[Cyberyeti]

MrAce, on 2011-July-25, 14:07, said:

Anyway, i am with Schafer on this one and i bid 5 ♦, my reasons are;

- You said pd is asking us to evaluate our hand vs ♥ void, and my 1♦ opener dropped to 5 hcp with a stiff Ace.

- You say if all he needed was a cue he had other way to do it.

- You say 1♦ showed 5+ ♦ unless 4441 shapes, so pd knows we cant be 4432 and might have stretched himself.

- We already wrong sided this. Perhaps wrong sided even for 5♦. Idk.

- All i know is there are hands that slam is cold, and there are hands it has no play. I will give priority to game since i have doubts. I just could not find a way to learn if we are lack of keycards without a RKCB, and if i RKCB i would not know how many ♠ we lose.

There is a very good chance that your ♥AK are not waste paper. If you cue 5♣, partner knows the spades are open, if he has the ace, your hearts are probably working. You need as little as Axx, void, Axxxx, xxxxx to make the slam good even if trumps don't split, KQxx, void, AKxx, xxxxx is fine too. If partner bids 6 it's probably with the odds, I'm more worried about going off in 5. The only advantage to 5♦ is not telegraphing the spade lead.

Partner is unlikely to have 5 spades as he might have bid them, so spade losers figure to go west on the hearts as long as he has control, if 3♠/4♣ are fit jumps over 2♥, the odds of partner having 5 diamonds (or 5 bad clubs insufficient for a FJ) improves and both of these are good news.

[mikeh]

Lump me in with the 5♣ contingent.

The hand is way too strong to consider 5♦, even tho we may already be doomed to a minus score. Doesn't that say something about this game of ours....consider partner with Kxx void AKQxx KQJxx and we rate to go down on a spade lead. Then place him with Axx void AKxxx xxxxx and we rate to make 13 tricks....and here we are, guessing

Since we are forced to 5♦ anyway and really need little more than Axx in spades to make slam, we have to make a move and we have to tell partner that our problem is in spades.

Yes, by doing so, we virtually ensure they make the best lead, but it's not like 4N is a panacea. What if partner bids 5♠ and gets doubled? And maybe West has a natural lead anyway.

Maybe we can't reach grand after 5♣, but were we ever reaching it with confidence after 4N, when partner could have Axxx void AKxxx Kxxx?

Meanwhile, by bidding 5♣ we avoid slam when partner has no spade control....and to me that is more important than worrying about whether we should bid 7. Btw, I worry about the Axxx in spades because the OP announced that partner might choose to suppress spades....and to me that would imply spades with no internal texture....probably with chumky clubs. After all Axxx void AQxxx KQJxx wants to play in diamonds, not spades, opposite Kxxx Axx KJxxx x.


Otoh, since partner rates to hold no more than 3 spades, I think the 'wasted values' thread name was a good piece of misdirection This hand is, so long as partner owns a spade control, extremely good, with zero wastage that I can see.

[jmcw]

Can partner really have a hand that is wide open in ♣?, as some have suggested. If you say yes, then I would like to know why he isn't just as likely to be wide open in ♠?. I say responder must have 2nd round control of ♣(with length) for his 4♥ bid to make sense. Similarly, responder needs to have a control in ♠ for slam to have a legitimate chance. In this scenerio, I expect partner's ♣ length to provide a parking place for my losing ♠'s, failing that the AK♥ might let me shake his losing ♠.

Practically, I would commit this hand to slam, I just don't see an intelligent way to stop short. I'm bidding 4NT and going to grand if he shows 3 keys.

[Phil]

aguahombre, on 2011-July-25, 15:21, said:

I don't understand. Don't we usually show them in the normal order? Ok, so the second one must have the diamond queen because of 13 cards.

I have never held a trump fit of:

ATxxxx

across from

J8xxxx

[cherdano]

I am more pessimistic than others. I think partner is quite likely to have 4 diamonds only - it is much more likely than 5-card support a priori, and with a void in preemptor's suit and 4-card support for the unbalanced diamond I think he should splinter often. Also, any spade finesse we may need is likely to be off. 5♣ seems ok but I would rather bid 5♦ than commit to slam.

[y66]

Not wasted opposite Axx --- KQxxx Kxxxx. 5C for me.

[jschafer]

Does 5422 qualify as 'unbalanced' for opening 1D?

[benlessard]

The more i think about it the more im convinced 5C (asking for a S control) is an awful bid.

Even if you can respond to keycard with 4S (showing 1 keycard and not total garbage) i still think 5D is better bid. Hoping partner got only 3 S and 2 keycards is a narrow target.

[dake50]

4H shows H-void splinter and no other control?
My splinter has H- D:Kxxx+ AND an A.
Otherwise he starts 2H Q-bid, then repeats.
My question is what controls/D-honors separate
lower Q-bid and repeat from 4H?

[mikeh]

cherdano, on 2011-July-25, 19:04, said:

I am more pessimistic than others. I think partner is quite likely to have 4 diamonds only - it is much more likely than 5-card support a priori, and with a void in preemptor's suit and 4-card support for the unbalanced diamond I think he should splinter often. Also, any spade finesse we may need is likely to be off. 5♣ seems ok but I would rather bid 5♦ than commit to slam.

a priori don't enter into it

partner has a heart void, which, a priori, is a rare occurence. He also usually lacks 4 spades and he's not trying to show us clubs. Add to this that we might be 4=4=4=1 or 1=4=4=4, and it seems odd that he'd be making a slam try with only 4 diamonds.

I agree that a 4 card suit isn't impossible....but it must surely be odds against.

[aguahombre]

benlessard, on 2011-July-25, 21:37, said:

Hoping partner got only 3 S and 2 keycards is a narrow target.

Unless your style precludes partner having more than 3 spades.

It is a given that pard has strength, so having some keys is not a big stretch.

[gnasher]

jschafer, on 2011-July-25, 20:28, said:

Does 5422 qualify as 'unbalanced' for opening 1D?

Yes, usually. With bad diamonds and two doubleton honours we might open 1♣.

[wank]

5♦ - partner knows we opened the bidding. he can still bid slam if he likes.

people placing partner with Axx of spades are insanely optimistic

[Cyberyeti]

benlessard, on 2011-July-25, 21:37, said:

The more i think about it the more im convinced 5C (asking for a S control) is an awful bid.

Even if you can respond to keycard with 4S (showing 1 keycard and not total garbage) i still think 5D is better bid. Hoping partner got only 3 S and 2 keycards is a narrow target.

Partner doesn't need to have only 3 spades or 2 key cards (KQxx, void, AQxxxx, xxx or even AKQx, void, Qxxxx, xxxx is 50/50 ).

Axxx, void, Axxxx, xxxx is playable, and that's pretty weak for the splinter, either ♣KQ, ♠KQ, both black Ks or K♦ make it laydown on top of this. Your opps have bid a vulnerable 2♥ with a suit no better than QJ9... even if partner only has K♠ without the Q, there is a decent chance the ace is onside anyway. If you're wide open in spades, you're wide open in spades and probably not making 5, so 5♣ doesn't cost.

[cherdano]

mikeh, on 2011-July-25, 21:43, said:

a priori don't enter into it

Just because you are unable to use a priori and conditional probabilities correctly doesn't meant that the rest of us are not allowed to use them.

[aguahombre]

mikeh, on 2011-July-25, 21:43, said:

a priori don't enter into it

partner has a heart void, which, a priori, is a rare occurence. He also usually lacks 4 spades and he's not trying to show us clubs. Add to this that we might be 4=4=4=1 or 1=4=4=4, and it seems odd that he'd be making a slam try with only 4 diamonds.

I agree that a 4 card suit isn't impossible....but it must surely be odds against.

Mikeh did not include 4-4 in the majors with only 3 diamonds for opener. For those of us who have that as a possibility, a four-card diamond fit by responder is more than odds against. Add a neg double with 4 spades to the mix, and this debate becomes almost totally dependent upon opening bid styles; the 5♦ advocates are probably right if 1♦ had unbalanced implications, and the slammers are probably right if Walshish.

[mikeh]

cherdano, on 2011-July-26, 09:15, said:

Just because you are unable to use a priori and conditional probabilities correctly doesn't meant that the rest of us are not allowed to use them.

Ok....it is entirely possible that I am mistaken. It was my understanding that a priori, in this context, referred to the assumptions one would make before experience modified one's view. On that basis, if we were to look at our hand and ask how likely it is that any raise by partner would be be based on 4 card support, rather than 5, it seems to me pretty logical that one would expect 4 more than 5. After all, we hold 6 of them, and we promise 4+, so partner will readily raise on 4. Alternatively and perhaps more accurately, in a technical sense, one would merely assume that one opened 1♦, with the posted systemic constraints, and then asked, without reference to the auction or even to our hand, how many diamonds partner will hold for a raise....will it be 4 more frequently than it would be 5.

However, at the time of decision, on round 2, we have gained experience, in the former interpretation, from the auction, and in the latter from looking at our hand and the auction, and we can use a posteriori reasoning, which, to me, suggests that the chances are now considerably higher, than we would have thought a priori, that he has 5.

I apologize if I have misused the terminology. I claim no expertise in philosophy. Anyway, I am sure that you will show me the error of my ways if I am mistaken

Anyway, the new and (slightly) improved mikeh doesn't fight anymore......tho I can debate, I trust...

[han]

Until we find a linguist, a philosopher and a summarizer I will read mikeh's debates from the sideline. While I do so, I will assume that cherdano meant that given our hand, there are more hands with a heart void and a 4-card diamond suit than hands with a heart void and a 5+ diamond suit. That's the a priori part. The way I read his post, he then reasoned that given our unbalanced 1D opening, partner would bid 4H often enough with the hands that hold only 4 diamonds so that 4H is still often based on only 4 diamonds.

Throughout the years I've noticed that if you start with the assumption that cherdano knows what he's talking about, his posts make a surprising amount of sense.

Quote

people placing partner with Axx of spades are insanely optimistic

Trying to guess what partner has in spades instead of letting partner speak for himself is needlessly needless. <will edit this when I find a better comeback>

[OleBerg]

At the table nothing but 5♦ would enter my mind.

Having followed this thread got me to think a little, but not enough to change my mind.

[hrothgar]

mikeh, on 2011-July-25, 21:43, said:

I agree that a 4 card suit isn't impossible....but it must surely be odds against.

FWIW, I threw together a simple MATLAB script to check on the expected number of Diamonds in partner's hand.
This script assumes that the Spade length in partner's hand is <= 3 (which obviously ignores suit quality)


Estimated Diamond length
Four = 0.5792
Five = 0.3349
Six = 0.0794
Seven = 0.0065

MATLAB Code

%% Shape

Shape(1:12,1) = 'C';
Shape(13:19) = 'D';
Shape(20:29) = 'H';
Shape(30:39) = 'S';


%% shuffle

simlength = 10000000

MC_Result = zeros(simlength, 4);

for i = 1:simlength
    
    index = randperm(39);
    Shape = Shape(index);    
    foo = Shape(1:13,1);
    
    MC_Result(i, 1) = length(foo(foo == 'C'));
    MC_Result(i, 2) = length(foo(foo == 'D'));
    MC_Result(i, 3) = length(foo(foo == 'H'));
    MC_Result(i, 4) = length(foo(foo == 'S'));
    
end

Voids = MC_Result(:,3) == 0;
MC_Result2 = MC_Result(Voids,:);

Spades = MC_Result2(:,4) <= 4;
MC_Result3 = MC_Result2(Spades, :);

index = MC_Result3(:,2) >=4;
Raises = MC_Result3(index,:);

Four = length(Raises(Raises(:,2) == 4))/length(Raises)
Five = length(Raises(Raises(:,2) == 5))/length(Raises)
Six = length(Raises(Raises(:,2) == 6))/length(Raises)
Seven = length(Raises(Raises(:,2) == 7))/length(Raises)