I have been advocating what I feel is the simplest and often the most useful form of counting: Counting the tricks available and needed in a hand.
Here is a recent hand from a club game.
J
AKQT8
J93
Q764
KQ8432
J9
K64
AT
You open the S hand 1
and the uncontested auction goes: 1
- 2
- 2
- 3
- 3NT All Pass.
The opening lead is the
Q . Given that you see the J in dummy you figure this is from Qx. Could be the stiff Q, or even Qxx perhaps, but probably Qx.
It's matchponits, but never mind the overtricks, you want to take 9 tricks. Where are they? You play low from dummy, RHO produces the 8. Standard carding. It seems likely RHO was dealt AJ8xx, as indeed he was.
Five hearts, one diamond, one club are in plain view, we need to more. We have six spades to the KQ and the stiff J on the board. Once the A is taken,we have two more top spades. Unless someone has a better idea, that looks like where we go for tricks.
OK, the plan is to develop spades. How, and what are the dangers?
The most obvious danger is that if you win the
at T1 and lead a
, LHO will go up with the A and lead a
, resulting in down 1.
Can you do anything about this? Well, yes. You can duck the first
. Does that solve the problem? Maybe, maybe not. Perhaps LHO continues ithe a second
, RHO takes his A and leads a third
, you lead a
and guess what? It was your RHO who held the
A so he takes it and takes the rest of his diamonds, down 1.
Here is the point: At T1, after you think through that you will come to 9 tricks by developing spades, you have a choice of how to go about it. Maybe you choose correctly, maybe not, but you have a plan and you make a choice. Under your assumption that the lead is from Qx you can figure that LHO holds 11 non-diamonds and RHO holds 8 non-diamonds so it seems there is a better chance that the
A is on your left, but there is no way to be sure.
Now for the follow-up: If you guess right, you want to collect. The actual declarer took the A on the first round, which could be right if the A was on her left, and then ran her hearts on the board. Uh oh. Suppose her right hand opponent started with Ax of spades and xx in clubs. After the
and the run of five
, he is down to four good
, Ax in
and x in
. As long as he plays ow when declarer now leads the
J , the hand is at least in danger. It's true that declarer could save herself by playing the Q on the J and then a low
dropping the A and establishing he K, but that's an extreme play that could easily go wrong.
Here is what would work, of the
A is to her right. Win the
then lay an immediate small
. RHO may well refuse to take his A. No problem, back to hand with a
. Lead the
K. Declarer takes 2+5+1+1=9 tricks.
The point is this: You count tricks and see that you need two tricks in
, or at least that appears a reasonably straightforward source for 9 tricks. So you go after them. You have to place a bet on where the
K is, but if you guess right they are not going to be able to stop you unless you destroy your transportation.
This is not all that sophisticated. It requires that you count to 9 at T1. Lacking any better idea, you go after the spades.
As it happened, the
A was on her left. If she ducks the first
she makes the hand.
Added: Of course LHO is not required to continue
if declarer ducks the first trick If he does, and I am confident that he would have, then 9 tricks are there. As the cards lie, no shift can hurt as long as declarer is up for thinking it all through. Anyway, I was the one with the long
, I played the 8, I seriously doubt partner was going to find a shift. A
shift is the most difficult to deal with. It would draw the J from me and the A from declarer. Now declarer's 9 tricks are not yet established and while I had no entry in
, I do now have one with the
K . The hand gets a bit tricky after that but if we assume the the defense is clever enough to switch we should probably assume that declarer is clever enough to cope. In the real world I think partner would have continued with another
. Just as I had asked him to with my 8.
