Post by Dave FlowerK 10 8 x x x x
Q J 10
-
A Q x
A Q
A K
A K Q 10 9 x x
K x
North opened 1S, and south bid 7NT
West led H8 (probably second highest from rubbish), East following.
South won, and cashed SA, West discarding a heart.
Take it from there
You can win at most three tricks in hearts and three tricks in clubs
(due to only having three cards in each). We know where all the spades
are, and they're offside, so unless you can force East to discard the
Jack of Spades somehow you can win at most three spades. Likewise, we
have no way to take the diamond finesse without putting the opponents on
lead (thus failing to make 7NT), so we need to either drop the Jack of
Diamonds or force the opponents to discard it if we're going to make
more than three diamond tricks without losing one.
If West has 4 or more diamonds, the contract is obviously unmakable;
West can simply hold onto diamonds (and East to spades) and we have no
way to dislodge that.
If West and East have 3 diamonds each, the contract is trivially makable
because the diamonds run.
The interesting case is if East has 4 or more diamonds. Making the
contract without the defenders being able to stop you then requires
playing six tricks in hearts and clubs, with East being unable to hold
both the guard in diamonds and spades. Unfortunately, at that point,
South will have /only/ diamonds and spades, leaving the spade suit as
the only possible entry between the hands. If East unguards diamonds,
you can make the contract as long as the Queen of Spades hasn't been
played: three hearts, three clubs, two spades (the Ace you cashed
earlier and a low card to the Queen as an entry), five diamonds. If
East unguards spades, though, you'd need to have played the Queen of
Spades already, because otherwise you can't run spades in the North hand
(because the suit is blocked). So sadly, you have to commit to a suit to
attack before East gets to discard, meaning that any possible squeeze
attempt can be broken up somehow.
Any possible winning line in the situation where East has length
would thus rely on denying East information about the hand; although
East has a winning strategy to beat any potential play by you, that
doesn't guarantee that East will be able to find it. However, that
doesn't seem to be possible on this hand either; you'll have shown out
of hearts and clubs by the time East has to make a decision, and West
has shown out of spades, so East will pretty much know the entire hand
by that point. The only thing East won't know is /which/ diamonds are in
your hand, but a good East should be able to work out that after three
clubs, three hearts, two spades (the Ace played on trick 1 and the
low-to-the-Queen as an entry) South will be stranded and be forced to
play diamonds repeatedly, so the only suit possibly worth guarding is
diamonds.
As such, I don't think you can do any better than 13 tricks if diamonds
split and 12 tricks otherwise. The way to do that is to play three
diamonds immediately, running diamonds if they split, otherwise cash all
the remaining honours in South's hand and use the club x as an entry to
North's (to make 12).
(Incidentally, I'm curious as to how this hand would work in a suit
contract. 7D seems like it might be makable (although obviously not
after the start above!). 7S won't make because the Jack of trumps is
offside, and the opponents have long trumps in hearts or clubs.)
--
ais523