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Plan for this grand?

#1 User is offline   dkharty 

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Posted 2010-September-06, 10:45

Scoring: MP


You are playing 7 after an uninterrupted auction. The lead is a low heart; trumps are 2-1, LHO having 2. What is your plan?

I am particularly interested in your decision process regarding club finesse vs. double squeeze, and what discovery process you use to help make the decision.
Diane, I'm holding in my hand a small box of chocolate bunnies...
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#2 User is offline   Dirk Kuijt 

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Posted 2010-September-06, 10:59

Interesting hand. We have a 6-4 fit, and find the outstanding trumps 2-2.

codo said:

It is a fact that most people here write as if their opinion is a dogmatic fact.

eugene hung said:

My opinion is that this ought to win the award for best self-referential quote of the new year.
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#3 User is offline   dkharty 

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Posted 2010-September-06, 11:06

Dirk Kuijt, on Sep 6 2010, 11:59 AM, said:

Interesting hand. We have a 6-4 fit, and find the outstanding trumps 2-2.

Oops :) I'll fix it...
Diane, I'm holding in my hand a small box of chocolate bunnies...
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#4 User is offline   manudude03 

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Posted 2010-September-06, 12:16

My gut says double squeeze is better, not sure how much better though (if at all).

I can play off all the trumps discarding a spade and a club. This also has the extra advantage that I make on some layouts where the club finesse was working all along (give East JTxxx x Txxx xxx). The main disadvantage of course is that I may have to guess the end position.

Depending on who discards what, I may be able to decide which squeeze to persue. Cashing 3 rounds of diamonds would complete the standard double squeeze (the A would drop a blank K if that guard was removed. Cashing the A and 2 spades ending in hand instead establishes a squeeze against either opponent guarding both minors.
Wayne Somerville
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#5 User is offline   gnasher 

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Posted 2010-September-06, 13:47

A priori, an automatic double squeeze would be better than the finesse, because if the hand with the shorter diamonds is more than 50% to hold K. However, unless diamonds are 5-2 I don't think there's any way to avoid having to guess who's guarding what.

I play a third round of trumps, because I can. Then I cash the top diamonds.

If RHO shows out on the third diamond, I can play the double squeeze and I know its good odds, so I throw Q, cash A, and run trumps, throwing a spade on the penultimate one.

If RHO follows to the third round of diamonds, I'll probably just take a club finesse.
... that would still not be conclusive proof, before someone wants to explain that to me as well as if I was a 5 year-old. - gwnn
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#6 User is offline   gnasher 

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Posted 2010-September-06, 14:42

gnasher, on Sep 6 2010, 08:47 PM, said:

If RHO shows out on the third diamond, I can play the double squeeze and I know its good odds, so I throw Q, cash A, and run trumps, throwing a spade on the penultimate one.

Actually, there's no need to be that committal. I can throw a spade and cash all but one trump before deciding. But I've more or less decided already anyway.
... that would still not be conclusive proof, before someone wants to explain that to me as well as if I was a 5 year-old. - gwnn
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#7 User is offline   dkharty 

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Posted 2010-September-06, 16:36

gnasher, on Sep 6 2010, 02:47 PM, said:

I play a third round of trumps, because I can.  Then I cash the top diamonds.

If RHO shows out on the third diamond, I can play the double squeeze and I know its good odds, so I throw Q, cash A, and run trumps, throwing a spade on the penultimate one.

If RHO follows to the third round of diamonds, I'll probably just take a club finesse.

Both opps will throw clubs on the trump, and on the play of the diamonds LHO plays 4-5-10, RHO plays 3-7-8.
Diane, I'm holding in my hand a small box of chocolate bunnies...
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#8 User is offline   lmilne 

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Posted 2010-September-06, 22:04

My first instinct is to play 5 rounds of trumps (to try to induce some to discard a diamond or a spade from 3, followed my 3 rounds of diamonds, the ace of clubs, and a spade to the king, to come down to


which seems to preserve all squeeze chances.

Now on the play of the last heart, we pitch the diamond (unless it becomes high!).
This wins if:
1) West has 4+ diamonds and East has the K (double squeeze);
2) West has 4+ diamonds and also 5+ spades (positional);
3) Either opponent has both 5+ spades and the K (automatic).

I'm having a bit of trouble thinking about all the possibilities - also, say diamonds are 4-3, do we get an indication (restricted choice) about who has the other diamond honour? Trying to get better at double squeezes and math :)

Edit: Also cashing the ace of clubs seems kind of counter-intuitive now that I look at this again. Not sure.
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#9 User is offline   lmilne 

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Posted 2010-September-06, 22:20

Having thought about this a bit more, it's frustrating because the contract is cold on any layout, we just have to pick which one to play for (I think - maybe there is a line I'm missing).

1) West has the K - finesse;
2) East has the K and 4+ - auto squeeze;
3) East has the K, West 4+ - auto squeeze if West has 5+S, double squeeze if spades are 4-3.

Now someone better at squeezes (and logic) than me can work out which line picks up the most (and more likely) combination of these.
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#10 User is offline   Mbodell 

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Posted 2010-September-08, 12:45

lmilne, on Sep 6 2010, 08:20 PM, said:

Having thought about this a bit more, it's frustrating because the contract is cold on any layout, we just have to pick which one to play for (I think - maybe there is a line I'm missing).

1) West has the K - finesse;
2) East has the K and 4+ - auto squeeze;
3) East has the K, West 4+ - auto squeeze if West has 5+S, double squeeze if spades are 4-3.

Now someone better at squeezes (and logic) than me can work out which line picks up the most (and more likely) combination of these.

Basic logic suggests 1> 3 > 2, no? Given we don't know anything asymmetric about the suits 1 is 50%. 2 is less than 3 and less than 50% because it requires the hand with long diamonds (less spaces) also to have a specific club card and starts of with the 50% problem of East having the K.
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