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All comments by Craig Zastera
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If East's T were a stiff (say he had one more and one fewer ), a shift at trick 4 by West would not defeat the contract assuming the same major suit honor lay-out.

Nor would leading the 4th succeed.

No, if this were the lay-out, only a shift from West at trick 4 would defeat 2.

After a shift is ruffed by East, say he exits with K. As long as South wins and plays J he succeeds. If East instead exits a low , declarer must play J to succeed.
Nov. 12, 2018
Craig Zastera edited this comment Nov. 12, 2018
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(a) Don't see why the discard on 3rd was bad.
East expects partner to continue with 4th on which
he can discard his other , ensuring set too I think.
Not sure *which* East should discard given their
signalling methods. I would discard T count, but
I play “standard” count signals, so perhaps with UDCA
the 2 should be the first discard.

(b) Once West shifts to a instead of playing 4th
(or a ), it is no longer possible to defeat 2
if declarer plays correctly.
It doesn't matter what East does on the 2nd high
from dummy–ruff high, discard, or ruff low–
declarer can make 2 in all cases.

© You are right that South's low exit did give the
defense a second chance–if West wins T and leads
a , the defense can prevail.
Nov. 12, 2018
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Chris,
I'm not sure I understand your comment–possibly you did not say exactly what you meant.

According to OP, West played a at trick *4* (not trick 3 as you say).

And a shift at trick 4 is by no means necessary to defeat 2, although it would be OK.

A 4th round of s (on which East must pitch her other ) also suffices to defeat 2.

Interestingly, both a opening lead or a switch at trick 2 would lead to defeating 2.

But a shift at trick 3 (after cashing two s) would NOT work!

Once West has cashed *2* rounds of s, he *MUST* continue with a 3rd (high) in order to defeat 2.

However, after cashing the 3rd round of s, the defense can prevail either by a 4th (provided East pitched a on the 3rd round) *or* by a switch.
Nov. 12, 2018
Craig Zastera edited this comment Nov. 12, 2018
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West should continue 6 (East pitching last ), so that after declarer ruffs and leads a to dummy's A, EAST will now have the opportunity to err by ruffing and exiting a *low* .
Nov. 11, 2018
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Partner *should* be at least 5=5 in the pointed suit for his 3 level responsive double. This is based on LOTT thinking:

1 overcaller shouldn't be supposed to hold a 4 card side suit (though, of course, that is not impossible).
Thus, to compete 3 over 3 without support, doubler ought to be at least 5=5 so he can reasonably *hope* for an (at least) 8 card fit or enough length (and hand strength) to sit for (3X).

How strong should doubler be? When partner has made a *2* level overcall, strength for doubler isn't as crucial as overcaller has promised more. So in that case, 6-7 HCP is probably an adequate minimum. The 5=5 shape, though is crucial.

But when partner has made a mere 1-level overcall, I think a 3 level responsive double should require a bit more strength, 9-10 HCPs.

But here he is a passed hand, so his range is narrow.
KQxxx-x-KQxxx-xx is probably an opening bid, so I'd figure not that good. Maybe KJxxx-x-KQxxx-xx would be the best we could hope for and perhaps a point (or 2) less.

It is debatable whether it is better for doubler to have a stiff and two s or the other way around.
The former allows for a perhaps playable 3 when overcaller doesn't fancy either pointed suit.
But the latter allows for a more likely to be successful pass of (3X) in such cases.
I'm in the camp of preferring the responsive double to be better able to stand a penalty pass (so 5=1=5=2 preferred).

If doubler isn't 5=5 (sadly, real world partners all too frequently are not), he is more likely to have 5 s and only 4 s. So if I'm going to bid, 3 seems better than 3 (also, our s are better and we can run to 3 in an emergency).

But our hand looks pretty good for defense. We have the top s, so 2 s and a ruff seems possible. Then a trick from partner on the side and our A beats (3X).
Why, we might even beat it 2.

On the other hand, we might easily be down in 3, losing perhaps a and two tricks in each pointed suit.

Seems close to me between 3 and PASS. If they were vul, I'd pass and try for +200. But here, +100 might lose badly to +110 our way, so I think I'd risk 3 if I can trust partner to have the shape he should.
And, who knows? Maybe they will bid again over 3.
Nov. 11, 2018
Craig Zastera edited this comment Nov. 11, 2018
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Tom,
I think it was Jeff Rubens who wrote an article a long time ago contrasting transfer methods (e.g. Rubensohl or transfer lebensohl) with “good/bad” methods (e.g. lebensohl).

The transfer methods solve the problem of suit ambiguity by transferring into the suit. So suit identity is known immediately.

But what is lost is any strength distinctions.

Responder (to the transfer) will occasionally have the fit and strength to do something other than just accept the transfer at the lowest level, but often he will not.

After a simple acceptance, the transfer bidder will be in the dark about partner's strength and degree of fit, so will often have to guess as to whether just to pass or to pursue a higher contract.

So I think choice between “transfer methods” vs. “good/bad” methods has to be made on a case-by-case basis (i.e. determined for each specific auction).

I tend to use a lot of “good/bad” in cases where game is a possibility for our side because then having ways to show different strengths unambiguously is crucial.

And I find that opponents often do not compete further over the “suit ambiguous 2NT” bids, so this theoretical problem with such methods often does not manifest in practice.
It is worth considering, though, whether 2NT is best used for “bad” (traditional) or “good” hands.

I do use a “transfer lebensohl” method when our 1NT openings are overcalled at the 2 level. The transfers always promise “game invitational or better” strength.
With lesser hands (but judged still worth competing to the 3 level), I use a suit-ambiguous 2NT.

But this auction type has a special property which favors the transfer methods (over straight “good/bad”). That is that opener's hand strength is known +/- one point, and his shapes (balanced) are also fairly narrowly defined.

Thus, responder can usually accurately judge from his hand whether he is “just competing at 3 level” vs. “game-invitational strengh” vs. “game-forcing strength”.
So a transfer method revealing suit and some strength info immediately works well.

But in other competitive auction types where there is more uncertainty about one or both partners' hand strength, a “good/bad” method to immediately reveal strength information may be more useful than a transfer method which obscures strength information in favor of immediate suit revelation.
Nov. 11, 2018
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They show that the improved spot cards have a significant effect on the probability that 12 tricks are makeable–goes from 43.19% to 51.51%.

That is a big change. It suggests that comparative analysis between the two versions (bad spots vs. good ones) should perhaps be done to account for what cases lead to the 8.5% improvement in the chances of making 12 tricks.

The large increase *might* suggest an alternative line with the better spots. Or, maybe not–perhaps the extra chances in s do not result in another line (besides playing s from the top first) becoming better.

The East hand type Michael suggests (4-1-Hxxxx-QJT) in no way account for these increases.
That particular East hand will occur on only about one deal in 10,000 (actually, one in 9,268). So that would only increase the percentages by perhaps 0.01%.

And I disagree about the value of simulations here.
First, they set an *upper bound* on what is possible.
For example, knowing that *double dummy* it is only possible to make 12+ tricks 44.7% of the time will spare you from trying to find a line that is better than that.

Also, if you find a line that makes 12+ tricks 36% of the time (about what the AKx line scores), then you have only to ask yourself under what conditions that line will fail.
If you can account for the missing 9% (or so), you will then have a good handle on all the possibilities and be able to consider whether any alternative line might be better (perhaps you can find a way to succeed on some of the missing 9% without sacrificing much from the 36% you already succeed).
Nov. 11, 2018
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I agree. Considering alternatives is good even if the conclusion is to reject the considered alternative.
Nov. 11, 2018
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Seems to me that I have bid my hand already with vulnerable 2 overcall. In my view, I am relatively minimum for this action.
Therefore, I think it is up to partner to compete further if he deems that appropriate.

For me to bid again over 3 strikes me as “bidding the same cards twice.”
Nov. 11, 2018
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Michael,
The original problem gives the N/S hands only the smallest spots (2,3,4 with North and 5, 6 with South).

Therefore, N/S can have no threats based on spot card value (only on length).

In this OP case, 7NT cannot be made when East has your suggested 4=1=Hxxxx-QJT.

However, one of the replies changed the N/S holdings to
give North : K752 and South :A98.
In *that* case, 7NT is indeed makeable when East holds
4=1=Hxxxx-QJT.

All my analysis is based on the OP problem lay-out where N/S has only the smallest spots.

I should note that using the “improved” spots changes the simulation percentages significantly to:
13 tricks: 1.68% 12 tricks: 51.51% 11 tricks: 46.81%
Nov. 11, 2018
Craig Zastera edited this comment Nov. 11, 2018
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Tom,
The only problem with your line for 13 tricks is that the given conditions specify T lead with, presumably, East following, so that shape is known not to exist.
Nov. 11, 2018
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Be careful in these analyses.

Just because there has to be some sort of a squeeze after ducking a trick in each minor, that does not mean that you can play the cards in such a way as to actualize all possible squeezes without seeing the E/W hands and knowing which to play for.

The problems that may occur have to do with perhaps needing to decide what to pitch (say, from the North hand on the 3rd ) and/or cashing winners in the proper order to avoid blocking a suit when the correct choice depends on the as yet uncertain lay-out of E/W cards.

It is not clear to me that you can describe in detail (showing what you pitch and exact order of play for each circumstance) a single-dummy line that will in fact always succeed in making 11 tricks.
Nov. 10, 2018
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Here are some interesting double-dummy statistics from a 50,000 deal simulation of these two hands vs. random E/W hands:
1. 7NT is makeable on 1.590% of the deals
2. 6NT is makeable on 43.190% of the deals
3. 5NT is makeable on 55.220% of the deals

You will notice that these add to 100%.

Therefore, it is *ALWAYS* possible to make at least 11 tricks with perfect (double-dummy) declarer play.

So one interesting problem is to find the best single dummy line to make at least 11 tricks. How close can you come to the double-dummy 100% without seeing the E/W cards?

A second problem is how to make 13 tricks.

The only possibility that occurs to me is to drop the doubleton :QJ.
The theoretical probability of finding :QJ in either hand is 3.2298%.

After that, one still needs either 3=3 spades or the same defensive hand holding 4+ cards in both black suits.
The theoretical probability of one of these black suit lay-outs is (assuming 4-2 diamonds and each defender having at least one heart): 47.05%

Thus, the combined probability of :QJ and either 3=3 spades or a black suit squeeze is: 3.2298% * 47.05% = 1.52%

Comparing that to the 1.59% from the 50,000 deal simulation where 13 tricks were makeable is strong evidence that this is, in fact, the only way that 13 tricks are possible.

This analysis of how 13 tricks can be made also sheds light on the liklihood of making 12 tricks by following a similar line.

That is, one could first try to make at least *3 tricks* in s (usually losing one).
Then, if that works, play for 3=3 s or a / squeeze to produce the 12th trick.

s can be played for 3+ winners about 69% of the time.
If we assume that 3=3 or black suit squeeze is roughly the same 47% as calculated above,
this would suggest that this line would make 12 tricks roughly 1/3 (33%) of the time.

Since the simulation showed that even using double-dummy declarer play to make 12+ tricks on every hand where it is possible gives less than 45%, this suggests to me that this is probably the best line for 12 tricks.
But I have not done a calculation (yet) for alternative line (e.g. ducking a ), so I'm not positive.
Nov. 10, 2018
Craig Zastera edited this comment Nov. 11, 2018
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Richard,
I'm not sure I buy your “s/he had no idea what responder's intentions were.”

To me, the 2NT response shows interest in playing at least game in opener's suit, perhaps only if opener has a good hand. So if opener has the good hand with the A, I do not know why s/he would fear to show it at the 4 level, still below 4.

Now one reason would be if partner is in the habit of “psyching” the 2NT response. To me, that means making that call on a hand that does not have serious game intentions.

I'm not commenting on whether I think such “operating” is good tactics or not, but I absolutely think that if your partner uses this tactic, and you know s/he does (either by agreement or because you've seen it before), that your alert and explanation to the opponents needs to be more than just “asking for a side A/K”, but should necessarily include “partner may make this call with some support and little/no game interest as a tactical call.”

In my partnerships, a 2N inquiry response over a weak 2 (actually, we use the cheapest bid for this purpose) is always “serious.”
With such an expectation, I would not hesitate to show my A (actually, we use Ogust not “feature ask”) with a good 2 opener including that card even if I had to show it at the 4 level in competition.

So my thinking is that opener's 4 need not show great length as long as the partnership treats the 2NT inquiry as showing serious game interest–in that case, 4 is just showing the A and confirming a good enough hand to accept responder's (at least) game invite.

The 4 call would need to be based on unusually long s (5?) only if opener believed responder's 2NT call might have been “frivolous.”
Nov. 9, 2018
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I was thinking of some similar remark.

I have long played “DONT” vs. strong NT.

From the voting, one would conclude that this is not considered one of the better conventional alternatives.

Yet, I get overwhelmingly more good results using DONT than bad ones, even though I “stretch” the limits of soundness routinely, e.g. showing a 2-suiter with some 4441 pattern and guessing well as to which 2 suits to admit to.

Perhaps the moral is just to “get in there” over their strong 1NT openers as much as possible with whatever tools you have available.
Nov. 8, 2018
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Calling opening 1 “completely normal” is an overbid.

I think West hand is a rather close decision between 1 and 1.

A lot depends on your “relative loathing” between:
(a) opening 1 then rebidding 2 on only a 5-bagger
vs.
(b) opening 1 then rebidding 2 with 4=5 minors
after partner's likely 1 response.

Personally, I think I would choose 1, planning to rebid 2 over 1, but I would not object to the 1 then 2 strategy which has the advantage of describing more about opener's hand at the expense of some ambiguity about relative minor suit lengths.

The 1 then 2 strategy also suffers from strongly suggesting 6+ s when opener has only 5 while also hiding the 2 suited nature of the hand.

Hmm, I'm almost talking myself into preferring 1 :-).

In any event, on the OP question, I said “all West” because I think West's double of (4) is awful.
One could debate whether West should have passed or bid 4NT, but there is certainly nothing about this hand to suggest doubling (4).
Nov. 8, 2018
Craig Zastera edited this comment Nov. 8, 2018
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In US I think that most play that 2 would be NF.

Usual agreement after 1X-(DBL)-?? is that new suit by responder at the 1 level is forcing but that at the 2 level it is NF.

Of course, after 1M-(X)-, some play transfer responses, so conceivably responder might be able to bid 2 here as a transfer and then follow-up in some reasonable way if the partnership had good agreements about continuations.
Nov. 8, 2018
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Why?
I agree that the 2NT bid is bizarre, hence I would not consider it.

But presuming that 2NT is supposed to ask for a side A or K as OP stated, then if opener has such, why wouldn't he show it? Presumably, his partner is interested in playing at least game if opener has said side card.

Thus, I think there is no inference whatsoever that North has length (much less 5) in s. Just the A.
Nov. 7, 2018
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I doubt if many would respond 1 with that (I wouldn't).

You could change it to Qxxx-K-JTxxx-xxx to overcome that objection.

In that case, responder would be likely to pass a 2 rebid with 4 a reasonable contract (although it is certainly not a lock).
So what?
He would also pass 3 or 3 I suspect.
Not clear what he would do over a 3 (overbid) jump shift.
Perhaps 3, as I would regard 4 as showing slam interest, hence not appropriate.
Nov. 7, 2018
Craig Zastera edited this comment Nov. 7, 2018
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I like the idea of winning (K) and shifting, but it is not clear to me why shift is best.

Not saying I think is wrong, just that I can imagine either a or a being right and I do not see a way to know which shift is best.

I expect partner to have about 13 HCPs, hence 12 or so outside s, likely scattered.

One example to show that a return doesn't have to be right:
Give partner: JTxx-KTxx-KT-AQx (certainly a possible hand)
Now after J-Q-K-x at trick 1, the defense can hold declarer to 4 tricks *if South returns any card EXCEPT A *.

Or, an example where the defense can score *10* tricks vs. 1NT, but only if South switches to specifically a after winning K:
North holds: JT-K9xx-AQx-KJxx

Don't want to suggest a shift is never good.
For example, if partner has: Jx-KQT9x-KQxx-Kx,
then after winning 1st , a shift is necessary to beat 1NT.
Of course, partner probably wouldn't have led J from that hand.

A more plausible example where a shift is good is when partner has:
JTx-QTx-KQxx-AJx
Now, after winning 1st , a is the only shift to beat declarer 3 tricks
(although it would be OK to continue A before shifting to *IF* partner drops his T under the A).

To support the idea of a trick 2 *shift* rather than a continuation, give partner:
Jx-AQTx-T8xx-AJx
Here, after winning K at trick 1, *any* shift will defeat 1NT but any continuation will allow declarer to make 1NT.
Nov. 7, 2018
Craig Zastera edited this comment Nov. 7, 2018
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