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Bridge Winners Profile for Kurt Schneider

Kurt Schneider
Kurt Schneider
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Basic Information

Member Since
July 19, 2012
Last Seen
Aug. 12
Member Type
Bridge Player
about me

Canadian living in NH

Country
United States of America

Bridge Information

Favorite Bridge Memory
Bidding and making 6CX with 10 combined HCP against perfect defense.
Favorite Conventions
Anything involving canapé
ACBL Ranking
Bronze Life Master
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Paul Hightower's bidding problem: KQ743 AJT5 --- KQJ9
I enclosed the loser reference in quotations for specifically that reason. But using any online reference would show that I need 18+ "points" (including useful distribution - which you have in three of the four suit contracts) for a jump shift. Bidding only 2 would miss solid games when responder ...
Rafael Sacramento's bidding problem: A2 A3 Q7 AKQJ982
9 tricks only in clubs... There is no standard bid available for this. By opening 2N, you promise at least 2 in each suit, 20-21 HCP, and keeps slam in the equation. Opening 2C gets you nowhere fast.
Paul Hightower's bidding problem: KQ743 AJT5 --- KQJ9
You have a "four-loser" hand - how can you not jump to 3?
Rafael Sacramento's bidding problem: A2 A3 Q7 AKQJ982
Open 2N - seems to describe the hand well.
Trust the Lead
Alll of this extra structure can be obviated by opening 4card majors. It won’t be Precision but it will work better with a strong club.
Stig Holmquist's bidding problem: AKQJ8754 Q KJ8 6
I con-kehr...
Stig Holmquist's bidding problem: AKQJ8754 Q KJ8 6
4 Namyats
What bidding systems are most advantageous to the play?
A relay system where dummy defines their hand would provide the opponents with a minimum amount of useful information while allowing you with all the information you would need to make sensible contract decisions. Relays can be easily attached to strong club front ends.
Kieran Dyke's bidding problem: KQ KQ753 KQ9 AJT
I expect partner to have at least 43 in the majors for the 4 bid. With 6 clubs, a GF and no interest in 3N, I would bid 4 - why would this NOT be forcing in this auction? IMO X would be the non-GF method of landing in ...
Frances Hinden's bidding problem: 53 53 Q9 Q976542
My point that is being missed here - opener is highly likely to have a doubleton club, significantly more so than a tripleton (singletons and voids have been eliminated due to the 2N bid). In that case, no matter partner's point count, 2N is in jeopardy. And if 2N makes ...
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