[inquiry]

You are playing UDCA.

Click the NEXT button to see the first three tricks. What do you play to trick four. Is the correct defense "clear". What can you speculate about the unseen hands?

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[MrAce]

Spoiler

I think declarer must have 5♦ since pd did not play ♦ back. Pd showing 3♣ unless he has J8 which then he would have 5-5 majors and probably bid. So Declarer has 5♦+3♣ and upto ♠ play by pd 2♠+3♥. If we play ♠ back we will be ok if pd's ♠ are AT9xx, but this will cost if he has ATxxx or A9xxx. IF declarer has AQJ ♣. Lets kill his entry to dummy which is ♦K so he cant use 4th ♣. In case declarer is something like J9 AQx JTxxx AQJ.

Second scenario that comes to my mind is, if pd has ♣A instead of ♠A, making declarer something like AJ AQx JTxxx QJx. If this is the case then we need to play ♠ now and establish pd's ♠. If we dont then declarer will make, lets say we play ♦ he takes in dummy plays another ♣, pd can duck but has to take 3rd ♣ and play ♥, now declarer can go up with the ace and play ♦J, making 2♠+1♣+1♥+2♦+another ♥ due to endplay.

Overall i think ♠ back works better and more often

[BunnyGo]

This spoiler contains my long and rambling thought process.

Spoiler

OK, let me think "out loud". Let's start by counting points. Partner can have 6-8, and has already shown up with an A. So he has 2-4 unseen points, and everything else is declarer's.

partner is likely to have a stiff diamond, since he'd likely continue my suit without many entries, but it is possible that he has Ax and didn't see profit in continuing diamonds and wanted a quick switch.

Partner has to have 4+ spades for his switch, but has at most 5. He has between 2-5 hearts

Partner seems to be signaling an odd number of clubs (If I read his 8 correctly). If he has 8x, he'd surely play x, With A8 or J8 he would play the 8. This most likely means declarer has 4-6 clubs (I guess partner could have stiff 8, but he passed?! no way) so declarer has 4 clubs.

So, either partner has J8x (x= 7 or 6), or 876. In the former, clubs are frozen. In the latter, he has some points elsewhere.


If partner is so good as to have AT943 of spades, he can just run off 4 spade tricks now. If I don't lead a spade now, then declarer has 1 heart, 1 spade, 1 diamond, and 3 club tricks before I get in again to lead a spade. No rush.

What if partner has AT543 of spades? Then, I don't think that we can more than the two tricks he has coming.

One other problem seems to be looming, and it looks like I could be subject to a possible redsuit strip an endplay. This would be unpleasant, as I *should* get 3 tricks in the red suits. Combined with the two we have in the bag, that would set the contract. I could really use a heart through declarer now.

If I were to lead a diamond, for example. I bet declarer will take his 4 clubs. If I pitch a diamond, he puts me in with a diamond. Then I exit a spade, and partner gets two spade tricks (assuming he has A 5th). If declarer has the spade A, then partner has to have a club and/or heart honor, so there's no endplay.

In short, I haven't figured out where the true danger of this hand lies yet. I'm just going to keep typing till I get it.

Leading the heart K is an interesting idea. If partner has the Q, we could be setting up his 4 hearts. If partner has no heart honors, then he has to have a black A.

Yeah, what if partner has the Qxxx of hearts? Then he has at most another jack or two. If I don't lead hearts now, then declarer has 3 club tricks, 2 spade tricks, 2 diamond tricks, 1 heart trick. I don't see it....Where are his 9 tricks coming from.

Let's consider that. Declarer has 1 diamond (and another if/when he drives out my queen...possibly two if I can be endplayed somehow), he has 2-4 club tricks (depending on if partner has J8x or 876 or J8) he has 1 heart trick unless I lead them to give him the Q (giving him AQ) he has at most 2 spade tricks unless we set up the 8 in dummy. This is at most (in suit order) 2 + 1 + 2 + 3 = 8. So he needs another by endplay or something *and* he needs the maximum from every suit.

So give partner J8 of clubs (this gives declarer 4 club tricks), and so partner *has* to have the heart queen--I know, why didn't partner bid? I don't care, this is just a case I want to consider. In this case declarer has

AJ, A, JTxx ,AQxxx

Strange, This is the only construct with 5 clubs that can possibly be. Then declarer has 2 spades, 1 heart, 2 diamonds, 4 clubs (if we give him time to set up the diamonds). In this case, any major suit lead will set the contract.

Ok, so what if declarer has only 4 clubs?

J?,??,JTxxx,AQxx

I don't have to give him the 5th diamond, I'm willing to believe partner had 2 diamonds and WANTED a switch. Strange though that he wouldn't save the diamond A to cover the jack though. If he does have 2 diamonds, he better have AT9xx of spades. So ok, unless partner has that, he has a stiff diamond.

Those 4 "?" have to sum to 7-9 points. Could be A of hearts and or spades, and/or Q of hearts. The only way this can work is if he has both Aces and no queen. I should lead hearts now, and not worry about whether partner has the AT9 of spades or not...he has Qxxxx of hearts. Lead the heart King. Why didn't partner bid 5-5? Don't care.

Finally, if declarer has 3 clubs:

J?,???,JTxxx,Q??

Ok, the points to spread around are the spade A, the heart A and Q, and the club A and J. Partner gets 2-4 of them, so in particular, if he has the J he has the Q. As discussed before, in this case I should lead the heart K.

If declarer has the club J, and both Aces, then it doesn't look good. He has 2 spades, 1 heart, 1 diamond, and 3 clubs. I either need to hit spades or hearts before he can set up more tricks. He can set up another in diamonds, and maybe a 9th in spades.

If declarer has both the J and Q, then partner has exactly one ace. If it's the A of clubs, then declarer has only 2 club tricks, 2 spade tricks, 2 diamond tricks (eventually) and whatever else he can scramble for. If I lead a heart, that's 2 heart tricks. I think the K of hearts works here too.

This one contains what I lead now

Spoiler

The heart K.

[wyman]

Spoiler

Pard has at most an ace remaining. I'm guessing he also started with a stiff diamond, since he couldn't know that I don't have 6 diamonds from the spots he saw, but I won't swear to it.

P could basically have:
SA
HA
HQ
CA
CJ & HQ

But partner played a low spade back, so he should really have a spade card: the ace.

How about shapes? If declarer has 5 diamonds, as I suspect, he has <= 3 clubs, lest he be off-shape for his 1N call. Sure looks like this is the case with pard's count card in clubs. So it looks a lot like declarer holds:

Jx / AQx / J10643 / AQJ

Of course, this means declarer false-carded on the diamond. Maybe partner had A3 and didn't return one. He'd have to have a pretty good reason, like ♠A1094x. Then declarer has something like:

Jx / AQx / J1064 / AQJ + 1 card somewhere outside of ♦. It'd be weird if it were clubs, since that would give partner J8 of clubs, which gives declarer only a 14 count. So probably not that. Maybe Jx / AQxx / J1064 / AQJ ? The extra heart is harmless, given our holding, and if it were Jxx / AQx / J1064 / AQJ, it would be a weird shift from partner from
A10xx / xxxxx / A / xxx but not so weird from
A109x / xxxxx / A / xxx

But it would be a guess for declarer to play the J then from Jxx opp K876; He'd probably not play east to hold SAQ, since he's already shown up with the diamond ace, and the heart and club honors are yet unseen. [East is a passed hand.] Probably better to play low to the first spade, catering to something like Ax or Qx on your left, as in the actual layout.

Back to basics: let's count declarer's tricks. He has 2 clubs + 1 diamond + 1 heart + 1 spade = 5. So there seems to be no hurry for us to cash out. If partner's spades are good enough, we'll get the diamond ace, club king, 4 spades, and a heart, whether I lead spades now or not (though a diamond now might hold declarer to 5 tricks). So let's assume partner has 5 noncashing spades -- like A10xxx. P has no entry other than the SA, so we can't set spades up, so we'll need to score 2 spades, 2 diamonds, 2 hearts, and the club to beat this. Notice that if I lead spades now, we'll only get 2 spades if p cashes them, but this establishes a spade in dummy. So we'll need to cut declarer from dummy by leading the low diamond now.

Now declarer can score another diamond per force (something I didn't consider earlier) but that still only brings him to 6.

♦2 for me

[Phil]

This one is probably a little over the B/I range I think.

Spoiler

My thinking closely echoes MrAce's (1st layout) and Wyman's. I think it would be very strange for partner to shift to a spade from AT9xx, since he does not have the entries to cash the suit, and he is potentially giving up tricks with the shift. Much more likely is partner holding T9xxx Axxx A 8xx and declarer holding AJ QTx JTxxx AQJ. I also think with AT9xx xxxx A 8xx, partner might balance over 1N w/r. BTW if declarer has AJ Axx JTxxx AQJ, he's blundered by playing the ♠J. Assuming declarer does have AJ QTx JTxxx AQJ, after I win the club, I don't think it matters much what I do.

[MrAce]

No need for too complicated analyis. We should never play pd for ♥A because if he has this declarer can not make.

We all seem to agree that declarer has 2353 shape.


When pd has AT9xx ♠, both ♦ and ♠ now defeats 1NT

When pd has A9xxx or ATxxx ♠, ♦ now wins

When pd has ♣A, only ♠ back now wins.

When pd has ♥ A, it doesnt matter we always defeat

When pd has ♥ Q, it doesnt matter, we can never defeat because declarer then has to have ♠A and making 2♠+2♣+2♦+1♥ and we cant cash 7 tricks before he does.

When pd has ♥Q + ♣J it doesnt matter we always defeat regardless of what we play again.

My choice is ♠ now as i said earlier, not only because of pd's choice of bidding, but also i am losing only if declarer has J9 JT doubleton ♠ which by the way declarer could have made it by simply playing another ♠ after K if he had one of those holdings.

[BunnyGo]

 MrAce, on 2011-September-06, 16:09, said:

No need for too complicated analyis. We should never play pd for ♥A because if he has this declarer can not make.

We all seem to agree that declarer has 2353 shape.


When pd has AT9xx ♠, both ♦ and ♠ now defeats 1NT

When pd has A9xxx or ATxxx ♠, ♦ now wins

When pd has ♣A, only ♠ back now wins.

When pd has ♥ A, it doesnt matter we always defeat

When pd has ♥ Q, it doesnt matter, we can never defeat because declarer then has to have ♠A and making 2♠+2♣+2♦+1♥ and we cant cash 7 tricks before he does.

When pd has ♥Q + ♣J it doesnt matter we always defeat regardless of what we play again.

My choice is ♠ now as i said earlier, not only because of pd's choice of bidding, but also i am losing only if declarer has J9 JT doubleton ♠ which by the way declarer could have made it by simply playing another ♠ after K if he had one of those holdings.

The moral of the story (for me at least) is to always pay attention to the contract. I was defending 3NT, and only just now realized it.

[inquiry]

The posters have rationalized the hand well, and found the killing return (your last spade). The rationale about partner has only one diamond is fairly clear, and his club signal is helpful. Of course, we should all quickly get the 6 to 8 points for partner. So you can work out fairly closely partner's hand. The key decision is to return a diamond to kill an entry to the "long club" in dummy, and return a spade to set partner's long suit up while he still has an entry. As you can see, the hand was just a Mr. Ace proposed that needed the spade return.

I think Phil's logic about partner not underleading spade ACE is useful in determining the best play.

Give yourself a lot of credit if a.) you worked out everyone's distribution, b.) you remembered to count partners hand out. This one I think is advanced defense level, but thought the problem would be challenging and educational for the upper intermediates.

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This how Defenders did against NT by East. Notice holding this to 6 tricks is not easy. Actually the two 6 tricks and 5 trick defenses benefited from poor declarer play, not great defense.

Contr   Ld   Tr  Score   Pts
3N	E D7	7   200   2.93    	
2N	E D7    6   200   2.93    	
1N	E D7    5   200   2.93    	
1N	E H7	6   100   0.07    	
3N	E D9	8   100   0.07    	
2N	E C3	7   100   0.07    	
1N	E D7	7   -90  -4.93 
1N	E D2	7   -90  -4.93    	
1N	E D7	8  -120  -5.87