Winkle/Stepping Stone Squeeze

Winkle and Stepping-Stone

There is not a lot of difference between the winkle and the stepping stone, so I will treat them together.

Elsewhere, I analyze winner squeezes. The purpose of a winner squeeze is to force an opponent to discard a winner, or a small card that leads to partner's winner. In the typical winner squeeze, this allows you time/space to set up a trick by force. In the winkle/stepping stone, it forces the opponents to lead to a good trick that you could not cash because of blockage.

So I would classify the winkle and the stepping stone in with the winner squeezes. But they are different from the typical winner squeeze.

Introduction: The Basic Stepping-Stone and Winkle

This is the basic stepping-stone squeeze:

	AJ
	x
	x
	--

Qx            --
--            --
A             --
x             AKQJ

	K
	A
	Kx
	--

Declarer wants two more tricks. Declarer has these tricks in spades, but the suit is blocked. However, Lefty has to protect against the threat that declarer overtakes the K of spades with the ace. That requires saving two spades. Lefty also has to protect against the diamond threat. So whatever Lefty's last card is, it must be discarded, reducing the position to this:

	AJ
	--
	x
	--

Qx            --
--            --
A             --
--            AKQ

	K
	--
	Kx
	--

Now declarer cashes the king of spades, then leads a diamond. Lefty must lead a spade to the good ace of spades on the board. In essence, declarer has used the ace of diamonds as a stepping stone to the good spade on the board.

However, there is another way of phrasing the situation, one that is more compatible with the logic of a winner squeeze. From the defender's point of view, the goal is to take the last two tricks after the king of spades is cashed. However, their ability to do this is fatally compromised by the need to also have the queen of spades be one of the last two cards.

To create the winkle, I just need to eliminate the defense's club winners and give the defense another diamond winner in Righty's hand.

	AJ
	x
	J
	--

Qx            --
--            --
Kx            A109
--            --

	K
	A
	Qx
	--

The heart winner forces Lefty to discard. Lefty cannot pitch a spade, and jettisoning the king of diamonds allows a diamond winner to be established by force. So Lefty has to discard the small diamond. Now the king of spades is cashed and a diamond is led. If Righty overtakes the king of diamonds, then the queen of diamonds is good; if Lefty is allowed to win the king of diamonds, the last trick goes to the ace of spades on the board.

Again, it is very useful to analyze this from the perspective of a winner squeeze. Lefty has to discard either a winning diamond or a small card that could be played on partner's winning diamond. Therefore, the need to keep a spade in the last two cards fatally compromises the defense's ability to win two diamond tricks. (It would, of course, have been simple for the defense to win two diamond tricks before Lefty was squeezed.)

A Variation

As Reese presents them, both the stepping stone squeeze and the winkle work to take 2 of the last 3 tricks. But a loser can be added to the winkle/stepping stone suit without any problem. I have never seen this variation, but it occurred in play:

On January 16, at a club game, playing matchpoint and a contract of 3NT, declarer cashed 9 winners in the red suits, eliminating all of the red cards and coming down to this position with the lead in dummy. Note that neither of the black suits have been played.

K
--
--
Jxx

AJ
--
--
Qx

If one defender has the Queen of spades and both the A and K of clubs, then the stepping stone squeeze has already worked: That defender has been forced to discard down to

Qx
--
--
AK

To complete the squeeze, Declarer cashes the king of spades, leads a club, and wins the last trick with the ace of spades.

If the club honors are split, then the winkle has already operated. The situation will be exactly as given above:

	K
	--
	--
	Jxxx

Qx            --
--            --
Kx            AJ10x
--            --

	AJ
	--
	--
	Qx

The only way to play this club suit to win a trick, before the winner squeeze has operated, is to lead to the queen of clubs hoping the AK is onside, then if that fails, play the J of clubs hoping that the other defender has AK. But in the 2-loser winkle, you duck a club from both hands. This reduces the situation to the one-loser winkle. In the situation above, if Lefty plays the king of clubs on the first club trick, a club is established by force; if Lefty plays the king on the club second trick and wins, the last trick goes to the ace of spade; and if Righty overtakes, the jack of clubs is good.

Again, the need to keep just one spade among the last three cards has fatally compromised the defense's ability to cash three club tricks. And if they cannot do that, perforce the ace of spades is winning the last trick.

How to Play?

Let's approach this question the other way. Suppose you were declarer, and you are faced with the standard two-loser winkle situation. What should you do?

K
--
--
Jxx

AJ
--
--
Qx

In actual play, declarer overtook the King of Spades with the ace, hoping the queen would drop. It did not, and declarer lost the last three tricks, for a bottom (instead of an average).

I rate this play as very unlikely to succeed, because the defender with the queen of spades is very unlikely to bare it. So best play is to first cash the king of spades.

The normal play in clubs would be to lead to the queen, hoping that the AK is onside. If that fails and Lefty wins, then Lefty has to lead small and you can play your jack, hoping that Lefty started with the AK.

That puts you up to about 50%. However, because a defender had to save the queen of spades, the odds are actually significantly higher. If Righty had to save the queen of spades, then Righty has only two clubs. Righty might win the jack of clubs, but then Righty has to lead a spade. This pushes the percentage for success to about 75%.

The odds on ducking the first round of clubs from both hands are much better. First, it is a winkle and succeeds whenever the club honors are split. Second, if the AK of clubs are together, it succeeds if that hand also has the queen of spades -- your stepping stone squeeze has already worked. That puts you up to the same about 75% (actually, 74%) of playing the queen of clubs on the first club trick.

That leaves the possibility that one defender has the AK of clubs and the other has the Q of spades. If the defenders defend correctly, you will go set about 25% of the time. But best defense is extremely unlikely.

First, a defender with the Q of spades is liable to come down to Qxx of spades and a club for the last four cards. First, that defender has little reason to save two clubs; second, that declarer is likely to worry that declarer's situation is this:

K
--
--
Jxx

AJx
--
--
A

If a defender saves Qxx of spades, then you will know that 2 of the defender's last six cards are spades. In this situation, you can lead to the queen, play the jack on the next lead (if the queen loses), and be absolutely assured of getting either a club trick or a spade trick.

I rate this as pushing the chance of success to almost 100%. (The same misdefence can occur for the first line of play. However, a defender is very unlikely to bare the club honor and save Qxx of clubs.)

But what if the defender with the Q of spades does come down to Qx (and hence there is only one spade out for the last three tricks)? The defense can still mis-step. If Righty has AKx of clubs as the last three cards, it is going to be difficult for Lefty to duck a club. For all Lefty knows, you will play your queen of clubs, it will win, then you will cash your ace of spades and he will get the bottom board instead of you.

Reprise

When I analyzed the "two-suited winner squeeze", I was astounded at the extremely high percentage of success. When I analyzed the two-loser winkle situation (above), I again was astounded at the extremely high percentage of success.

I am apparently not alone in this. I gave both problems to a good player, and he did not run a squeeze in the first situation or try the winkle in the second situation.

In retrospect, the two situations have the same reason for (1) seeming to have a low probability of success yet in fact (2) having a high probability of success. In both, the chance of success before the squeeze is low. In both, the chance for success becomes very high because the squeeze (usually) forces out a needed losing card. It is just very difficult to appreciate, without analysis, the problems created for the defense by the loss of this losing card.