Veering Squeeze Card Variant

[UPDATE: lots of posts the past couple of weeks; trying to settle in to a regular weekly schedule again starting now, though. I have several things lined up, so you should be able to rely on that for a while.]

Here's an 8 card double-dummy problem demonstrating a secondary compound squeeze with a veering squeeze card in the ending:

		♠	A Q 9 2
		♥	A K 5 4
		♦
		♣
♠	J 10 8			♠	K 4 3
♥	Q J 10			♥	9 8 7
♦	K J			♦	A Q
♣				♣
		♠
		♥	3 2
		♦	4 3 2
		♣	A K Q

Clubs trump. South to lead and take 7 of the last 8 tricks.

Solution below the fold.



Cash a trump. West must pitch a diamond (or set up spades or heart-spade strip vs East), then East is triple-squeezed. If a spade is pitched, use the 2 heart entries to ruff out the king and then enjoy the queen. If a diamond is pitched, duck a diamond to set up that suit. So, East must part with a heart.

On the next trump, West is triple squeezed. On a spade pitch, you can again cross to ruff out the king while pinning West's honors. A heart pitch sets up that suit. So West must pitch another diamond.



Now, cash the hearts and lead a 3rd one in this position:

		♠	A Q 9
		♥	4
		♦
		♣
♠	J 10 8			♠	K 4
♥	Q			♥
♦				♦	A Q
♣				♣
		♠
		♥
		♦	4 3 2
		♣	Q

This is the "familiar" veering squeeze card ending. If East pitches down to one diamond, ruff and throw him in to lead into dummy's spade tenace. If East blanks the ♠K, pitch a diamond. West wins and leads a spade so you can drop the king and get home.

Verified with Deep Finesse.
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Which squeeze? My solution

I posed this problem yesterday:

♠	A K Q 4 3
♥	K 10 4
♦	K 7 6
♣	6 4
♠	6 5
♥	A 9 3
♦	A 9 3
♣	A K Q J 10

You're in 7N facing the ♠J lead. How do you play?

This might not be right, but let's say that you start with 3 rounds of spades and RHO shows out on the 3rd round. Take over from there (don't forget to pitch something on this trick!).

Yesterday I suggested pitching a diamond, then running 4 clubs (pitching a spade and a heart) for a compound squeeze. The problem is that this is a restricted squeeze and you have to make a key guess before cashing the last club. There's a way to make it unrestricted, though, which gives you much better odds of reading the position. Answer below.



Pitch a diamond on the 3rd spade, run 4 clubs pitching a spade and the HT dummy, then cash DK, DA, and play the last club in this position (planning to pitch a spade unless it's good):

♠	4
♥	K 4
♦	7
♣
♠
♥	A 9 3
♦
♣	10

This is a compound guard squeeze that can work if LHO has either missing heart honor. You still have to read the position, but you'll almost always be able to. Your next play is the HK, and, unless the diamond is good, a low heart towards your hand. When RHO follows low, only then, halfway through trick 12, do you face your guess (and even then, only if you've seen a heart honor from LHO. Either RHO's last card is the other heart quack and you finesse (this is why you unblocked HT), or LHO has it along with a spade and you drop it.

For starters, if heart honors are split and you play for it, you'll always make: a bit over 52% of the time. You'll also pick up a bunch of scattered things, such as when RHO starts with the sole diamond guard (or LHO unguards diamonds unwisely) and both heart honors.

On the other hand, once you play this way, they can always beat you if LHO guards diamonds and RHO has both hearts.

Overall, not that big an advantage, but it's there and the play is cooler.

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Which squeeze?

I'm interested in problems where there are multiple squeezes you might play for and selecting the best one. There are some of these in recent posts. Here's one that I just constructed:

♠	A K Q 4 3
♥	K 10 4
♦	K 7 6
♣	6 4
♠	6 5
♥	A 9 3
♦	A 9 3
♣	A K Q J 10

You're in 7N facing the ♠J lead. How do you play?

This might not be right, but let's say that you start with 3 rounds of spades and RHO shows out on the 3rd round. Take over from there (don't forget to pitch something on this trick!).

Some thoughts below, my full answer tomorrow.



If you've been reading the blog at all recently, you've probably noticed some compound squeeze talk. One line is to pitch from one red suit, run 4 clubs pitching a spade and the other red suit from dummy, coming down to something like this with 6 tricks to go:

♠	4
♥	K 10
♦	K 7 6
♣
♠
♥	A 9 3
♦	A 9
♣	10

LHO is known to hold a spade guard, so can't guard both red suits. Play off the king and ace in the red suit they've unguareded, then the ♣10 effects a double squeeze.

So, you can always make it on this line if you read LHOs shape properly. Say he starts with 2 clubs, though. He'll always be able to pitch 1 from each red suit and you'll have to guess which suit he started with a majority of -- a straight 50-50 guess.

I think there's a different line with much slightly better odds.

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Another NY Times compound

This deal is from a recent NY Times column:

(When I checked, the full deal appeared wrong, but the end position was right. Some spots in the full deal may be misplaced in my diagrams):

		♠	A 6 4 2
		♥	Q 10 5 4
		♦	K Q 10 8 4
		♣
		♠	K 9
		♥	A J 8
		♦	A 6 2
		♣	A K Q J 4

Trying for a BAM win, you land in 7N. At the table, declarer ran some diamonds and some clubs, discovered RHO had a club guard, unblocked the HQ and came down to:

		♠	A 6
		♥	10
		♦	8
		♣
		♠	9
		♥	A J
		♦
		♣	4

Declarer played a spade to the Ace and when RHO showed out he was able to play it as a double squeeze: cash the last diamond forcing East down to 1 heart, pitch the last club, and now West is also down to 1 heart.

This is pretty enough, but is not really any better than the heart finesse or stiff K with West. If West had had the HK, East would not unguard spades and expose West to the major suit squeeze.

Could declarer have done better? ...


In practice, I don't think so. But in theory, if he knew early enough that East had the club guard (for example if West shows out on the 3rd round), then I think a different line of play would be better. Details below the fold.




I think declarer wants to reach this position with the lead in the North hand:

		♠	A 6 4
		♥	10 5
		♦	8
		♣
		♠	K 9
		♥	A J
		♦
		♣	J 4

By not cashing the 4th club winner (impractical single dummy if both follow to the first 3) there's room for an extra heart with North. This will also require a spade entry to South in some variations. Presuming RHO still has a club guard, that only leaves room for 4 major suit cards, thus he can no longer guard both majors. If you judge that hearts have been unguarded, cash the last diamond pitching a heart, HA, CJ for a double squeeze around spades. If spades are unguarded, SK, CJ (pitching a heart), SA, diamond for a double squeeze around hearts.

This is an alternate threat compound. Cashing the CJ too early wrecks it by squeezing North.

What about the odds? Say RHO shows up with 6 clubs and 2 diamonds. He's going to pitch 1 club and 1 major before you make your decision and you should play for the major he pitches to be unguarded. The layout where you rate to go down is when he's 4=1 or 5=0 in the majors and pitches a spade, which happens about 19% of the time by my math (less a couple percent for stiff HK). Or, if he's 2=3 in the majors and pitches a heart, another 32% or so. So, in fact, this is a virtual toss up compared to the finesse.

On the other hand, there's some scope for misdefending on the compound line. I think, though, that restricted compounds rate to be hard to read in general and not much better than finesses. Of course, if you don't have those nice middle heart honors, the compound is still about 50% and there's no other line.

How could this be unrestricted, allowing you to cash an extra diamond before committing to deciding which major East has unguarded? A good start would be to move the SK to North (and a small spot to South). Now you don't need that 2nd heart in dummy, can cash all the clubs and then finish diamonds coming down to:

		♠	A K 4
		♥	10
		♦
		♣
		♠	9
		♥	A J
		♦
		♣	4

Now you'll have seen 2 major suit pitches from East. If he pitches 2 from the same major, you only lose to 5=0 or 1=4, both very unlikely. If he pitches 1 from each, you have to guess whether he was 4=1 to begin with or 2=3. You'll guess the latter, and wind up making something like 80% of the time.

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Squeeze Articles

I've occasionally thought I might write a squeeze book someday. I've got drafts of some chapters and I'm starting to put them up here. There's also a link on the right tool bar.

Please use this as a spot for leaving comments on these articles.

UPDATE: The Compound Squeeze article and the Compound Guard/Clash squeeze article are both in good shape. The latter also delves into Saturated squeezes (very briefly) and what for now I've dubbed "Cover" squeezes (similar to guard squeezes).

UPDATE July 2010:  I've filled in some more Compound squeezes where a ruff is used for an entry.  I've also added "Ruffing Communicator Squeezes" discussing positions where the ruffing menace also provides compensation for flaws in other menaces (sort of similar to a clash squeeze).  

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Maastricht Challenge

I'm a big fan of Tim Bourke's _Maastricht Challenge_. The introduction states "Many of the quizzes reflect my interest in developing better technique in trump, criss-cross and multi-loser squeezes." If this sounds appealing, you should definitely check it out. Many of the problems also rely on estimating the chances of various distributions, so I've been re-reading it to practice some techniques from the latest Rubens book.

This is problem 9:

♠	5 2
♥	A K 10 7
♦	K 8 6
♣	A K 4 2
♠	K 6
♥	Q J 5 3
♦	Q 7 5
♣	Q J 9 3

West opens 1♠, North doubles, East passes, and you wind up in 4♥. Opening lead is the ♣8 with runs to the ten and your jack. Everyone follows to the first round of trumps. Now what? [You may want to stop here and think about it.]

As the solution points out, if trumps split you can duck a spade from both hands and later exit the ♠K to force West to lead away from the ♦A. (Presuming that West has both Aces for his opening bid.)

What if trumps are 1=4, though? [Again, consider stopping to consider.]

Now you run trumps and clubs coming down to:

♠	5 2
♥
♦	K 8 6
♣
♠	K 6
♥
♦	Q 7 5
♣

You have 2 options here: if West has 2 spades, duck a spade and again he'll have to lead away from ♦A. If he has 3, then you can lead a diamond to the K and duck one on the way back. Bourke says if you're unsure of West's distribution, you should play East to have the most even possible distribution in spades and diamonds, or exactly one more spade. This is good general advice, but ignores an advantage to the 2nd line: if West started with 2 or more of the jack, ten or 9 of diamonds then a diamond to the K and covering East's jack or ten on the way back will work even if you misguess his shape.

If West started with 10 spades & diamonds, this advice is good enough. If he pitches 2 diamonds, then playing diamonds is guaranteed to work (assuming he didn't open 1S on 4=1=6=2). If he only pitches 1 diamond, then the key cases (for RHO's initial holdings in spades=diamonds) are 2=4 and 3=3. By my math (similar to below but not shown) this is 1:4. Playing diamonds will work in the first case or half the 2nd case (again see below), for 60%, so play spades and make it 80% of the time.

But, say West started with 3 clubs and so 9 spades and diamonds. Again, if he pitches 2 diamonds you can't go wrong. But if he pitches 1, the relevant cases (again for East, though you could also think of these as diamond layouts) are 3=4 and 4=3. This time, the odds (ignoring the auction) are 1:2 (I'm placing West with 2 Aces and the SQ but allowing all the other spades & diamonds to be distributed randomly). Bourke's advice is to go with 4=3 and so play on spades. But, playing on diamonds will work in the 3=4 case and half the 4=3 cases and so it's essentially a toss-up. (I get "half" since each defender has 3 non-Ace-diamonds and West needs a minority of 3 middle diamonds to defeat you with perfect defense).

So, this isn't really an error. However, I do think that going into more depth suggests that the alternative line (when West starts with 3 clubs) is slightly better. When you consider other possible inferences, everything points to the alternative: Perhaps East would have acted over 1♠-X with 4=4=3=2? Perhaps with AJxx or ATxx in diamonds, West will fail to unblock the low honor?

With no Jacks, West might not have opened (e.g. if he follows the "rule of 20"), or he might be more likely to open a 6133 hand than a 5143 hand, and/or a hand with good diamond spots (the 10-9).

What if we assume a religious rule of 20 West and place him with 1 or both key Jacks? Since I've come this far I'll try to work it out:

1. (40% of the time) West has SJ, East has DJ. Diamonds started 4=3 vs 3=4 is 2:1 (2=4 spade spots half as likely as 3=3 while diamonds-below-J are 3-2 in either case). Ducking a spade works 2/3, playing on diamonds works 1.6/3 (1/3 + original AT9x) + as much as another .6 vig if West might fail to unblock from ATxx.

2. (60% of the time) West has DJ, SJ in either hand. 4=3 vs 3=4 is again 2:1 (7 spades 3=4 or 4=3 equally likely while 5 diamonds twice as likely to be 2=3 as 1=4). Here, the recommended line is a 2/3 shot, while the alternative is 2.4/3. The possible vig of an unblock failure gets it to 100%.

Ignoring the vig, this makes the recommended line 2/3 and the alternative 2.08/3, for a small advantage. More intuitively, the lines started as a tossup, but West should be a bit more likely to have the DJ than open spaces suggests and that helps the alternative line by making it more likely that West can't (or won't) be able to get off lead even when he shows up with 4 diamonds.

Finally, if West has both Jacks but no SQ another diamond spot could be important to whether he opens. This again points to the alternative line.


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More squeezing

		♠	A 6 5
		♥	7 6 4
		♦	A Q J 9 5 2
		♣	K
♠	Q J 8 7			♠	10 4 3
♥	K J 10 9			♥	8 3 2
♦	8 7 3			♦	10 4
♣	7 6			♣	Q J 10 3 2
		♠	K 9 2
		♥	A Q 5
		♦	K 6
		♣	A 9 8 5 4

Chris showed me this deal. South opens a 14-16 1NT and winds up declaring 6D on a trump lead. What is supposed to happen?

More below, more to follow in a future post.


At the table, declarer won in dummy, unblocked clubs, crossed in trumps to ruff a club and drew trumps. East had a problem and discarded a heart. At which point, declarer lost a heart finesse but eventually got home on a double squeeze. Could the defense have done better? Read more!

Compound squeeze defense

Jonathan already beat me to the correction to todays NYT column.

		♠	K Q 4 3 2
		♥	J
		♦	A Q J 9 3
		♣	A 8
♠	J 10 9 8			♠	5
♥	Q 7 4			♥	10 8 6 5
♦	10 6			♦	7 5 2
♣	Q 5 3 2			♣	J 6 4
		♠	A 7 6
		♥	A K 9
		♦	K 8 4
		♣	K 10 9 7

After a long relay auction, South placed the contract in 7♦ and got the ♦10 lead. According to the report, declarer drew trumps (West pitching a club), tested spades, then played 2 more trumps effecting a compound squeeze: West shook hearts, then ♥A, ♥K squeezed him in the blacks, and ♣A, ♠Q squeezed East in the rounds.

Of course, once a club was pitched and the spade break revealed, declarer could ruff a club, either establishing a trick or uncovering who held the sole guard and arranging either a black suit squeeze vs West or a double squeeze organized around hearts.

But, I'm more interested in figuring out whether as a pure compound squeeze how the defense should have gone. Let's imagine that the extremely revealing relay auction happened to make North declarer so that the defense knew what they were facing. Let's further imagine that the hand is in NT so we can ignore the club threat possibility and the order of discards. Picking up after testing spades and diamonds, declarer should run diamonds and come down to this ending:

		♠	Q 4 3
		♥	J
		♦
		♣	A 8
		♠
		♥	A K 9
		♦
		♣	K 10 9

West is known to have a spade guard and to have started with 7 round cards to East's 9. West must be down to only 4 round cards and so can't guard both clubs and hearts. Declarer's strategy is to cash dummy's winner(s) in the unguarded suit (perhaps pitching a spade on the 2nd heart if that suit is still guarded), then CA, SK will squeeze East. In fact, West must unguard a round suit on the 4th diamond, but the position allows cashing a 5th before attempting to read which suit he unguarded (this is not always the case in compound squeezes).

The defense's strategy is to present declarer a guess as to which suit West unguarded. To solve this fully, you need to consider all club-heart breaks, what West should do in each (possibly mixed), and the resulting guess (if any) this presents to declarer. For now I'm not going to tackle that, but I think we can get pretty close to the right answer:

West in fact pitched 2 hearts and 1 club. Unless this was hopelessly unguarding both suits (e.g. from 4=3), this is most likely from an original 3=4 in hearts=clubs (as it was in practice) or 5=2.

I think in practice declarer should assume that it is: from extreme holdings like 1=6 or 6=1, I think pitching 3 from the long suit is likely to succeed, so we can probably rule those out. From 2=5, pitching 1 heart and 2 clubs is probably best: likely to look like an original (and more likely 4=3) and lead declarer astray.

So, for starters, declarer should work out that the odds of West holding 3=4 vs 5=2 is 10:9 (if you don't know how to verify that, check out Jeff Rubens's latest book -- it's 9c3 * 7c4 : 9c5 * 7c2). You can ponder the psychology of the order of discards, but a proper expert should identify the 3 discards needed and make them in a random order. I suppose in practice most experts would try to unguard early (so pitch 2 hearts first) rather than late. Perhaps West was playing a deep game by discarding in the "easy" order (of course, in practice West had to consider many possible hands for declarer).

So, if West was playing a perfect game-theoretic game, his best bet was to discard (as he did) 2 hearts and 1 club and hope that declarer miscalculated. East could have helped this illusion by pitching clubs (as if he started with 5) instead of hearts, though theoretically his discards should be ignored (so long as he guards the suit that West unguards).

This also means that West can use some 3=4 "losers" to "protect" his 6=1 winners -- occasionally pitch 3 hearts (or clubs) from that holding so that declarer can't conclude that 3 pitches is always from an original 6. This will reduce the 10:9 odds, but not enough to change declarer's best strategy. (Of course, the HJ makes that a bad idea with this particular holding.)

So, basically I think the defense is slated to lose when West starts with 3=4 or 4=3, but succeed otherwise. By my math, that's about 64%.

In practice, I think if East had pitched a club and a heart or 2 clubs, declarer had a realistic chance to go wrong: 5=2 (West hearts=clubs) is almost as likely as the actual 3=4, but in practice there must be a 10% higher chance that East pitched clubs from an original 5 instead of an original 3, which could be enough to tip declarer to the wrong decision.

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Pyrrhic victory

This was not a success, but still kind of a cute
hand
.

The key diamond suit was remarkably similar to this deal.



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The Monthly Trout?

Sorry for the long absence.

I've been reading Jeff Rubens new book, _Expert Bridge Simplified_. I quite liked it. A lot of it I was already familiar with (but anyone that isn't would likely benefit a lot from that material), but the stuff that was new to me (e.g. estimating the chances that 2 suits will break such that the same opponent is short in both) was really cool. Even for the stuff I was familiar with, it really cemented some important stuff in my mind. How often this material will actually help me, I'm not sure, but I'll try to pay attention.

The subtitle is "arithmetic shortcuts for declarer" and it focuses a lot on choosing approximately the best line. Many problems go to a lot of trouble for an ultimate answer of "who knows" or "you could work a lot longer on this to get a precise answer, but I [JR] am not interested". Also, often it's about convincing yourself you have the best line without necessarily knowing how good it is overall or how much of an improvement it is over the alternative.

So, there were a couple of deals where I thought there was a bit more of interest than JR brought up, but it was within the spirit of the book not to bring it up. Here's one example:

♠	A K Q 10 3
♥	Q
♦	10 9 8
♣	10 9 8 7
♠	5
♥	A J 3 2
♦	A K Q J
♣	A K Q J

After starting 2♣ you wind up in 7N and get a diamond lead. How do you play?

The solution from the book, and my own are both below.



The books solution was to point out that you should play for spades to come in and if they don't you have to choose between a finesse and a squeeze. The squeeze requires the 2 card psuedo-suit of outstanding-major-suit-honors to split 2-0, a bit less than 50%, while the finesse is a straight 50-50 shot and so better. He explicitly says he's not bothering to consider possible info learned from the minor suit distributions because it's too complicated.

In a variant on this problem, that was true, but I think here it's not that hard. You can learn about both minors and most of the spade info before making your final decision: run diamonds (shake a spade) and 3 clubs, then 2 spades (shake a heart) and play a 3rd spade. After seeing East's play, you have to decide whether to unblock clubs and play for the squeeze, or pitch another heart and play for the finesse.

Case 1: RHO shows out, LHO has 5 spades. If the 11 outstanding minor suit cards break 4=7 it's a pure guess based on open spaces. If they break worse, play for the squeeze, better play for the hook.

Case 2: RHO follows. Assume LHO is about to also follow small -- if he shows out then the finesse and squeeze have the same odds (both require RHO to hold HK). Now we know 17 cards (all the minors and the spade spots). If LHO has 6 open spaces (minors break 4=7), then the squeeze is 6/9 * 5/8 + 3/9 * 2/8 = 1/2, while the finesse is only 1/3: play for the squeeze. If LHO has 5 open spaces, then 5/9 * 4/8 + 4/9 * 3/8 = 4/9, the same as the finesse.

This seems to be worth only 1 or 2%, but is not that hard to work out at the table.

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