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What's the rate of difficulty of this problem?  

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  1. 1. What's the rate of difficulty of this problem?

    • Easy
      2
    • Intermediate
      5
    • Hard
      5
    • Very hard
      3


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I know I am probably wrong, but here is what I think.

 

If declarer is 4-2-5-2, on the 5th diamond I will be squeezed, even if I choose to throw the AC, since the 5th diamond will eliminate my strip card or set up his QC.

 

KQxx

Kx

AQTxx

Qx

 

However, if for some reason partner holds the QS with Qx, my spades are protected if I go down to the JTx, therefore, I will eventually pitch a low club and a low spade.

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Hi,

 

Sorry, ASkolnick and Hanoi5.

 

You got that wrong:(

 

I don't think 4252 usual NT opening.

 

How about to focus those :

 

NT range is 15-17. When I count my hand I see 8 HCP, Dummy shouts 15 HCP. We reached total 23 HCP. 40 HCP in deck so if rest 17 HCP with declarer nothing to do. He brings th contract home with 3S+2H+4D+3C while his C holding is at least QJx.

 

Then you might question what if p has only 1 HCP with a "good" shape?

If missing club knave accompanied by ten and small three times with your partner can you set the game? I think : "yea":rolleyes:

 

Omg, i told a lot again. How careless I am, why i do that;-)

 

Soon, I will try to clarify what's wrong about which discards.

 

Best regards

Hamdi

 

p.s. I believe it is more easier for us to remember the correct move if we are aware of the reasoning behind it. Do you agree?:ph34r:

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Well, it would be nice to know a bit about the opponents' bidding.

What strength was the 1NT opening, for example?

What did 2C mean?

What did 2S mean?

What did 3C mean?

What did 3D mean?

 

If we assume that 2C was Stayman, that 3C and 3D were natural, and that 1NT was 15-17 balanced, then I'd still like to know if opener can be 4=2=5=2, but I would assume not, as that isn't a 1NT opening for most people I know. Or can declarer have five spades? What about 5=2=4=2?

 

There are three possible interpretations of partner's 2 of spades at trick 1:

- he wants to give count, but can't because he has Hx

- he knows that we know the spade count, and is giving suit preference to show us that he has a club honour

- he is playing a card at random so as not to help declarer, as he knows that we know the spade count.

 

The whole "he knows that we know the spade count" depends on declarer not being possibly 5=2=4=2, or 5=3=3=2 and bidding a diamond fragment. But let's assume that is the case.

 

That means that declarer one of the following hand type(s)

 

KQxx

Kx(x)

AQ10x

(Q/J)x(x)

 

Kxxx

Kx

AQ10x

QJx

 

if he's got the second hand he's just cold (and he's played it a rather odd way) via a major suit squeeze, given that clubs are 3-3.

 

If he has

KQxx

Kx

AQ10x

Qxx

 

we are triple squeezed without the count. If we let the club go, he discards a heart from dummy and ducks out the ace of clubs and has 12 tricks. If we unguard spades, he discards a club from dummy and cashes the spades, catching us in a heart/club strip squeeze. If he unguard hearts, he discards a club from dummy and has 12 tricks with 4 hearts, 3 spades, 4 diamonds and a club.

 

What about

KQxx

Kx

AQ10x

Jxx

 

(carefully without the C10)

we can discard the ace of clubs now. But he just crosses to dummy and leads a club towards the jack, making 3 clubs, 3 spades, 2 hearts, 4 diamonds.

So there's no point worrying about that hand.

 

That leaves

KQxx

Kxx

AQ10x

Qx (or Jx)

 

If we just discard a low club he's stuck.

 

So I must be missing something: either it's cold, or we make the obvious discard of a low club.

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For FrancesHinden :

 

If he has

KQxx

Kx

AQ10x

Qxx

 

we are triple squeezed without the count. If we let the club go, he discards a heart from dummy and ducks out the ace of clubs and has 12 tricks. If we unguard spades, he discards a club from dummy and cashes the spades, catching us in a heart/club strip squeeze. If he unguard hearts, he discards a club from dummy and has 12 tricks with 4 hearts, 3 spades, 4 diamonds and a club.

 

 

Sir are you sure please?

 

Sincerely

Hamdi

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This is a much harder problem than those you posted earlier, and it definitely belongs in this forum.

 

As Frances says, it's a good idea to provide an explanation of the bidding. This forum is read by people from all over the world, and things that are standard in one country may mean something completely different in others.

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OK, let's have a rethink

 

.......................Ax

.......................A10xx

.......................KJx

.......................K9xx

 

 

J109xx..........................................xx

QJxx.............................................9xx

x..................................................xxxxx

Axx..............................................Jxx

 

 

 

.......................KQxx

.......................Kx

.......................AQ10x

.......................Qxx

 

After the ace of spades, a diamond, a club to the king and two more diamonds, we are here:

 

.......................x

.......................A10xx

.......................-

.......................9xx

 

 

109x..........................................x

QJx.............................................9xx

-..................................................xx

Ax..............................................Jx

 

 

 

.......................KQx

.......................Kx

.......................A

.......................Qx

 

on the last diamond:

- if we let a spade go, he has 11 tricks and an easy rounded suit strip squeeze

- if we let a heart go he has 12 gtricks at once

-if we let a low club go, he ducks out the ace of clubs and has 12 tricks

 

- I missed out if we let the ace of clubs go, but actually we are still stuck even then. He can't set clubs up, but what he can do is cash the queen of clubs and throw us in with the last spade to give 3 heart tricks (or finesse against the CJ if he started with Q10).

 

But if declarer instead started with the jack of clubs instead of the queen, my comment "But he just crosses to dummy and leads a club towards the jack, making 3 clubs, 3 spades, 2 hearts, 4 diamonds."

isn't right, because partner has a diamond to cash.

 

So what if declarer started with Jxx in clubs? This is the end position

 

.......................x

.......................A10xx

.......................-

.......................9xx

 

 

109x..........................................x

QJx.............................................9xx

-..................................................xx

Ax..............................................Qx

 

 

 

.......................KQx

.......................Kx

.......................A

.......................Jx

 

actually now we can happily throw a low club.

 

So I'm still back to throwing a low club on the 4th diamond, playing partner to hold the queen.

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.......................x

.......................A10xx

.......................-

.......................9xx

109x..........................................x

QJx.............................................9xx

-..................................................xx

Ax..............................................Jx

.......................KQx

.......................Kx

.......................A

.......................Qx

 

- I missed out if we let the ace of clubs go, but actually we are still stuck even then. He can't set clubs up, but what he can do is cash the queen of clubs and throw us in with the last spade to give 3 heart tricks (or finesse against the CJ if he started with Q10).

I don't think that's right. When we're in with the fourth spade, we can exit with a low heart.

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.......................x

.......................A10xx

.......................-

.......................9xx

109x..........................................x

QJx.............................................9xx

-..................................................xx

Ax..............................................Jx

.......................KQx

.......................Kx

.......................A

.......................Qx

 

- I missed out if we let the ace of clubs go, but actually we are still stuck even then. He can't set clubs up, but what he can do is cash the queen of clubs and throw us in with the last spade to give 3 heart tricks (or finesse against the CJ if he started with Q10).

I don't think that's right. When we're in with the fourth spade, we can exit with a low heart.

That's true.

 

So throwing the ace of clubs gets you into the papers when that is the layout, but makes you look remarkably stupid when declarer is 4=3=4=2

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[hv=d=e&v=b&n=sa2hat74dkj8ck965&w=sjt954hqj63d6ca74&e=sxxhxxxd97xxxcjtx&s=skqxxhkxdaqtxcqxx]399|300|Scoring: IMP[/hv]

 

FrancesHinden Posted on Nov 4 2008, 08:58 PM

"That's true.

 

So throwing the ace of clubs gets you into the papers when that is the layout, but makes you look remarkably stupid when declarer is 4=3=4=2"

 

----------------

Uh Oh, everything in life has a price; defending tough contracts is no exception.

 

Congratulations for "good work".

Hamdi

----------------

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Uh Oh,  everything in life has a price; defending tough contracts is no exception.

Nor is playing them, of course. Whereas it seems only natural caution for declarer to play a club to the king before establishing East's long diamond, this is actually a mistake. Details left as an exercise for the reader, but commendation to the original poster for finding an example of a triple squeeze without the count - possibly one of the rarest positions in bridge.

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