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Bridge Winners Profile for Andrew Gumperz

Andrew Gumperz
Andrew Gumperz
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Basic Information

Member Since
Sept. 18, 2010
Last Seen
July 21
Member Type
Bridge Player
about me

Andrew Gumperz is a part-time bridge professional based in the SF bay area. He has numerous regional victories, but his proudest bridge accomplishment was upsetting the CAYNE team at the 2009 Spingold in Washington DC. In his spare time, Andrew enjoys musical theater, especially when his daughter is performing.

United States of America

Bridge Information

BBO Username
ACBL Ranking
Gumperz and Brenner
2 over 1
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Gumperz CC
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Gumperz and Wagner
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Jannes van 't Oever's bidding problem: QJ43 J4 A642 T95
After partner removes my double, I plan to raise 3 to 4 and 3 to 4.
Yehudit Hasin's bidding problem: QTx x ATxx Q8xxx
i am in the camp that thinks penalty double is on the hungry side at IMPs. To show a worthwhile profit against 3X, you need to beat them 2+ tricks. The penalty doubler has 2.5 defensive tricks (admittedly with some chances for an extra). Double is implicitly asking ...
Avon Wilsmore's bidding problem: AJT53 KJ43 T64 5
What is more likely, exactly 3-3-1-6 or partner holding one of: 2-4-2-5, 2-5-2-4, 1-5-2-5, etc. So, yes, double can come up empty, but so can 2 and it may well do so more often than double.
Avon Wilsmore's bidding problem: AJT53 KJ43 T64 5
Even if 2 is an inverted raise, game is still possible for our side. Give partner as little as: Kxxxx, Qxx, x, xxxx and we can make a surprise 4 game. (And remember, partner could have an ace more than that). So even though entering the auction ...
Jess Cohen's bidding problem: 9 JT AQJT976 AJ9
3 is much more attractive at IMPs.
Kevin Fay's bidding problem: 97532 Q9 A62 973
There are game chances and vul at IMPs I don't need great odds to justify bidding it. I want to offer 4 and 5. However. 4 leaves partner room to bid 4, if his are excellent, which I will pass.
Jeff Lehman's lead problem: A654 84 T73 QT82
Excellent point.
Restricted Choice
“The statement that restricted choice produces 2-to-1 odds is based on the implicit assumption that P(x)=P(y)=P(xy) which does not always hold.” It's not an assumption. It always holds (the second equal sign should be an “approximately equals” sign) for a fair and honest shuffle ...
Niall Igoe's bidding problem: JTxx QJTx xxxx x
Both are vul. Even if we do have a fit, we are no lock to make a 2-level contract.
Paul Gipson's lead problem: 75 K742 A532 QJ5
I need less from partner in hearts than in clubs to establish 2 tricks for the defense (and i don't expect partner to hold much). I might rethink this aggressive lead if my opponents are known to be weak players.

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