, a singleton 2 suffices, provided the other trumps split
6-6. Declarer makes seven tricks in the side-suits.Last newsletter, I set the problem of finding the worst combined trump holding with which it is still possible to make a given number of tricks. The defence is assumed to be optimal, except that a trump lead was disallowed. `Worst' is defined by total length and then by highest card, so that 432 is better than A2 which is better than KQ.
Here are the solutions and best entries. It may still be possible to improve on these, but I think they're optimal.
(a) To make 1, a singleton 2 suffices, provided the other trumps split
6-6. Declarer makes seven tricks in the side-suits.
(b) To make 2, declarer must make a
trump trick. A stiff ace of spades is enough against any lead,
but if a trump lead is barred, a stiff K suffices. Declarer cashes seven tricks in the side suits
and then leads a suit in which he is void towards the king. If
RHO has the A, the king scores en
passant.
(c) To make 3, the AK in one hand would suffice, with trumps breaking 6-5.
However, the worst possible holding is K8 split between the
hands. Consider the following layout:
|
|||||||||||||||||||
After cashing 7 tricks declarer leads a heart from dummy,
ruffing with the 8, while LHO has to follow. A club from hand
allows the K to score en
passant once more.
If a trump lead is permitted, then
A8 in one hand will do.
(d) To make 4, declarer needs at
least three trumps between the hands. Many entrants missed the
optimum holding on this one, as curiously the best layout for
declarer involves an uneven trump break!
|
|||||||||||||||||||
After the seven side suit winners declarer ruffs a diamond
in hand and a club in dummy, with LHO under-ruffing twice, and
then scores the K as before.
If a trump lead is permitted, K102
in one hand suffices, on a 5-5 break.
(e) To make 5 requires 3 trump
tricks. AQ9 in hand is optimal,
whether a trump lead is permitted or not. Declarer scores eight
tricks then leads towards the AQ9, intending to ruff low and
exit with a side suit. RHO, holding
KJ10xx must ruff with an honour, but declarer discards. RHO
must then lead a trump which declarer wins and exits with a
side suit.
(f) A small slam requires 4 trumps between the hands, with a
5-4 break. Best seems to be AQ10 in
hand and a stiff 7 in dummy. After 8 winners declarer scores
the 7 while LHO, with his familiar holding of 65432, wishes
he'd led one. Then a ruff in hand and an endplay as before.
A trump lead is very damaging to this plan. Giving declarer
AQ108 in one hand is enough on a trump
lead, but not if LHO tries a side suit. I think AKJ7 in hand,
with Q1098 on the right is the weakest possible holding.
(g) And finally, what do you need for a Grand? Clearly AKQ87 in one hand is enough (fortunately the J109 drop). But the weakest holding, found by only two entrants (and myself!) is A8 opposite KJ7, provided The Q109 are under the jack. For instance:
|
|||||||||||||||||||
Eight side tricks ending in dummy, a 
ruffed with the 8, a 
with the 7
and a with the ace completes the
trump coup. This was the hand on the front cover. I hope you
elected to lead a trump? Bad luck! Confused by the auction,
partner leads a trump out of turn and declarer disallows a
spade lead. Note that no other grand slam makes on any
lead!
Allowing a trump lead does not require much strengthening of declarer's holding. With Q10xx onside KJ9 opposite A2 is unstoppable.
I received a number of very good entries, including some
from outside the county. Graham Brightwell, who is an ex-county
member, submitted a perfect set of solutions. Next best was
Julian Wightwick, who missed only the possibility of a 5-3
split in 7. These two will receive
mystery prizes at the editor's expense. A perfect set of
answers from Alistair Flutter and Paddy Boyle was submitted
after the closing date. Only slightly sub-optimal solutions
came from Chris Chambers (Suffolk) and Alan Mould (Manchester).
These last two will receive a free copy of this newsletter. The
County's generosity knows no bounds!
KJ10xxx