Just when you thought you had seen it all...

[fred]

[hv=d=s&v=b&n=sj98hd982caj108763&w=sa107hj8daj104ck952]266|200|Scoring: IMP[/hv]

 

RHO opens 4H and all Pass.

 

You lead the Jack of hearts.

 

Declarer, who started with AKQxxxxx, plays 7 of his 8 trumps.

 

He discards a spade, 2 diamonds, and 4 clubs from the dummy.

 

Your partner says he likes diamonds but doesn't like spades. His 4 discards are 3 spades and 1 diamond.

 

What are your 5 discards?

 

This problem is very hard even as a double dummy problem. I am not sure if it makes a good single dummy problem or not. If you would prefer to try to figure it out looking at all 4 hands, declarer's actual hand appears in hidden text below.

 

If you can figure out even the double dummy problem without using GIB or Deep Finesse then you are good!

 

 

 

K5

AKQ76532

63

4

 

 

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[marie__L]

About the Club suit.

 

If my partner doesn't discard a Club it must be :

 

Or because he is void( then declarer has 10 tricks!)

 

Or because he has the Queen bare.If he had the 4 bare, he should discard it.

 

Maybe also my part gets two cards in Club.

 

Then I keep two cards in Club :P

 

What about the Spades? I should have the count of the suit at this point I think.

How many Spades got my part?

[benlessard]

HIDDEN

 

 

The way i see it is

 

1--if declarer got 2 clubs we are screwed.

 

2- Partner doesnt have KQxxx of spades so declarer got a S honnor.

 

3- If opener S honnor is the K then he probably doesnt have a club void or a D honnor (he would open 1H)

 

4- If opener S honnor is the Q then he still cannot have the DQ (since partner said he dislike S and like D then his D are better then his S)

 

So its likely that declarer got.

 

Hxx

AKQxxxxx

x

x

 

or

 

Hx

AKQxxxx

xx

x

 

or

 

Qxx

AKQxxxxx

xx

 

and less likely

Kxx

AKQxxxxx

xx

----

 

Kx

AKQxxxxx

xxx

---

 

So i think I need to keep 2 clubs, AT of S and the H4 of D or ♠ATx,♦4,♣Kx. But im not sure why i can/cannot discard a S. But im pretty sure that if declarer got 3S he alway make unless hes got a C void (plus impossible partner would discard 3 of them holding Hxxx)

 

If the declarer got Kx,AKQxxxxx,xx,x partner ending will be,

 

 

Qx

---

KQ7

Q

 

If declarer play his last trump i can pitch a diamond and partner too. So i still dont see the endplay.

 

If declarer exit with a D partner can return the club

 

If declarer play C to the A and D partner will play low ill take and play the K of C.

 

 

 

[nige1]

Beautiful stuff, Fred!

I guessed declarer's hand :)

But still misdefended :P

HIDDEN

If you reduce to ♠AT ♦AJ ♣Kx (as I did) and your partner keeps ♠Qx ♦KQx ♣Q, then declarer crosses to CA. Now, dummy's ♣J catches your partner in a vice/submarine squeeze:

• If your partner discards a ♦ then declarer ruffs and exits in ♦ to endplay either of us iin ♠.
  
• If your partner bares ♠Q, then declarer discards a ♦. You win ♣K. You can try 2 rounds of ♦. But deckarer ruffs and leads ♠K, dropping ♠Q and endplaying you.
  

The solution is for you to keep a singleton ♦ -- not ♦A.

 

[qwery_hi]

Hidden, since I saw declarers hand

 

 

After seeing the declarers hand

 

By elimination, I cant discard 2 spades, or 3 clubs.

 

So I discard a low spade, 2 diamonds, and 2 clubs.

 

[Fluffy]

I was trying to defeat ♠Kx ♥AKQxxxxx ♦xx ♣Q, but there is no way to do it.

 

On the given hand I must admit I don't see the problem yet, keeping what qweri_hi said and being careful enough to hop up with ♦A in order to play a club and keep ♦J as comunication seems to beat it.

 

Since Fred suggests it is complicated, I think the winning defence is to discard ♦A and keep 3 clubs to be able to escape latter.

[Fluffy]

Now I see how declarer is gonna make this (I'll keep you the surprise, but it is some sort of balancing squeeze!), I wonder if keeping 3 clubs is good enough to defeat the contract I think it is.

[cloa513]

One question is when do the signals come i.e. In what order?

[cloa513]

How are signals done? People seemed to assumed that clubs are no discarded

[benlessard]

After a good night of sleep...and reading declarer hand

 

 

 

 

I think its possible to guess declarer hand because he cannot have 3S (partner discarded 3) and with a C void he will go down (partner like D so hes got the K).

 

Im sure the threat is C to A and C squuezing partner. If he keep 2 diamond he will be endlayed, while if he keep 1S ill be endplayed (he will pitch a D on the clubs ruff the next D and crash the Q of S with the K). Like Fluffy said keeping 3C break the squeeze.

 

[fred]

One question is when do the signals come i.e. In what order?

Your partner's first discard is a diamond that says "I like diamonds".

 

Your partner's second discard is a spade that says "I don't like spades".

 

Your partner's next 2 discards are spades.

 

Here is some additional information: your partner's signals on his subsequent spade discards suggest that he was dealt 5 cards in spades.

 

So he is either 5332 or 5341. If 5332 probably his first discard would not have been a diamond so you can assume 5341.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[mikeh]

I'll give it a try

 

 

Assuming rho is 2=8=2=1, with Kx AKQxxxxx xx x, we need to cater to various H's in the minors... say declarer is Kx AKQxxxx Qx x... we have to be very careful... we cannot, for example, pitch the diamond A.

 

So reduce to A10 void A4 Kx. declarer to Kx x Qx Q and partner to Qx void Kxx Q with dummy J9 void 9 AJ10

 

If declarer cashes the last trump, we stiff our spade A.. he can establish the spade K, but can't enjoy it.

 

He can't lead spades.. we cash out.

 

If he leads a club and ruffs out the suit, partner pitches his small spade and we pop the diamond A when declarer exits that suit.

 

If he exits a diamond, we pop the Ace and lead a club

 

 

I will post and then look at the hand to see how wrong I was :)

 

 

wrong :) B) as I suspected, I didn't look deep enuf, and noway Fred would have posted the problem if my play worked B)

[sceptic]

if declarer has singleton Q he has to over take or lose his second !C trick , so I see no advantage if declarer has Qx or Qxx of clubs he can make anyway. so only need to keep KX

 

pard likes diamonds so if he has k spades he cab get in and play to my diamond holding

 

so I discard First signal card high Diamond to say I like Diamonds, then another diamond the low one then 2clubs and a spade discard

 

 

also, if pard has Qx clubs and declarer plays clubs first, ( no idea how likely that is) he knows to lead diamonds

 

I suppose mps or imps change things, if kx clubs loses first and opp makes all them over tricks, you would be screwed, but I cant see another way to set it (by discarding anything else)

[fred]

Here is what I think the answer is:

 

Assuming you decide to play declarer for 2821 with the King of spades, and your partner for the King of diamonds (all reasonable given your partner's discards)...

 

1) It doesn't matter if declarer's singleton club is the Queen or a small one (you will see that the size of South's club never comes into play in any of what follows)

 

2) If declarer has the Queen of diamonds the contract can't be defeated (you will see why if you read the rest of what follows)

 

So let's give declarer:

 

Kx

AKQxxxxx

xx

x

 

(which turns out to be exactly what declarer actually had).

 

As West, it feels obvious to discard a spade, 2 diamonds, and 2 clubs, but if you do then this will be the ending:

 

[hv=d=s&v=b&n=sj9hdxcaj10&w=sa10hdajckx&e=sqxhdkqxcq&s=skxhxdxxcx]399|300|Scoring: IMP[/hv]

 

Now declarer plays a club to the Ace and leads another club.

 

If East discards a diamond, declarer ruffs the club and exits in diamonds (where the defense is now 2-2). The defense is endplayed into breaking spades.

 

East must therefore discard a spade, stiffing his Queen. Careful now declarer! He can't ruff the club and bang down the King of spades - the defense would then run 3 diamonds.

 

Instead, declarer must discard a diamond on the 2nd round of clubs. West wins the King of clubs and has to play diamonds, but South ruffs the 2nd round and now plays the King of spades. Winning trick 13 with the Jack of spades is probably not what declarer expected when the dummy came down :)

 

So, in the end position above, the declarer can prevail against any defense.

 

West's counter to this is to discard a winner (a diamond - the Ace in fact) in order to keep a loser (an extra small club). Here is the ending that will be reached:

 

[hv=d=s&v=b&n=sj9hdxcaj10&w=sa10hdajckx&e=sqxhdkqxcq&s=skxhxdxxcx]399|300|Scoring: IMP[/hv]

 

If declarer exits in diamonds, the defense can cash their 2 diamonds and play a club - declarer will have to lead spades himself.

 

And it won't do any good for declarer to lead his last trump in the above position - the defense has 4 winners.

 

So let's say declarer tries what worked for him before - leading a club to the Ace and then another club.

 

In the 1st end position diagram, East was squeezed on this trick - he could not afford to discard a diamond so he came down to a singleton Queen of spades.

 

But in the 2nd end position diagram, East can afford to discard a diamond (if you understand the difference without reading further, you have done very well!).

 

If declarer ruffs the 2nd round of clubs, the defense has 4 winners for the last 4 tricks (1 spade, 2 diamonds, 1 club).

 

If instead declarer discards a diamond on the 2nd club, West wins the King and now 2 rounds of diamonds will endplay declarer in his hand. The King of spades won't save declarer this time because East was able to retain Queen doubleton of spades.

 

Note that if West had kept the Ace of diamonds (instead of the Jack) in the above diagram, he would have been unable to get out of his own way.

 

That is why there is no defense if declarer was dealt the Queen of diamonds. If South had that card, in the 2nd end position diagram he could lead a club to the Ace and a diamond toward his Queen.

 

Even if you had to read all of that in order to understand why West's third club is so important, if you do understand that now, you are a lot closer to good then you are to bad!

 

If you still don't understand, I do hope you blame the tough hand itself (as opposed to my possibly unclear and/or possibly faulty analysis of it!).

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[benlessard]

Great hand. Im wondering what happened at the real table.

 

Im also thinking about making a software that create double dummy problems of 6-9 cards endings.

 

Computers can create unusual endgame positions with a double dummy analyser easily but its the rules that pull out the interesting hands or rules that discards the uninteresting hands that arent obvious.

 

Any ideas where i should start ?

[nige1]

[hv=d=s&v=b&n=sj32hd432ca765432&w=sa9hj9daqjt6ckjt9&e=sqt8765ht87dk87cq&s=sk4hakq65432d95c8]399|300|Scoring: IMPs

Simplified version of

Fred's wonderful Vice/Submarine Squeeze,

illustrating the power of the Curse of Scotland.

4♥ makes on a trump lead.

But if you swap ♦8 with ♦9,

then the defence can prevail.[/hv]

[P_Marlowe]

Great hand. Im wondering what happened at the real table.

 

Im also thinking about making a software that create double dummy problems of 6-9 cards endings.

 

Computers can create unusual endgame positions with a double dummy analyser easily but its the rules that pull out the interesting hands or rules that discards the uninteresting hands that arent obvious.

 

Any ideas where i should start ?

Assuming you have a double dummy solver, you also have

the solution tree.

One possible way would be to ask, how many exist ways / path

exist in the tree to get n-1 tricks.

 

If the number of ways to get n-1 tricks, but not n tricks is high

enough, you may have the indication that the problem is not

very straighforward.

 

I would also assume, that the overall freakness of the deal

should not be too high.

Lets say, if you give everyone n cards of the same suit, the

problem is trivial.

 

<Added Later>: Another option would be to look at the stability

of the problem, try to exchange one single card, and see if the

problem is still similar, solvable.

My guess would be, that a fairly stable problem would also be

a fairly boring problem.

The adv. of this approach is, that you dont need the solution tree,

you can use the double dummy solver as black box.

 

With kind regards

Marlowe

 

PS: If you make money from this, please mention my name ...

[qwery_hi]

Even if you had to read all of that in order to understand why West's third club is so important, if you do understand that now, you are a lot closer to good then you are to bad!

 

If you still don't understand, I do hope you blame the tough hand itself (as opposed to my possibly unclear and/or possibly faulty analysis of it!).

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

I can't seem to get this hand out of my head even after a couple of days - it reminds me again why I love to play this game.