculate that either we can make 7NT or your partner were underualified for the 1♦. Quote Link to comment Share on other sites More sharing options...
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