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Bridge Winners Profile for Martin Lindfors

Martin Lindfors
Martin Lindfors
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Basic Information

Member Since
Sept. 24, 2017
Last Seen
an hour ago
Member Type
bridge player

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None
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Winston Chang's bidding problem: QT954 A963 8 A72
It does. Do the opps have a hidden 10 card diamond fit? Unlikely. But possible.
Matej Ivancic's bidding problem: T32 93 QJ42 AK87
Cue and correct 2 to 3. Invitational with 44 minors. Seems accurate.
Martin Lindfors's bidding problem: KT 8 AKT9543 T84
I agree. Would you be more likely to preempt if the spade king is changed to the queen?
Jenish Shah's lead problem: K86 KJ3 A76 Q732
The ace is equally dangerous to lead. In order for the ace to be unlikely to lose us a trick, we need partner to have the king. In order for a club to be unlikely to lose, all we need is for partner to have the club king or jack ...
Martin Lindfors's bidding problem: KT 8 AKT9543 T84
Alas, partner bid 3, making us miss a cold major suit game. I just wanted to check how popular the choice is.
Martin Lindfors's bidding problem: AQ63 A753 J KQ75
No, they aren't playing canapé openings. It's a 4 card major strong club system.
David Corn's bidding problem: AK7 Q A853 AQJ93
Double. 1NT over 1, raise 1.
Mark Raphaelson's bidding problem: AJ7 KQT8 A963 53
I don't want a spade to be lead through my holding when we might play a notrump contract, and with the second flaw of a doubleton club, double is out of the picture.
Winston Chang's bidding problem: AQ764 --- QT43 JT98
I have no problem because I opened 2.
What does 2S mean?
Even if the singleton queens are not counted, the hand should be opened according to the rule of 19. I would probably open it 3 though.
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