Re: Swiss teams decision 07/13/2008
- From: wheaties <dwheat1961@xxxxxxxxx>
- Date: Mon, 14 Jul 2008 12:36:22 -0700 (PDT)
On Jul 14, 9:57 am, "Bud H" <budh9534REMOVES...@xxxxxxxxx> wrote:
"Tim DeLaney" <delaney.timo...@xxxxxxxxxxx> wrote in message
news:fde0a23b-a963-4b73-8a43-84ccec63cb9a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You are in contention in a club Swiss team event. You figure to be a
bit better than the opponents at both tables, but you are behind the
leaders, who you lost to earlier. Last round (Playing something close
to SAYC)
Both vul IMPs You are West
Q62
A94
K987
KQ5
West North East South
P P
1D 1S 1NT 2S
?
Whatm is your action, and what are your thoughts?
Tim
With the possibilities being
1. Turn +100 into +200 (or +200 into +500)
2. Turn -110 into -670.
what are the chances of 2S making?
Either LHO overcalled a 4-bagger, RHO raised on a doubleton, or partner bid
1NT with a doubleton - which in itself is VERY unusual. With the 8 to bad
11 HCPs she should hold, I would expect her to not have a minimum and a very
strong doubleton honor. Even K10 would likely not be strong enough for me.
AK or AJ doubleton would be more likely.
And partner didn't raise diamonds. With four diamonds in her hand, a 2D bid
would be far more inviting than a 1NT bid. Now that I think about, if I
held a good THREE card diamond suit, a diamond raise might be more
attractive on many hands. This also means your 8-card fit, which you must
have in some suit if they have an 8-card spade fit, is going to be in clubs
(since no negative double from partner).
I would expect you hold about 23 HCPs between you and consider the chance of
them making 2S at less than 33% and a two trick set to occur almost half the
time. I would expect partner would lead trump with AK doubleton, so if a
diamond is led, I'll expect AJ doubleton in her hand and likely will be
leading trumps when I first get the lead through declarer's king.
Bud H- Hide quoted text -
- Show quoted text -
Just for the record, an 8 card enemy fit does NOT preclude the absence
of an 8 card fit with partner as in:
3=4=4=2 opp 2=3=3=5 or 3=3=4=3 opp 2=4=3=4
I know you mentioned no negative double and probably meant to say that
the combination of an 8 card enemy spade fit combined with the lack of
four hearts in partner's hand will yield an 8 card fit and that is
true (since 26 cards minus 5 spades minus (at most) 6 hearts = 15
cards which must be divided into 8+ and 7-)
Daivd W
.
- References:
- Swiss teams decision 07/13/2008
- From: Tim DeLaney
- Re: Swiss teams decision 07/13/2008
- From: Bud H
- Swiss teams decision 07/13/2008
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