This is a revised version of some quick comments I made yesterday.
On Wednesday there was an open discussion on squeezes. This hand came up. The lead was the
K and the squeeze works, but the question was raised about what happens on other leads.
Contract is 6
.
Let's first look at the hand with the
lead (a very likely lead!)
Q52
A82
Q864
KQ2
A
J74
AKJ975
A74
Opening lead, the
K. We assume W also has the Q for this play.
We count 1+1+6+3=11 obvious tricks on top, we need a 12th.
If W holds the
Q (highly likely) and the
K we can make this by ducking at T1, unless (most unlikely) the defense can get an immediate ruff at T2. So we duck.
A
continuation gives us a
trick, so he leads something else. We take it, we draw trump, we cash the black winners, and then we run the rest of the
.
After 10 tricks the position is
Q
A8
J7
5
We are assuming W was dealt the
K so, of course, he still has it. So he now has only two
.
We now play the last
, aka the squeeze card.
W plays what? If he plays the
K we can see what to do. So he plays a
, leaving him with only 1. He has to hope his partner has the
J but alas for him, we have it. We pitch the
Q from the board, a
to the A drops the last
on the left, the
J is good.
We pause to note that the duck at T1 served two purposes. It rectified the count (12 more tricks to be played and we have 12-1=11 winners) and, very important, it preserves the
A as an entry to the board. The
Q is to be a threat, but if there is no entry it is a toothless threat.
Ok, that takes care of the hand as presented. The question was asked whether the hand could be made on other leads. Well, Gib says yes so, in a way, that answers the question. Those bots are clever. But how? I first worked out how on a club or diamond lead and wrote that up, but now I also see how to do it on a spade lead. And it can all be done the same way. So: Assume that the lead is something other than a
. No matter what it is we can cash five rounds of
, throwing the
2 on the 5th
, then the
A, and then three rounds of
ending on the board. So we are all down to four cards:
Q5
A8
J74
5
The lead is on the board. Let us suppose we somehow know which four cards W has held onto. If his only
is the K, we ruff a small
dropping his K and go back to the board on
and take the
Q. But if he has kept the
Kx, then he must have only two
. Fine. Play
A and then the 9 from the board and the 7 from hand. We are now left with a trump and the good
J.
You might well ask: How can we tell which to do and the answer is we can't. But, perhaps, we guess right. Better than nothing.
Much better if we get the lead of the
K. We duck and then we run the squeeze. No ambiguity. It might not work, it won't work if the
K is with E, but there is no doubt about what we should try. In the more complicated case where the lead is a non-
, we just have to do our best. At the end, even if it has worked, we have to guess in which way it worked. Life is like that sometimes.
I was going over this in my head while drinking my morning coffee and I realized that this squeeze is not positional. That is, as long as one opponent, either one, has all three of the
KQ and the
K it will work. This makes it at least a little more likely that it could show up in a real situation. As the cards actually lay, probably most defenders would start with the
K at T1 holding KQx and this was what happened. But on another day, with Rho holding
AK and
K, Lho might well hold
JT9x and start with the
J. Now this squeeze earns its keep.
