Which squeeze? 2

Another installment in "which squeeze should you play for?"

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

Contract: 7♦
Lead: ♦10

You draw trumps (East pitches a club), cash 2 hearts (West pitches a club). Now what?

Answer below the fold.



You could play for a compound squeeze: cash another diamond (pitching a spade) to get to a 7 card ending. East must keep 2 hearts and can't keep 3 of both black suits. Whichever suit he unguards, cash those winners ending in hand (crossing back in spades if needed) then play the last trump. This will squeeze East out of the other black suit, then the final heart honor will squeeze West in the blacks.

The problem with this line is that you'll not see many pitches before guessing which suit East unguarded. Basically, he'll pitch one from each black suit, and you'll have a near 50-50 guess as to which one he started with 4+ of and still guards.

The alternative line (suggested by Jonathan Weinstein) is to ruff a spade. If someone shows out, you'll have either a marked major suit squeeze (vs East) or a marked double squeeze pivoting around clubs. If everyone follows to 3 rounds, you'll still have a guess for which of those 2 squeezes to play for. But, you'll be able to use restricted choice to make that guess -- whichever player shows up with a spade honor will be more likely to be out of spades than to have started with QJxx (about 3:2 for West, and about 3:1 for East). The odds of success are definitely higher.

And, occasionally the ♠10 will just set up.

Perhaps testing hearts early is a mistake. Too complicated for me today, though.

[This hand was taken from a recent post & slightly disguised to make a cleaner entry in this series]

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Another Maastricht Challenge

I found another problem from _Maastricht Challenge_ where I disagree with the book's solution:

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

East overcalls 2♣ on your way to 6♦. The problem specifies that West leads a 3rd highest ♣2, so you can take as given that clubs are 3=6. It also points out that finding the trump queen or ruffing 2 clubs is sufficient, but warns of a trap.

I've put the book's solution and my own below the fold.



Say you play a trump to the ten. If it loses, it's relatively straightforward to ruff another club and draw trumps.

If it holds, then what? The book points out that against a tricky defender, you might play a spade to the Ace and see East show out. It asserts that Qxxx/x is a 38% chance and xx/Qxx is only 29%, so against someone who is up to ducking the queen you should finesse the jack. A pretty cute play, finessing one way and then the other.

I agree that those are the relevant holdings. I get that their odds are 56:45 (12c6 * 4 : 12c4 * 6), which is very close to their numbers (not quite within rounding but I'm going to ignore that). But, it seems that after the ♠A and East shows out, you still have play: ruff a 2nd club and try shaking your other 2 on diamonds. The chance of West having 2 or fewer diamonds (which is required for this backup plan to fail) is under 30% (of the 38%), so the winning case for ♠J is more like 11%, a clear loser.

More generally: this sort of defensive play, ducking when declarer takes a loosing finesse, comes up from time to time. There was a nice hand in an early daily bulletin from San Diego with a similar theme. What should I look for so as to better recognize such situations on defense in the future?

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Remembering suit division odds

I always struggle to remember suit division odds. Rubens had one suggestion I liked: remember the probable ones and back out the improbable ones. If you can remember 3-3 and 4-2 are 36 & 48, then you can probably figure out that of the 16 left most of it is 5-1 and a small amount is 6-0. Similarly, 3-2 and 4-1 are 68 & 28, that leaves 4% for 5-0.

I also found that re-ordering the tables from most probable to least probable, as below, reveals some obvious patterns, many of which I hadn't noticed before (perhaps this shows how unobservant I am). I omitted 2 card suit divisions for 2 reasons: they break the pattern, and you should be able to calculate them on the fly.

1. x-y divisions are grouped by the difference between x&y:

1 off (e.g. 3-2, 4-3) is around 2/3.
2 off is around 50-50.
Even is 1/3+.
Off by 3 around 30%, etc.

(this is increasing except for exactly even)

If you forget 4-3, it's going to be closer to 3-2 than any other number you remember. Also, these groups don't overlap (excluding 2 card suits or 11+ card suits, and 5-5 is barely less likely than 6-3)

2. For the close divisions (not more than 2 off), the more cards there are the more likely you are to be off by more than 2, so the groups go from short suits to long suits (e.g. 4-2 is more likely than 5-3). After that, longer suits are more likely than shorter suits (e.g. 5-2 is more likely than 4-1).

3. As suits get longer, the odds become more similar. For example, 6-3 is virtually the same odds as 5-2. In contrast, 2-1 is 10% more likely than 3-2. If you remember 2 or 3 odds for same difference combinations, you can generally extrapolate the others. Generally, the differences cut about in half each time. For example, 3-2 / 4-3 are 68 / 62, so 5-4 is probably around 62 - 1/2 (68-62) = 59% (in fact it is).

Some more stuff below the table...

2-1	78%
3-2	68%
4-3	62%
3-1	50%
4-2	48%
5-3	47%
2-2	41%
3-3	36%
4-4	33%
5-2	31%
4-1	28%
3-0	22%
6-2	17%
5-1	15%
4-0	10%
7-2	9%
6-1	7%
5-0	4%

4. Void probabilities are relatively easy to derive from one another just by using open spaces:

1-0 is 100%,
2-0 is 12/25 of that, 48%,
3-0 is 11/24 of that 22%,
4-0 is 10/23 of that, a bit under 10% (9.6%),
5-0 is 9/22 of that, 3.9%,
6-0 is 8/21 of that, about 1.5%.

5. The singletons are harder but doable in a pinch (adjust for open spaces then number of possible singletons):

2-1 is 78%
3-1 is 11/23 * 4/3 of that, about 50%
4-1 is 10/22 * 5/4 of that, about 28%
5-1 is 9/21 * 6/5 of that, about 14%
6-1 is 8/20 * 7/6 of that, about 7%

So...

If you can figure out short suits and maybe some voids on the fly and remember these:

3-2 68%
4-2 48%
3-3 35.5%
4-1 28%
4-0 10%

You can probably work out the rest. 4-3? Well, 2-1 is 78 and 3-2 68 so subtract about half that difference from 68 to get 63 (actually 62). 5-3? All the off by 2s are right around 48%, should be a bit less (in fact 47). 5-2? 3-0 is 22, 4-1 is 28, so add half the difference to 28 and get 31 (in fact 30.5). 4-4? 2-2 is 41, 3-3 is 35.5, subtract half the difference and get almost 33 (in fact 32.7).

6-2 is maybe a bit harder. We memorized 4-0 as 10%. 5-1 is about half of 4-1 (9/21 * 6/5 to be precise), call it 14% (we can check this also by noting that 3-3 and 4-2 add to 84, so it should be a touch less than 16%). 6-2 is probably around 16% (in fact, 17.1. 4-0 is a bit less than 10% and 5-1 is 14.5, so my extrapolation suffered from some rounding).

You could also get there by noting that 5-3 and 4-4 add to 80%, so 6-2 ought to be most of the rest. Delving a bit further, you could consider there are 28 6-2 layouts vs 8 7-1. The latter are about half as likely, also (the last card has 12 open spaces vs 7), so call it 7:1, so the 7-1s are about 1/8 of 20% and the 6-2s are the rest (8-0 never happens, right?): 17.5%

The harder of these are not that practical, but trying to estimate these numbers and occasionally checking your results will result in better memorization. Read more!

More squeezing, follow-up

Following up on this post from last month.

		♠	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?

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?

Let's wait on East's pitch, and assume that declarer pitches a spade on the 3rd trump, then crosses in spades to cash and ruff a club (finding out what's happening in that suit) and plays his last trump in this position:

		♠	A 6
		♥	7 4
		♦	2
		♣
		♠	9
		♥	A Q 5
		♦
		♣	9

Note that the H6 has been unblocked.

East must keep a club, so Declarer will pitch his last on the D2. In the 4 card ending, the opponents have 7 major suit cards. Declarer would like to take a marked endplay if possible, or a heart finesse followed by an attempted squeeze (which may have already developed). Say there are only 3 hearts outstanding -- declarer can take the finesse and either win it or establish a long heart (ignoring, for now, the possibility that West kept 3 hearts and East none). If there are 5 hearts (and thus only 2 spades) then declarer might as well play SA, maybe establishing a trick, maybe establishing an endplay, and otherwise playing for the HK (now blank) to be onside. [Note that the defense would prevail if East kept all his spades and declarer played this way -- this requires an immediate heart pitch, though, which did happen but declarer played along different lines.]

If the defense keeps 4 hearts, it's kind of interesting. If they're 2-2, then in practice declarer can't go wrong -- SA and a spade will work, or a losing heart finesse followed by spades splitting will work. So what if East keeps 3 hearts and West stiffs his K? Then the HJT9 will all have been pitched and declarer can play the 6 to the Q. When this loses, a spade will come back, East will show out, and hooking the 5 will be marked. If West keeps 3 hearts, again East had to pitch hearts immediately (to keep a spade guard). Now declarer (if he plays in this counterfactual way) has to guess, but could go right: cash SA, then play a low heart and duck it.

In practice, the key is how likely declarer thinks East is to pitch a heart on the 3rd trump from an original 2=4=2=5.


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New squeeze article

The start of an article on compound squeezes with 2 "helper" menaces (such as a clash menace and a guard menace, both of which are necessary) now here. As always, an index of all articles in progress is at the top of the right column.

It still needs work, but a number of positions are identified.

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More squeeze defense

Jonathan made a good point in comments about defending this end position:

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

West is known to have the spade guard and, if he started with 3 clubs, must have pitched 1 card from an original 6 red cards (the defense started with 7 of each red suit). His goal is to present declarer with a straight 50-50 guess as to whether the suit he pitched from is still guarded or not. When he's 3-3 in the reds, he has no control over that, whichever suit he pitches is unguarded. This happens 40.8% of the time (7c3^2 / 14c6, see below for a faster approximation). So, when he has a long suit, he'll want to pitch from the long suit (still having a guard, 50/59.2 of the time (84.5%)) and from the short suit the rest of the time.

Given that the red lengths are symmetrical, it's easy to work out that with 3-3 in the reds you should discard each suit 50% of the time. With unbalanced red suits, you'd ideally like to do the above arithmetic to figure out the optimum discard frequencies. The problem is, this might take some time, potentially revealing that you aren't easily able to identify the optimum.

Is it therefore ethical with 3-3 to do the same calculation you would want handy if you had been 4-2?

Maybe it's not that hard. Throw out voids and having 1-5 cards in a suit is 1:3:5:5:3, so 3-3 is 25:(15+15+3+3), so with 61 holdings you want to pitch from a long suit 30.5 times and you have a long suit 36 times, or a touch more than 5/6. As evidence that this approximation is effective: 50/59.2 is 84.5%, 30.5/36 is 84.7%.

In case that last paragraph was too fast: having 1 diamond is 7c1 = 7 combos. 2, 3, 4, & 5 diamonds are 7c2, 7c3 etc, which are 21, 35, 35, and 21. There's a factor of 7 in all those numbers, so the odds simplify to 1:3:5:5:3. The ways to hold 1-5 hearts have the same combinations, but 1 diamond implies 5 hearts, 2 diamonds implies 4 hearts, etc, so flip those odds to get 3:5:5:3:1 and pairwise multiply to get the total number of ways to have both 1 diamond and 5 hearts (1x3), 2 diamonds and 4 hearts (3x5), 3d3h 25, 4d2h 15, 5d1h 3. Add them up to get 61 equally likely possibilities, of which 25 are 3-3.

In this weighting, the voids are 1/7 each and the 6 card suits are 1. (30.5 + 1/7) / (36 2/7) would be the precise calculation.


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Alternate threat compound with variations

This is from Problem 6 from the August (2002, I think) Pav problem set. I think it was played in the European Championships sometime before that:

Axx
AKx
AKx
Axxx
 
QJx
QTxxxx
xxx
x
 

LHO opened 1C and you land in 6H. CK led. How do you play?

Duck (creating that Ax/void basic menace that's often useful), ruff the 2nd, play SJ forcing LHO to cover (or you'll have a black suit squeeze), then run trumps coming down to:

 
x
-
AKx
Ax
 
Qx
x
xxx
-
 

Clubs is the basic menace (vs West), spades is the alone ambiguous menace, and diamonds is the ambiguous menace with the basic menace, except that South has an alternate threat. If West has unguarded spades, the extra entry in diamonds is needed to get the CA out of the way before coming back to hand and cashing the last heart for a simultaneous type R.

If West unguards diamonds, the diamond threat in dummy is not useful because of entry problems, but this can be overcome by using the alternate diamond threat in hand. Note that the recessed club menace is key to make room for that alternate threat. In fact, that's a general feature of alternate threat compounds: the basic menace needs a winner that is not useful as an entry such as shown, or AKx/x, or Axx/K, etc.

This deal has an alternate solution – cash all the red winners coming down to:

    Axx
    -
    x
    Ax
Kxx      T98
-        -
Q        Ji
QJ       x
    QJx
    xx
    x
    -
 

This was an alternate threat squeeze where clubs was the basic threat, diamonds alternate, and spades the alone ambiguous threat. Get rid of the diamond entries, though, and it still works because West can’t keep pitching spades (the alternate threat is still important, though – in fact, the original threat is idle). Play 2 hearts pitching a diamond and then (unless it’s good) a club. West can’t bare the SK or the spades will run, so he must pitch a diamond and a spade. Since he can’t duck the SQ, it’s basically a B1 with the R (diamond) winners already out of the way. This spade menace is fairly strange, neither guard nor clash. I think it's a "shortstop" menace.

Switch the SQ and SK and it’s just a simple black suit squeeze. Do that but also switch the SJ and ST and it’s a compound guard squeeze (Type-R Guard-O in my classification -- R since the basic threat is alone (the diamond and 3rd spade with North are both idle), O since the guard menace is opposite the basic threat).


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