[Echognome]

Scoring: MP

1♦ - 1♥
1N - 2♥
3♥ - 4♥

---

Bidding is all opponents way.

T1: ♣J - Q - K - x
T2: ?

What's your play to take 4 tricks?

[vuroth]

Novice opinion:

I can see all clubs from the 9 to the ace. Partner's lead of the Jack stands out - he wasn't leading from KJT, or JT, or anything like that. He's either single or double in clubs. South, therefore, either has 2 or 3 clubs.

Cashing the Ace and leading back a third club looks like it has a decent chance of giving partner a ruff. Partner may have as many as 3 hearts (I peg south for at least 6).

Partner can easily have 7-8 points here. If he has the king of diamonds (or other diamond wastage), we probably need a ruff to set the contract, if we can set it at all.

Since we have absolutely nothing we want returned to us, we could always lead a low club third round, trying to signal a diamond return. Partner isn't too likely to take this seriously if he can set the contract immediately or if he has a diamond honour, and it might cause declarer to guess incorrectly.

That's about all I can think of.

V

[ralph23]

This seems to happen to me all the time and it is soooo annoying....

Let's assume for the time being that partner and declarer each started with a doubleton ♣. That is more likely a priori than partner having a stiff and declarer having 3.

Let's also assume declarer has 6♥ (very likely) and the Ace of ♠. And the rest junk.

I can lead a ♣ at trick 3 and promote (I hope) a trump trick for partner, but that's only 3 tricks.

On the assumption that declarer has the ♠ Ace, then Declarer has ex hypothesi 4 pointed winners (AKQ♠ and A♦), and he will just discard his pointed loser on the 3rd round of clubs. Partner must trump because of the ♣ ten.

Partner gets a trump trick, but that's it. Unless partner started with ♥Kxx, I've not accomplished anything. And the same is true, even if partner led a stiff ♣... then declarer has only 4 pointed cards, and if he has the Ace of ♠, he has the rest. The 3rd club won't help, again, unless partner has either Kxx or Ax(x).

So using CPA (Card Placement by Assumption), I think partner needs to have the Ace of ♠ for us to have any chance here....

If he does have it, then why not just lead another♣ at trick 3 and let partner get his trump or trump promotion? Well, what if declarer started with

♠x
♥AKxxxx
♦Qxxx
♣xx

Due to the fortuitous placement of the ♦, declarer has no ♦ losers; the obligatory finesse procedure will drop partner's King on the second round of that suit. So on trick 3, declarer can just toss his ♠ loser, and partner wins the trick with a trump, but now his Ace of ♠ gets ruffed.

I think it's better to count on partner's having the ♠ Ace, and lead a high spade at trick 2, let partner win his Ace, partner returns a club to me at trick 3, trapping dummy's ten, then I lead the highest ♣ at trick 4. Declarer must trump high, or he is down. Hopefully this promotes a trump trick for partner.

Yes, if partner started with the stiff ♣Jack and the ace of ♠, and declarer has 1+♠, this defense will get a lot of laughs, but it still seems right... partner didn't want to lead away from his Ace of ♠, or his ♥Jack, or his ♦ honor... so top of the club doubleton seems reasonable.

This post has been edited by ralph23: 2007-August-14, 10:24

[ralph23]

Jlall, on Aug 14 2007, 11:17 AM, said:

Why do you think declarer has specifically a stiff spade?

I don't think (necessarily) declarer has a stiff ♠ at all. And I certainly didn't say that I thought he had such.

What I said was, I think partner must have the Ace of ♠ for us to have a chance to set this contract.

So I am giving partner that card by CPA.

Now, if partner has the Ace of ♠, we will always set the contract, yes?

Because I'll lead ♣ ♣ at tricks 2 and 3, and either (1) partner will trump round 3 while declarer helplessly follows suit, and partner will cash his winning Ace of ♠ at trick 4; or (2) round 3 of ♣ will work a trump promotion for partner, giving us 4 tricks in all: the first two ♣, the promoted trump in partner's hand, and the ace of ♠.

Perfect. What could go wrong???

Well, the only think that can wrong is, that declarer has a stiff ♠ and tosses it on the third round of ♣.

So I'm not guessing or thinking that declarer must have a stiff ♠.... I don't know what he has.... I'm assuming (CPA) that partner has the ♠Ace and then asking, what could go wrong in that case?

What's the most likely layout, i.e. which method has a greater chance of success?? I dunno... maybe it's more likely that partner led a singleton. But I think a doubleton ♣ is more likely....

[cherdano]

Ralph, your line wins over Justin's when partner has a club doubleton AND declarer has a spade singleton. Justin's line wins whenever partner has a club singleton. I don't think it's close that Justin's line wins more often, Jx is not an attractive lead AND AJxxxx is an neattractive 1S bid.

Of course it depends on partner, if partner is Justin your line will never win, other partners might be reluctant to overcall on AJxxxx and out AND like the lead of Jx. Still the odds favor club-club-club I think.

[vuroth]

ralph23, on Aug 14 2007, 11:53 AM, said:

What I said was, I think partner must have the Ace of ♠ for us to have a chance to set this contract.

If partner has Axx♥, K♦, then we can set, but only if we return a club. Ditto if partner had Kxx♥

I think I'm in over my head on this discussion, but I don't quite see why we absolutely need partner to have A♠

[skjaeran]

vuroth, on Aug 14 2007, 07:08 PM, said:

ralph23, on Aug 14 2007, 11:53 AM, said:

What I said was, I think partner must have the Ace of ♠ for us to have a chance to set this contract.

If partner has Axx♥, K♦, then we can set, but only if we return a club. Ditto if partner had Kxx♥

I think I'm in over my head on this discussion, but I don't quite see why we absolutely need partner to have A♠

Please remember this is MP, not IMPs.
It's not a given that our goal is to set the contract.
Our goal is to maximize our tricks; if we get three tricks and the majority only two, we've got a great score.

Our best chance to maximize our tricks is to continue ♣A and another ♣.

[SoTired]

Even if partner has the ♠A, if partner has a singleton club, we don't set the contract unless we return a club NOW. We have no entry. Declarer will set up diam or spade for club pitches.

This is a very basic and simple hand.

[Echognome]

People have made some excellent points. I'm now going to post a slightly different problem to see how your answers might change. (Hopefully that's not annoying.)

Scoring: IMP

1♦ - 1♥
1NT - 2♥
3♥ - 4♥
All Pass

---

With the change in the diamonds and the change in the scoring, do you defend any differently?

I will post the full deal at the table later.

[ralph23]

skaeran, on Aug 14 2007, 12:31 PM, said:

Please remember this is MP, not IMPs.
It's not a given that our goal is to set the contract.

The question posed, was "How do we take four tricks?"

So I ignored the form of scoring in the bidding box and assumed that the poster meant what he asked.

[cherdano]

Echognome, on Aug 14 2007, 11:44 AM, said:

With the change in the diamonds and the change in the scoring, do you defend any differently?

No.

[ralph23]

vuroth, on Aug 14 2007, 12:08 PM, said:

I think I'm in over my head on this discussion, but I don't quite see why we absolutely need partner to have A♠

Well, give declarer the Ace of ♠ and see what happens.....

Under what circumstances will you prevail?

Assumption 1: Partner and declarer each started with 2♣. Declarer has the Ace of ♠.

You will need for partner to have both diamond honors, and a top ♥.

Proof: Declarer started with 8 rounded cards (6♥, 2♣). He therefore has only 5 pointed cards. If those 5 pointed cards include a diamond honor, either K or Q, then he has zero pointed losers and 5 pointed winners. In that case, Declarer will just trump trick 3 with the ♥Jack.

If your partner overruffs with his K or A, it is the last trick you will get. Declarer remains with the A (or K, whichever) of trumps in his hand and the Queen of ♥ in dummy.

If your partner discards instead of overruffing, then declarer will just play a ♥ to the Queen, a ♥ to his ten, and you can take your large honor then. But you won't get any more tricks.

Assumption 2: Partner started with the stiff Jack of ♣. Declarer has the Ace of ♠.

Here, as noted earlier, you only need for partner to have a natural trump trick... either the A or K of ♥ (but not Kx of ♥). Partner will trump with his low ♥ and have his other trump to come.

Nothing else will matter, because under this Assumption, declarer has 9 rounded cards and only 4 pointed ones... and those include the ♠ Ace. He has therefore no pointed losers.

So, yes, there are layouts where we can succeed without the Ace of ♠. I don't know how likely they are, but they don't seem all that likely.....

[ralph23]

cherdano, on Aug 14 2007, 12:05 PM, said:

other partners might be reluctant to overcall on AJxxxx and out AND like the lead of Jx.

Why must partner have the Jack of ♠? Can't he have A65432?

[jdonn]

ralph23, on Aug 14 2007, 01:24 PM, said:

cherdano, on Aug 14 2007, 12:05 PM, said:

other partners might be reluctant to overcall on AJxxxx and out AND like the lead of Jx.

Why must partner have the Jack of ♠? Can't he have A65432?

Now you are just nitpicking because your answer is being strongly challenged, or are you saying your entire defense is based on the possibility that declarer has a doubleton club and the singleton jack of spades?

[ralph23]

jdonn, on Aug 14 2007, 01:27 PM, said:

ralph23, on Aug 14 2007, 01:24 PM, said:

cherdano, on Aug 14 2007, 12:05 PM, said:

other partners might be reluctant to overcall on AJxxxx and out AND like the lead of Jx.

Why must partner have the Jack of ♠? Can't he have A65432?

Now you are just nitpicking because your answer is being strongly challenged, or are you saying your entire defense is based on the possibility that declarer has a doubleton club and the singleton jack of spades?

No, I'm just wondering if declarer can have the Jack of ♠. Everyone somehow seems to know that partner has it. I don't know how that is known.... and you misunderstand the point.

It's really not that complex... but for the 3rd (and final) time ....

As I said earlier in 2 earlier posts, I don't know that declarer has a stiff spade. Let me repeat that: I don't know that declarer has a stiff spade.

I never said that I thought he did. Why would anyone believe such a thing? There's no evidence for it, is there ?

I also don't know that he has 2 spades. He might even have 3♠!

So we are agnostic about his number of spades.

But I do believe, for us to have a realistic chance of setting this, that partner must have the Ace of spades.

You can agree or disagree with that belief. It is certainly not an absolute certainty. But it is a belief.

Suppose by some miracle (I doubt it though) you actually agree with that belief, and go forward from there. Just pretend like you agree with it, maybe.

So I make the assumption -- Card Placement by Assumption, CPA -- that partner does have the Ace of ♠.

Then ask: Won't we always succeed if partner in fact has the ♠Ace? ♣ ♣ trump promotion, then partner wins his Ace at some point. Four tricks.

No, we will fail if declarer has a singleton spade and ♦Kxx, or ♦Qxxx, and AKxxxx of trumps.

Is there a downside? Yes of course, as pointed out in my initial post, if partner had a stiff Jack of ♣. But even if that is true, all is not lost... for consider that declarer might have this:

xx
AJT987
Qx
xxx

or even this

xxx
AJT987
K
xxx

Partner will get his two major suit tricks and declarer can pitch a club on a good spade or the ace of diamonds, yes; but he still has a club loser.

[jdonn]

It's like arguing with F..... never mind I won't say it lol.

[ralph23]

jdonn, on Aug 14 2007, 03:16 PM, said:

It's like arguing with F..... never mind I won't say it lol.

That's amazing, because I was just thinking the same thing vice versa !! lol

[ralph23]

FWIW, not that anyone is interested in these things, but trying to estimate the likelihood of a singleton or a doubleton lead with no other information except that declarer has 6♥ {and 7 open spaces} and partner 3♥ {and 10 open spaces}:

a. Split 2-2= 39.7% of the time
b. Split 3-1 with declarer having the 3 = 14.7% of the time
c. Split 3-1 with partner having the 3 = 35.3% of the time (more likely than b. a priori since declarer has so many ♥, but ruled out because partner would not lead the Jack from such a holding here, w/out the ten)
d. Split 4-0, one way or the other = 10.3%, but also ruled out on the facts.

Is it valid to ignore c. and d. and just take the ratio of a:b, which yields 73%?
(39.7 / [39.7+14.7])

Putting aside arguments that "partner would never lead from Jx....".

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