[mycroft]

I thought it was "is this the failure case for 1♠ DN?" Which is why I coded in the "obvious" pulls by the short hand. I will trust you on the variance w/active opposition; I believe it "IMHExperience", and am too lazy to work out a decent set of numbers. Having said that, the "results against par" go some way to "faking" opposition bidding.

[DavidKok]

I don't like the idea of 1♣-1♠ as a double negative and have been following the discussion up to the point where the simulations were presented. At that point I lost track of what the question was and how the simulations answer that, sorry.

We're limiting the simulation to hands where we give opener a 1♣, responder a double negative, possibly no interference by the opponents, and then which losing case on 1♣-(P)-1♠-(P); ?. And then we compare that against 1♣-(P)-1♦-(P); ? with which continuation system?

[mycroft]

I think they are arguing that after 1♣-1♠ DN, you're frequently forced into a 1NT rebid that scores badly, in ways that hurt more than the benefit of removing the DNs from the rest of the responses (1♦ full positive, others semi-, or however they style it).

The simulations came in with someone saying that 15-19 opposite a DN scores very badly in 1NT; I can't imagine that not being the case.

PrecisionL wondered about what they were talking about: 17-19 (I assume playing 14-16 1NT and 16+unbal, 17+bal Precision style), so I started coding. And then things got weird, because I know I'm never sitting for NT with a sane runout even on a yarborough, no matter how strong the NT hand is. So I kept coding. And ended up with (with no interference) going slightly plus in "the normal contract" on average, which frankly I was surprised at (yes, I know, 17-24 combined is basically 20=20. But "playing out of your hand" is traditionally very bad).

Maybe this is a side-track too far. If so, sorry.

[foobar]

mycroft, on 2024-October-13, 14:06, said:

I think they are arguing that after 1♣-1♠ DN, you're frequently forced into a 1NT rebid that scores badly, in ways that hurt more than the benefit of removing the DNs from the rest of the responses (1♦ full positive, others semi-, or however they style it).

That matches my understanding of the trajectory so far as well.

Edit: I think that there are two or three independent things being discussed.

1) The first is that 1C - 1S (DN) denies a cheap entry to RHO, which is self-evident.
2) By the same token, 1C - 1S (DN) denies us of two bids at the one-level (vis-a-vis Meckwell) if RHO doesn't bid. If RHO bids 1M over 1D, it's unclear that it's especially damaging to our side.
3) There's the question of 1C - 1S (DN) - 1N vs. 1C - 1D - 1N *if* RHO doesn't bid. The simulations seem to be targeting the last question, but I think that the two situations aren't exactly identical.

mycroft, on 2024-October-13, 14:06, said:

PrecisionL wondered about what they were talking about: 17-19 (I assume playing 14-16 1NT and 16+unbal, 17+bal Precision style), so I started coding. And then things got weird, because I know I'm never sitting for NT with a sane runout even on a yarborough, no matter how strong the NT hand is.

The one nuance is that 1C - 1D - 1N playing Meckwell is more likely to be balanced than say 1C - 1S (DN) - 1N, which might include more 5431 shapes for example. So, the resulting part score comparison might not match the former completely (in addition to the slightly lesser playing strength of 15+).

[DavidKok]

So if I understand correctly, the argument is that if 1♠ shows a DN, the start of the auction 1♣-1♠; 1NT scores poorly. To simulate this I'd like to better understand the following:

• Which hand types are included in the 1NT rebid, and what is responder's rebid structure over this? Earlier in the thread there was mention of Polish-style transfers, what are the exact requirements to bid or to pass?
  
• We are assuming that LHO has no overcall over 1♣ and RHO has no overcall over 1♠, what are the assumptions on their requirements to bid?
  
• I think we are comparing against a more standard strong 1♣ structure, right? Mecklite? Some other one? Either way, now we likely start with 1♣-1♦ (still conditioning on no interference by LHO) - do we further restrict RHO to not have an overcall here?
  
• Are we also interested in continuations of 1♣-1♠ DN where opener's rebid is not 1NT?

[foobar]

 DavidKok, on 2024-October-13, 15:24, said:

So if I understand correctly, the argument is that if 1♠ shows a DN, the start of the auction 1♣-1♠; 1NT scores poorly. To simulate this I'd like to better understand the following:

• Which hand types are included in the 1NT rebid, and what is responder's rebid structure over this? Earlier in the thread there was mention of Polish-style transfers, what are the exact requirements to bid or to pass?
  
• We are assuming that LHO has no overcall over 1♣ and RHO has no overcall over 1♠, what are the assumptions on their requirements to bid?
  
• I think we are comparing against a more standard strong 1♣ structure, right? Mecklite? Some other one? Either way, now we likely start with 1♣-1♦ (still conditioning on no interference by LHO) - do we further restrict RHO to not have an overcall here?
  
• Are we also interested in continuations of 1♣-1♠ DN where opener's rebid is not 1NT?

My guesses appear below and nullve or kwiktrix can add more nuance.

1) Punt to the 1S DN experts
2) Good suit and decent values for overcall
3) We should assume Meckwell structure over 1C - 1D. It's probably best to model no overcall by RHO because it's unclear that a 1M is particularly bad for our side. For example, over the example hand of AKxxx x AKxxx Kx, over 1C - 1D, (1H) actually makes it easier to get the 5-5 distribution across easily. The 1C - 1D - 1S (F1) - 1N (DN) - 2D would show 54 either way.
4) We should take it into account since we are trying to see how it affects partscores.

[PrecisionL]

foobar, on 2024-October-14, 19:16, said:

My guesses appear below and nullve or kwiktrix can add more nuance.

3) We should assume Meckwell structure over 1C - 1D. It's probably best to model no overcall by RHO because it's unclear that a 1M is particularly bad for our side. For example, over the example hand of AKxxx x AKxxx Kx, over 1C - 1D, (1H) actually makes it easier to get the 5-5 distribution across easily. The 1C - 1D - 1S (F1) - 1N (DN) - 2D would show 54 either way.
4) We should take it into account since we are trying to see how it affects partscores.

Interesting. Our Strong Club Rebid Design shows 5M and 4+ of unknown minor:

1♣ (16+) - 1♦ (0-7/8 hcp) - 2♠ (5+♠ & 4+♣/♦, N.F.) - 2NT (asking for minor) - 3♦ - ?

1♣ (16+) - 1♦ (0-7/8) {1♥} - 2♠ (5+♠ & 4+♣/♦, N.F.) - ?

[foobar]

 PrecisionL, on 2024-October-15, 10:09, said:

Interesting. Our Strong Club Rebid Design shows 5M and 4+ of unknown minor:

1♣ (16+) - 1♦ (0-7/8 hcp) - 2♠ (5+♠ & 4+♣/♦) - 2NT (asking for minor) - 3♦ - ?

1♣ (16+) - 1♦ (0-7/8) {1♥} - 2♠ (5+♠ & 4+♣/♦) - ?

Meckwell uses 1C - 1D - 2S for the GF-hands with 5♠+4m (with 1C - 1D - 2H as either GF with 5+♥+minor OR balanced GF). The canape GF goes via 1C - 1M (F1) - 3m.

How does your scheme work?

[PrecisionL]

foobar, on 2024-October-15, 11:57, said:

Meckwell uses 1C - 1D - 2S for the GF-hands with 5♠+4m (with 1C - 1D - 2H as either GF with 5+♥+minor OR balanced GF). The canape GF goes via 1C - 1M (F1) - 3m.

How does your scheme work?

Yes, I am aware of Meckwell's design.

We have been using a modified version of Johnson-Berkowitz's scheme where 1♣ - 1♦ - 1M = one round force:

1♥ = (a) 4+♥, or (b) 2+♥ and 20+ hcp, usually balanced.

1♠ = 0-4 hcp (No Ace) double negative,
1NT = 4+♠ & 5-7 hcp
2♣ = 5-7 and 0-2♥
2♦ = 5-7 and exactly 3♥
2♥ = 3-5 and 4♥, no singleton
2♠ = 4+♥, 5-7 and unknown singleton (2NT asks for the singleton)

1♠ = 4+♠ and forcing 1 round. This allows for a canape into a longer minor.

Similar follow-ons as above

[foobar]

PrecisionL, on 2024-October-15, 13:19, said:

Yes, I am aware of Meckwell's design.

We have been using a modified version of Johnson-Berkowitz's scheme where 1♣ - 1♦ - 1M = one round force:

1♥ = (a) 4+♥, or (b) 2+♥ and 20+ hcp, usually balanced.

1♠ = 0-4 hcp (No Ace) double negative,
1NT = 4+♠ & 5-7 hcp
2♣ = 5-7 and 0-2♥
2♦ = 5-7 and exactly 3♥
2♥ = 3-5 and 4♥, no singleton
2♠ = 4+♥, 5-7 and unknown singleton (2NT asks for the singleton)

1♠ = 4+♠ and forcing 1 round. This allows for a canape into a longer minor

Similar follow-ons as above

Folding the Kokish 2♥ into the 1♥ sounds like interesting idea. However, I am still unclear about the 1♣ - 1♦ - 2♥ / 2♠ bids. Are the immediate 2♥/2♠ showing 5+M and 4m GF?

Also, do you use the 1♣ - 1♦ - 2N for something else?

[PrecisionL]

foobar, on 2024-October-15, 13:50, said:

Folding the Kokish 2♥ into the 1♥ sounds like interesting idea. However, I am still unclear about the 1♣ - 1♦ - 2♥ / 2♠ bids. Are the immediate 2♥/2♠ showing 5+M and 4m GF?

Also, do you use the 1♣ - 1♦ - 2N for something else?

No, they are not GF. GF or GI hands rebid 1M over 1♦ response = forcing 1 Round unless Opener jumps with his 3rd bid.

1♣ - 1♦ - 2NT is minors 5-5 usually and 1 round force.