Play 009 - repeat of old hand

[inquiry]

[hv=lin=md|1sakhat43da43cakjt,shdc,sqj432hk92dkj2c32,shdc|mb|6N|mb|p|mb|p|mb|p|pc|ST|pc|S2|pc|s7|pc|sA|pc|SK|pc|s5|pc|s3|pc|c5|]399|300|This is a repeat post of an excellent problem posted a decade ago by Mike Lucy. I updated it by giving spot cards for the "x's" and place it in handviewer mode. I will quote the original author below He did not provide the discard on trick two. You can click the next button to see the first two tricks if the description isn't clear below. [/hv]

I just watched a pretty neat hand played as I kibitzed. Declarer is a solid player as are the other three players at the table.

 

The result of deal in question was 6N down 1. In my opinion declarer did not give much thought to hand and played on "auto-pilot". After hand was over, the opps also said that hand requires double dummy play. The dummy player and myself think otherwise. Play the hand as the bottom hand is the declarers hand and top hand is dummy;

 

LHO leads the 10 of spade to your King. When you cash second high spade RHO shows out. How to proceed from here (obviously the club hook loses or this would not be a problem).

 

As you cash clubs you find that LHO started with 4 clubs. How to proceed?

• You may elect not to cash the 4th club from hand if you find it necassary to work another suit before finding the club distribution
• The hand can always be made when LHO has 4 clubs.

Additional "stuff":

 

• If someone has the time and energy perhaps they could analyze the hand when LHO has 3 clubs, and 2 clubs.
• Is there a better line of play from trick 2 forward? maybe find a better or optimal line of play from the start
  

 
The original quote gives you that west had four clubs... that was the layout on the original hand. The solution we are looking for has more to do with the "additional" stuff, where you have no idea how clubs divided. What is the best line starting with trick three.
 
After the problem was "solved", Mike wondered...

Furthermore, I wonder if a "world class" declarer would fine such a line of play. If they were able to find this line, would they be able to do this within a practical time limit

[gszes]

This post was getting entirely too long so I decided to stop

and since wooden plays yielded almost a 96% chance to make

I would probably go with that rather than waste too much time

working out if a squeeze was better==========================

 

Ill be interested in seeing the real answer ====

[shyams]

Trick 1: ♠A. If spades split kindly, I have 12 assured tricks and would be wasting too much mental effort for nothing. Therefore,

Trick 2: ♠K -- which reveals the 5-1 split.

 

In theory, a double squeeze can always be executed on this deal. The problem is deciding which suits are controlled by West.

The complex technique is beautifully explained in Richard Pavlicek's article on Pure Squeezes (link here). The rest of the play goes...

 

 

Trick 3: ♣J from hand. West wins and...

Trick 4: ...exits with a safe ♠. Win with dummy's ♠J and discard a ♥ from hand.

 

To continue from Pavlicek's article, I need to:

Trick 5: Cash ♣10

Trick 6: Cash ♣K -- discarding dummy's ♥2

 

Let us say (based on carding so far) I correctly assume that East is solely controlling the ♥ suit. If so,

Trick 7: ♥ to ♥K

Trick 8: ♥ to ♥A

Trick 9: ♣A squeezing West in ♦ and ♠, forcing him to reduce to 2♦+2♠. Dummy's long spade has done its job and can be discarded on this trick.

Trick 10: ♦4 to dummy's ♦K and then

Trick 11:♠Q squeezes East in ♥ and ♦.

 

PS: The entire solution is based on the above article. I would never manage to replicate this at the table!

PPS: Although this may be technically superior, imagine losing to East's ♣Q at trick 3 and then misreading the situation in the red suits!

[helene_t]

If spades are 5-1 I would just take the diamond finesse and then the club finesse.

[Fluffy]

If Ben published this, its because there is a 100% squeeze line ducking a club, Perhaps not a 100% single dummy, but starting with ♣10 probably succeeds double dummy against any lie out. The kind of play Montercarlo simulation would suggest?

[nige1]

 

 

[hv=pc=n&s=sakha654da54cakjt&w=sT9876hjdQT9cq987&n=sqj432hk32dk32c32&d=s&v=0&a=6nppp&p=sts2s5sa{Declarer finds the bad news in !S}sks9s3h7{Now declarer sets up a !C as his 11th winner}cac9c2c4ckc7c3c6{In a different layout, on the third round of !c LHO would have to abandon a red suit. if he had both of them e.g. if he had been dealt a 5332 or 5341 hand}cjcqh2c5{LHO belatedly switches to !D}dtdKd8d4h3hthahj{Declarer cashes the 4th !C to establish the the distribution of the suit}ctc8d3h8{Crossing to dummy with !HK squeezes LHO out of a !D}h4d9hkh9{Now declarer cashes dummy's !s to squeeze RHO}sqd6h5s6sj{Finally, RHO must give up a red trick}]400|400|Thank you, Mark Lucy and Inquiry.

Fluffy seems to be right.[/hv] On the simplified layout above, press Next to see a double-squeeze solution -- which assumes only that LHO has at most two hearts.

An opening ♦ lead threatens to break up the squeeze but on any other lead, it appears that you can make the contract.

It's not "sure-trick" but it's a "kibitzer-make" (i.e. at double-dummy the contract is makeable against any defence and distribution).

• As soon as you find out that LHO has 5+ ♠, a compound squeeze should be available. In the line of play illustrated above, if LHO had both red suits, he would have to abandon one of them, on the third round of ♣.
• When you discover LHO also has four ♣, this simplifies to a double squeeze, provided you can guess which is LHO's longer red suit (probably ♦).
• If LHO is 2-2 in the red suits, then you have a simple squeeze against RHO but you can successfully play a "double" squeeze.