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Bridge Winners Profile for Carl Mathiesen

Carl Mathiesen
Carl Mathiesen
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Basic Information

Member Since
March 20, 2013
Last Seen
July 11
Member Type
Bridge Player
about me

Occasional commentator on BBO - "Sharkey"

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Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
My hand was AJ109xx - A9xxx AJ. Yes - finesse was on, but were 3-0 so 12 tricks only. Played at ACBL afternoon game - more than 200 tables. One pair bid slam.
Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
Nikos, the negative x was "normal" - usually showing 4+ H: Nothing unusual about it
Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
Agree with your strategy, Henry, 5 is not a favorite MP contract when game is an alternative. You omitted my choice, 5, which I hope pd will interpret as a VERY good raise to 5. With Q, A and !K10xx I don't think 5 ...
Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
If that was part of our system it would have been stated
Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
We had no agreements as to whether 3 forcing or not. If pd stretches with xx KQJxx Jxx xxx hard to blame him. Should he really bid over 3 from pd holding AKJxx x AKxxx Kx? Regarding Richard's comment on bidding 2 with KJ10xxxxx don't ...
Carl Mathiesen's bidding problem: Q8 AT64 KT63 T75
We had no agreement in this regard. Paul's point is well taken - 4 may be construed as "to play" in this context.
Carl Mathiesen's bidding problem: 9 KQJ97 Q752 K94
I think 5 may be a stretch - A would be much preferable to KQJ as a ruff is looming. Pd and I obviously didn't have a specific agreement here - difficult as pd may in some other scenarios may Pass with xx QJ9xxxx xxx xx. Perhaps I should ...
Carl Mathiesen's bidding problem: 9 KQJ97 Q752 K94
On the actual hand pd bid 2 - I bid 2NT with Jxx 10x AKx AQJxx as I "assumed" pd had at least a partial stopper.. Pd continued with 3 and I closed with 3 NT which was not the winner. LHO had 4 and did not support ...
Carl Mathiesen's bidding problem: 9 KQJ97 Q752 K94
In my mind 2 would be to play. Think this is a difficult problem where agreements are needed
Carl Mathiesen's bidding problem: AK64 A53 AQJ62 5
At the table I x hoping my extra strength would make up for the lack of the 4th . If pd bids 3 correct to 3. In this case 2 NT and x would be successful. Pd's holding was J10xx-
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