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All comments by Craig Zastera
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Why is it “at best moot”?
Points are at best evenly divided. So unless you are arguing against the LOTT, a call that will likely result in our playing at the 3 level without sufficient “total trump” is likely to be a losing action.
Nov. 13, 2018
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you are correct. My bare-bones example does not produce a 50% play for 4.

On the other hand, I actually expect a somewhat better hand than this for partner's negative double.

Still, I suspect that 3 is probably the “percentage” action, which is why I think it would be my choice at matchpoints.

We all know that at IMPs, though, it pays to bid games aggressively. Even when these games turn out not to be percentage, the result is a “push” at -1 or even -2.
Nov. 13, 2018
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Why does our side have to hold at least half the deck?
I mean, we could,…. but also East could have 13 and West up to 11, so we certainly don't have to hold half the deck.

Another inference is that partner knows (2) is terminal for their side. So if he had anything (particularly with fewer than 3 s), he likely would have done something (e.g. double) over (2) since he would know that we might be in a tough position in pass out seat with some HCPs but no shape.
Nov. 12, 2018
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But playing a 4th will fail on my example
(East has K8xx-A9xxxx-T-98).
Must shift to a to succeed.
Nov. 12, 2018
Craig Zastera edited this comment Nov. 13, 2018
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Here is a possible lay-out that absolutely refutes any suggestion of “worst play ever.”

Suppose East's hand were:
K8xx-A9xxxx-T-98
Leaving declarer with:
AQJ9xx-KJ-xx-JT7

I believe this lay-out is reasonably consistent with the bidding.

After West starts with K, A, the *ONLY* defense that will beat 2 is:
1. Continue with Q on which East *must* pitch T.
2. Switch to a to give East a ruff

That's four tricks for the defense with the A and K still to come.

Since this is a possible lay-out for which West's (and East's) plays including trick 4 swtich constitute the only winning defense (after first two s have been cashed), West's switch cannot be the “worst play ever” as to contend for that title would require at the least a play that could not possibly be needed to defeat the contract on any conceivable lay-out.
Nov. 12, 2018
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I agree with your assessment of “close.”

I based my aggressive 4 choice more on the form of scoring–at matchpoints I would more likely choose 3.

I expect partner to hold a decent hand for a double at this level.

It doesn't take much to make 4 a reasonable spot.
Say: xx-xxx-Kxxx-Axxx.
Nov. 12, 2018
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Your “doesn't come up enough” is, if true, an argument for playing this double as something other than responsive (for example, it could be used to show s so that 3 would show a good raise and 3 a lesser raise as a supplement to a “transfer advance” scheme).

For a responsive double at the 3 level (as here) to be sound, it must conform to LOTT analysis. It is clear that our side can't have a great HCP majority (if we even have our half), hence we must have sufficient trump length to justify 3 level competition.

If (responsive) doubler is less than 5=5, it becomes unlikely that our side possesses a sufficiently long trump suit to justify 3 level competition. Even when he is 5=5
(say 5=2=5=1 in OP example), the responsive double may not work out (if overcaller is e.g. 2=5=2=4), but at least when he is 5=5 there is a reasonable chance that his side will have at least an 8 card fit somewhere.

Long ago Jeff Rubens also observed that hands truly suitable for a responsive double after partner has *overcalled* (as constrasted to when he has doubled for take-out) do not occur very often. This is particularly true at the 3 levl.

For that reason, Rubens suggested an alternative use for these doubles–“cue-bid doubles”–so that after e.g.
(1)-1M-(3)- a “double” would be used to show a good raise of overcaller's major while 3M would be a weaker raise.
Opportunities for a “cue-bid double” occur relatively frequently, perform a very useful function (allowing advancer to distinguish invitational raises from competitive ones), and conform to “LOTT” thinking in that they are used in cases where a suitable fit is known to exist.

So if you believe that hands suitable for responsive doubles after partner overcalls occur rarely (and may not work well even when they do), you might consider adopting the “cue-bid double” treatment as a more useful alternative.
Nov. 12, 2018
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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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