Basic topics (1)
i.
To
compete for the part score
ii.
To
help partner with the lead
If your partner
has already passed is it worth overcalling with 10 points and Qxxxx or worse in
a suit? The answer is ‘rarely’.
So you should have between say 8 and 17 points for an overcall. Try to overcall with at least 5 cards in the
major or 6 cards in a minor (especially if vulnerable).
Should you
overcall a 4 card suit? Again the
answer is ‘rarely’. However, over 1 of
a minor, consider a lead directional overcall of 1♠ with a 4 card suit if
say it is KQJx or AKQx. Best not to
overcall 1♥ with only 4 cards.
What about if you
are a passed hand? The answer is that
if the opponents stop in a partscore, say 3♠ or below, play double as
take out still. You should agree this
with your partner first!!! See example
at the end.
Example:
♠ K1096
♥ K1076
♦ 2
♣ KJ105
♠ Q8543 ♠ J2
♥ A53 ♥ 82
♦ 87 ♦ KQJ9654
♣ A87 ♣ 32
♠ A7
♥ QJ94
♦ A103
♣ Q964
How it should go.
N E S W
P 3♦ P! P
X P 4♥ All pass
You expect a
maximum of 24 high card points but now you know quite a bit about the
hand. Your partner is short in ♦ and you have the ace. Indeed your 2 aces and 2 9s plus a 10 make a
jump to 4♥ very attractive. Even if you bid 3♥
at pairs, your partner should raise with such perfect cards. Note that on best defence 3♦ X is one down.
This Law is vitally important to understand when pre-empting or competing
for the partscore. Simply put, you are ‘safe’ to bid to the level of your longest fit.
This means that if you know that you have a 10 card fit with your partner
it is safe to bid to the 10 trick (4) level.
No mention of points here. This
is not a trick!
If your partner opens a weak 2 and you have 4 card support, consider rasing
to the 4 level even if you have no points.
Be a little careful when you are vulnerable and they are not. Also take care when you are 4333 shape treat
it as 3333 shape.
Why does this work? If, the example
above, you have a 10 card fit and so do the opposition, the LAW says that if
you can make 10 tricks so can they... if you can make only 8 they can make
12... in other words, the total number
of tricks available = sums of the fit in the two longest suits. This is nearly always correct however there
are good books on the subject.