[Fluffy]

---

♦10 lead (natural)

Advanced Ops.

[FrancesHinden]

If clubs run I am virtually certain I can make this double-dummy. Does that help?

[Free]

I guess we'll need ♣ to come in. So I'll take the ♦Ace and lead ♠ to my T (Q might be better, but RHO had the opportunity to take the K if he had it). If it holds we have 12 tricks, otherwise I'll test ♦ for 3-3 or go for a double or triple squeeze. Don't play ♥ to the J at trick 2 because they might play another ♥ and ruin my communications.

[nige1]

Fluffy says "♦10 lead (natural). Advanced Ops"

---

Hoping that ♣s are good for 5 tricks, you might win the ♦ lead in dummy and lead ♥3.
If ♥J loses to West's ♥Q, you can cash winners reducing to this 3 card-ending.
On ♣4, you discard ♥6 unless it's good. Unless ♠Q is good, you rely on dummy's ♦s

---

[BillPatch]

I lke both free's and nige1's solutions, but find it difficult to make a decision between them. I think I was headed towards Nige1's squeeze position, and I am relieved that free's warning about entries in that line was wrong. Any other thoughts?

[Free]

nige1, on 2015-September-05, 19:56, said:

Fluffy says "♦10 lead (natural). Advanced Ops"

---

Hoping that ♣s are good for 5 tricks, you might win the ♦ lead in dummy and lead ♥3.
If ♥J loses to West's ♥Q, you can cash winners reducing to this 3 card-ending.
On ♣4, you discard ♥6 unless it's good. Unless ♠Q is good, you rely on dummy's ♦s

---

BillPatch, on 2015-September-05, 20:35, said:

I lke both free's and nige1's solutions, but find it difficult to make a decision between them. I think I was headed towards Nige1's squeeze position, and I am relieved that free's warning about entries in that line was wrong. Any other thoughts?

That line fails dramatically if West has ♠K, East has 4 ♥s and ♦s were 3-3 all along. That's why you need to test ♦s first, and because of that you'll still need ♥ as an entry.

[Fluffy]

Since oyu need an extra trick when majors break you can either play a heart to hand, or a spade to hand, heart is certain to isolate the heart menace while an avanced op might reveal the spade position when he is not prepared to play low with ♠K. Both have merits and I couldn't decide.

[BillPatch]

Free, on 2015-September-05, 23:48, said:

That line fails dramatically if West has ♠K, East has 4 ♥s and ♦s were 3-3 all along. That's why you need to test ♦s first, and because of that you'll still need ♥ as an entry.

I don't see how it fails. Cashing winners to reach the end position seems automatic. Then on the squeeze trick, either the last heart falls, the Q♠ falls, or we test the diamonds.

[BillPatch]

Fluffy, on 2015-September-06, 01:24, said:

Since oyu need an extra trick when majors break you can either play a heart to hand, or a spade to hand, heart is certain to isolate the heart menace while an avanced op might reveal the spade position when he is not prepared to play low with ♠K. Both have merits and I couldn't decide.

I am glad to have company in my uncertainty.

[Lovera]

There is already a triple on last club. The ♦ 10 can make thinking chat J falls but 9 is in W.

[gnasher]

People don't lead from 10x on this auction, so if anyone has a diamond guard it will be LHO. Hence there's less reason to isolate the heart menace.

I'd play a spade to the ten at trick two, then guess who to play for ♠K. If it's LHO, I have to cash the top hearts before the last club; if it's RHO I have to cash the diamonds.

If LHO plays back a spade at trick 3, I can delay this decision until after I've cashed four clubs. If he plays a diamond, I have to decide after two clubs.

[rmnka447]

North is the 2 threat hand. Only one of the opponents can guard either the ♦s or ♠s. Both opponents can guard the ♥s.

If a double squeeze exists, it must a (Love) type R squeeze. That means East must have the ♠ guard when the squeeze card is played. If West holds both possible ♠ guards, then the double squeeze fails for lack of an upper hand threat versus East.

So I'd opt for playing a ♠ 7 from dummy at trick 2. Unless East yields a tell, which spade you play from hand is a guess. But if West wins the trick with the next highest honor, you've moved the ♠ guard to East if the ♠ honors are split. Of course, if you guessed right on which honor to play, you won't need a squeeze as you'll have established your 12th trick.

Say West wins the trick with the next higher honor (J if you play 10, K if you play Q). You win the return and play a ♦ to hand (if it wasn't the suit returned). Assuming East played low to the first ♦ trick, the second ♦ trick should reveal something about the suit. If West stops ♦, the second trick should crash East's J. If not, you can play for a ♦/♠ squeeze against East which also keeps 3-3 ♦s in play. If the ♦ J crashes, the conditions for the double squeeze are satisfied, but you'll have to decide between the double squeeze or a ♦/♠ squeeze against West. With West presumably holding 4 ♦s to East's 2, I think the odds slightly favor the double squeeze (L ♦ 7, R ♠, B/C ♥ 6).

[nige1]

The decision is close but the technical approach might be better if East is a good player because

• There's no reliance on "tells".
• The heart menace is isolated (After 3 rounds, only one defender can control hearts).
• The defender with long hearts is more likely to hold the queen. Hence, if the knave loses to the queen, then the likelihood that West has sole heart control increases.

-- I now see that Fluffy and Gnasher have touched on this before..

This post has been edited by nige1: 2015-September-07, 10:57

[BillPatch]

A lead from 10x in diamonds is likely on this auction if the opening leader has quacks in the other suits. Opening leader should do it slightly more frequently than that result, to avoid making the lead from 10x always give honest info about the other suits.

Edit: Responding to Gnsher's "People don't lead from 10x...." I'm proud to be an unperson.

[Lovera]

I would remember that when only an opp can stop a suit but we cannot say whom is (as is the case of heart) in this eventuality the indetermination of unilateral is treath like a double(=indeterminate position) menace being this squeeze called "simple as double" (Love term) or "alternative" (Romanet term) and is a squeeze with one loser. About ♠/♦ squeeze : if one loser (simple) acts at 10 trick but there is the delayed duck criss-cross that is a two loser squeeze acting a trick before.(Lovera)

[BillPatch]

Further remarks about my unpersonality. A lead from JTx is not safe, so unlikely to be right on this auction. If forced to do it I haven't analyzed the falsecarding position fully, but surely I would play the false Ten at least as often as the Jack.

[phil_20686]

Throughout i will assume that only lho can guard the diamonds
If clubs are 4-1 I can still make if the person who has the fourth club has the spade K and no more than two hearts and the diamonds are 3-3. If after two top clubs the clubs are 41 i could, say, cash 6 red suit tricks discarding a spade from hand and play club club hopefully endplaying someone into giving me a spade trick and an entry to the 5th club.

I can test the clubs if I am prepared to play a spade to the QT instead, suppose that I win the diamond Q and cash two clubs, if they are 3-2 then I can cross to the diamond ace and play a spade to the Q, if that loses whatever they return I can still test diamonds unblock the spade ace and then run the clubs using hearts as the double menace. This works whenever RHO has one of the K or J of spades given that I am assuming that diamonds are guarded on my left. I think this is far stronger than the heart to the J variant since I cannot see how to fit in any chance of a 4-1 club break. Its a bit dependent on the inference that only lho can guard the diamonds.

[nige1]

Dealerscript

#
# An attempt to demonstrate that 
# If West has the queen of hearts then
# He is more likely to have heart control even if
# You also assume that he has 4 diamonds.
#
predeal north SA7,HAK63,DAQ74,CT63
predeal south SQT4,HJ4,DK63,CAKQ74
predeal west HQ
predeal diamonds(west) == 4
action average "West average number of hearts" hearts(west)

Results

West average number of hearts: 3.88012
Generated 10000000 hands
Produced 10000000 hands
Initial random seed 1441761535
Time needed    5.144 sec

[phil_20686]

nige1, on 2015-September-08, 19:22, said:

Dealerscript

#
# An attempt to demonstrate that 
# If West has the queen of hearts then
# He is more likely to have heart control even if
# You also assume that he has 4 diamonds.
#
predeal north SA7,HAK63,DAQ74,CT63
predeal south SQT4,HJ4,DK63,CAKQ74
predeal west HQ
predeal diamonds(west) == 4
action average "West average number of hearts" hearts(west)

Results

West average number of hearts: 3.88012
Generated 10000000 hands
Produced 10000000 hands
Initial random seed 1441761535
Time needed    5.144 sec

Since the hand with more hearts is more likely to have the Q, it must be the case that the hand which has the Q is more likely to have long hearts.

[BillPatch]

phil_20686, on 2015-September-08, 08:19, said:

Throughout i will assume that only lho can guard the diamonds.... Its a bit dependent on the inference that only lho can guard the diamonds.

I don't understand why we should make the assumption that only lho can guard the diamonds. LHO led the ♦After the first trick we can eliminate 6-0 breaks in either direction. With a 3-3 break neither defender defends diamonds, so we also eliminate that. A lead of a singleton against a NT slam may be rare enough to be disregarded. 4-2 splits are another matter. While one generally prefers to lead from length from a weak holding against a NT slam, one usually wants to conceal non-establishable honors. Since the ten from ten-x would drop on the next round, and it gives count info, it would always be led from this holding if this suit is led. From Txxx a spot card would always be let. From T9xx and JTxx a low spot also is mandatory. From T98x against lower contracts it would be apropos to lead high and normally an honest T, to give more info to partner, and to prepare for a possible unblock to prevent a throw-in. Since a high card lead will more likely come from three it is less likely usable info for him than declarer, who can usually count his potential tricks easily and know which tricks matter. At a NT slam it is less likely that a throw-in on a ten high suit matters. From the JT9x and JT8x a low lead may give up a trick to a low card, and the ten false card is frequent. Even among the most random false carders, restricted choice limits T98x and JT9x leaders to 1/3, and JT8x to 1/2 of the high honor leads.

A priori there are 4 possible holdings of the T and one lower card, from which we would expect a ten lead 100%. 4*46.88/30 = 6.25%
There are 2 possible holdings of from T98x, 1 possible JT98, 2 JT9X, and 2 JT8X. 7 possible but restricted choice lowers the relative chance of the 10 leads to 2.58 cases. 2.25*46.88/30 = 4.03%

There are 4 possible 5-1 holdings 2 JT98x 1 JT9xx 1 JT8xx but restricted choice reduces this to 1.33 5-1 cases. 1.33*18.75/12=
2.08%
A priori given a Ten lead and non-J from 3rd hand 6.25/6.11 1.023/1 in favor of RHO being the stopper in diamonds

Since the sequence leads are more attractive relative to T-x leads we do not believe that the actual odds are close to the a priori odds, so we know that LHO probably holds the stopper by odds of 2 or 3 to 1. I doubt that adopting the squeeze technique optimized for RHO never holding the stopper will be optimal for the actual case.