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Kit, indeed, your comparison of 2♠ vs. 2NT , i.e. natural bidding vs. relays, is most impressive. That is what I have been trying to understand over the years. For unbalanced hands, I think I have a good grasp of when to relay and when to bid naturally. But for balanced hands, I always believed that relays are better - until now. You convinced me that 2♠ and the subsequent 5NT were the best calls.

I have a couple questions, maybe you can clarify. When opener bids 2♥, he can have:

3 cases for 43♥33 3 cases for 443♥2 2 cases for 5♣3♥32 2 cases for 5♦3♥32

So there are 10 possible distributions. If you use 2NT to relay, there are only 8 sequences leading to 3NT. Am I missing something?

What if responder has a different shape, say 2=4=4=3? Do you still bid 3♦? If you do, you may miss an 8-card club fit?

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How about taking the first-round spade finesse at T2?

If it wins, and spades are not 4-0, lay down ♠K and test 3 rounds of diamonds. If diamonds fail to break 3-2, go back to dummy with ♠A, and take the club finesse.

Compared to Steve's line at the table, this line gains when East has ♠Qxx and 3 diamonds exactly; and it loses when West has singleton ♠Q and ♣Q is on side or when West has something like 1-7-4-1.

Another variation is: after the first-round spade finesse, you test diamonds immediately before laying down ♠K. This has the advantage that it makes whenever diamonds are 3-2 and East has ♠Qx or ♠Qxx. However, it loses to the line above when diamonds are 4-1 and ♣Q is on.

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In the 1♦-1♠ structure, I use 2-step sequences for both minimum and invitational strengths, as described earlier. And for the 1♦-1♥ structure, the immediate transfer for me is an invitational 3-card raise; while the 2-step is the 3-card raise with minimum strength. When comparing two 1♦-1♥ structures, I believe yours is probably better. I need to switch them. :) Is your 1♦-1♠ same as mine?

I began to think about this last year when a friend of mine asked me to play this unbalanced diamond/balanced club system with him. I liked the transfer rebids in unbalanced 1♦ result so much that I pushed the idea to balanced 1♣ and 1M openings.

In 1♣ opening where transfer responses are used, if you let opener's acceptance of the transfer describe the weak NT hand with less than 4-card support, you also free up the 1NT rebid to serve as the starting point for transfers, as in unbalanced diamond. As a result, you have the same benefits: 2 ways to describe 3-card supports and no need to use reverse Flannery.

Some players prefer to use the 1NT rebid over the transfer responses as strong balanced hands (say 18-20). I found that you could incorporate that into the acceptance of the transfer. That is, 1♣-1♦-1♥ is now 12-14 or 18-20, balanced without 4 hearts. The level is low enough to accommodate dual meanings.

I believe I first read about the idea of transfer rebids was Feldman-Bramley in Challenge The Champs many years ago (not in this unbalanced setting though).

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Yuan- “The simple raise would then also promise 4.”

What do you do with 3-1-5-4 then? You would transfer to clubs via 1NT, then bid 2♠ for min? What about the same shape with extras?

To be consistent with the 3♠-6♦ hand type, we play

1♦-1♠ 1NT(4+clubs)-2♣/2♦ (to play) ? 2♥: typically 3-1-5-4, 15-17 2♠: typically 3-1-5-4, 11-14

And leave the immediate 2♦ rebid to show 5♦-4♥ with minimum opening values, so that you don't need reverse Flannery from responder. (I believe Dale also suggested this in another post.)

Yes, our supports are slower. But opponents are unlikely to interfere, given that they already passed earlier.

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Josh - if opener has 4-1-6-2 minimum opening values and responder has a game-forcing 2-suited 5♣-4♥, responder could punt with 2♥, then opener bids 2♠, and responder 3NT, as he described in Notes 1. The danger is that opener may cross you up by bidding 3♠ (Notes 5) over 2♥. Then bidding becomes murky.

Regarding 1♦-2♣-2♥-3♣-3NT is 3-4-4-2 12 count or 3-4-5-1 16 count, my guess is that opener's 3♠ over 3♣ could be latter and 3NT is just minimum, 3-4-4-2 or 3-4-5-1.

1♣ (balanced club) - 1♥ (4+ spades) ? 2♣ = clubs and hearts, non-forcing 2♥ = clubs and hearts, forcing to 3♣

(Here, we assume opener will accept the transfer when he has 12-14 balanced with 2-3 spades, so that 1NT can be the starting point of transfer rebids, as in 1♦ opening. Note that we are not using 2♣ here for clubs and diamonds, since hearts are more valuable.)

With these transfer rebids, we don't have a need for reverse Flannery - they are problematic anyway when opener has 1-suited minor.

So, we are OK (meaning that we can always locate 4-4 major-suit fits) when opener has a (43)(51)and responder bids a major, using transfer rebids above. The remaining issue is when responder does not have a major. When 1♣ (balanced club)is opened, if responder has 4+ clubs, he can bid 1NT (see the structure I describe in another comment below), we will be also OK (meaning that we can find 3NT or 5♣).

The only two sequences that I have trouble with are: 1. 1♦ - 1NT (4+ clubs); and 2. 1♣ - 1♠ (4+ diamonds)

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Victor, your structure has a lot going for it. The 3=3=4=3 GF hands is not a big problem. You can easily incorporate it into the 2♣ (weak or strong diamonds) transfer. So,

1♣-2♣ 2X-2NT : GF balanced hands with 4-5 diamonds

and

1♣-2♣ 2X-3♦ : GF 1-suited long diamonds

In fact, this 2♣ transfer can (and should, imo) cover 5♦-4M GF hands as well (responder just bids his major after the transfer.). You did not explicitly say how you would bid these hands. I assume you would start with 1♦ or 1♥, which may not be best. Starting with the 2♣ can get your relative suit-lengths sorted out correctly.

The structure I play reverses your 1♠ and 1NT. That is,

In terms of effectiveness, we are probably about the same. But the structure above somehow is more natural to me.

Neither of our structures could solve Yuan's Death Hand ((41)=3-5, 15-17) completely. Over my 1NT or your 1♠, which shows 4+ clubs, opener with the Death hand can explore above 2♣ (say 2M), knowing that we are at least safe at 3♣. So, 3NT or 5♣ can still be found. However, if responder shows diamonds (my 1♠ or your 1NT), opener still has a problem. A possible partial solution is to extract hands with 8-10 HCP and 4-5 diamonds from the diamond bid, and add it to the club response. So, in my structure, 1♣-1NT is now either 4-5 clubs, 6-10 or 8-10 balanced. Opener with the Death Hand can still bid above 2♣ safely (stopping at 2NT or 3♣) when responder bids 1NT. And when responder bids 1♠ (diamonds), he will downgrade his hand and rebid 2♣, knowing that responder does not have 8-10 balanced. The trouble with this structure, in addition to its complexity, is that when the opponent enters the auction with 2M, our side may not be able to compete effectively, if opener has the unbalanced club hand.

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I thought Rubin's 2-level bids 2X shows either strong in X or weak in X+1. So, if he opens 2♥, he is either strong in hearts or weak in spades. Am I mistaken? One problem with this either-or treatment is that responder sometimes preempts opener (e.g. with a 3♠ bid) when opener has the strong type (strong 2♥ bid).

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Just for your reference, Marshall Miles proposed (in The Bridge World)several years ago, over 1M-(X), to use XX = forcing 1NT hands, and pass for XX hands.

He also recommended, over 1M-(2M/2NT), using X = negative X hands and pass = penalty-oriented hands.

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I assume so. In double dummy, small diamond lead wins mainly due to the case of 5-3-2(partner)-3 distribution around the table (not ♦Jx from partner and ♦Qxx from declarer); while DA wins mainly because opponents have a finesseable or droppable ♦Q. I don't know which is more likely. Maybe that's why the actual results are almost inseparable.

I was also wrong to say the simulation should be run with 15-17. After South's acceptance, it can almost only be 16-17.

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Dean, this is a wonderful quantification of Kit's marvelous article.

Let me try to understand the numbers better. Are these two numbers, EVns(pass): 69.92 ± 334.5 and EVns(double): -1.20 ± 511.36, the expected values of all the 10,000 deals you ran, or the expected values of the 4183 deals when a small diamond is led?

I suppose if the simulation is run with 1NT = 15-17, instead of 16-17, the results would be even better. I am a bit surprised by the fact that the ♦A lead is only slightly inferior to the small diamond lead (by less than 1%).

Kai-Ching Lin

I have a couple questions, maybe you can clarify. When opener bids 2♥, he can have:

3 cases for 43♥33

3 cases for 443♥2

2 cases for 5♣3♥32

2 cases for 5♦3♥32

So there are 10 possible distributions. If you use 2NT to relay, there are only 8 sequences leading to 3NT. Am I missing something?

What if responder has a different shape, say 2=4=4=3? Do you still bid 3♦? If you do, you may miss an 8-card club fit?

Kai-Ching Lin

If it wins, and spades are not 4-0, lay down ♠K and test 3 rounds of diamonds. If diamonds fail to break 3-2, go back to dummy with ♠A, and take the club finesse.

Compared to Steve's line at the table, this line gains when East has ♠Qxx and 3 diamonds exactly; and it loses when West has singleton ♠Q and ♣Q is on side or when West has something like 1-7-4-1.

Another variation is: after the first-round spade finesse, you test diamonds immediately before laying down ♠K. This has the advantage that it makes whenever diamonds are 3-2 and East has ♠Qx or ♠Qxx. However, it loses to the line above when diamonds are 4-1 and ♣Q is on.

Kai-Ching Lin

Kai-Ching Lin

I began to think about this last year when a friend of mine asked me to play this unbalanced diamond/balanced club system with him. I liked the transfer rebids in unbalanced 1♦ result so much that I pushed the idea to balanced 1♣ and 1M openings.

In 1♣ opening where transfer responses are used, if you let opener's acceptance of the transfer describe the weak NT hand with less than 4-card support, you also free up the 1NT rebid to serve as the starting point for transfers, as in unbalanced diamond. As a result, you have the same benefits: 2 ways to describe 3-card supports and no need to use reverse Flannery.

Some players prefer to use the 1NT rebid over the transfer responses as strong balanced hands (say 18-20). I found that you could incorporate that into the acceptance of the transfer. That is, 1♣-1♦-1♥ is now 12-14 or 18-20, balanced without 4 hearts. The level is low enough to accommodate dual meanings.

I believe I first read about the idea of transfer rebids was Feldman-Bramley in Challenge The Champs many years ago (not in this unbalanced setting though).

Kai-Ching Lin

What do you do with 3-1-5-4 then? You would transfer to clubs via 1NT, then bid 2♠ for min? What about the same shape with extras?

To be consistent with the 3♠-6♦ hand type, we play

1♦-1♠

1NT(4+clubs)-2♣/2♦ (to play)

?

2♥: typically 3-1-5-4, 15-17

2♠: typically 3-1-5-4, 11-14

And leave the immediate 2♦ rebid to show 5♦-4♥ with minimum opening values, so that you don't need reverse Flannery from responder. (I believe Dale also suggested this in another post.)

Yes, our supports are slower. But opponents are unlikely to interfere, given that they already passed earlier.

Kai-Ching Lin

1♥ - 1♠: like forcing 1NT, up to 4 spades

?

1NT : 3+♣

2♣ : 3+♦

2♦ : 1-suited ♥

2♥ : minimum, 5♥-4♠ (Your Flannery hand)

2♠ : usual reverse

1♥ - 1NT: 5+ spades

So you catch all 4-4 and 5-3 spade fits without Flannery.

Kai-Ching Lin

1♦-1♠ (4+♦ // 4+♠)

2♣-2♦ (1-suited 6+♦ // to play)

?

2♥: 15-17 3♠-6♦

2♠: 11-14, 3♠-6♦

So you get to show your 3-card spade support, with BOTH hand-type and strength identified, at or before the level of 2♠.

Just noticed that Dale J. said something similar above.

Kai-Ching Lin

Regarding 1♦-2♣-2♥-3♣-3NT is 3-4-4-2 12 count or 3-4-5-1 16 count, my guess is that opener's 3♠ over 3♣ could be latter and 3NT is just minimum, 3-4-4-2 or 3-4-5-1.

Kai-Ching Lin

1♦ (4+ diamonds, unbalanced) - 1!♠ (4+ spades)

?

2♦ = 5♦-4♥, non-forcing (probably 11-16)

2♥ = 5♦-4♥, forcing to 3♦ (probably 17+)

Similarly,

1♣ (balanced club) - 1♥ (4+ spades)

?

2♣ = clubs and hearts, non-forcing

2♥ = clubs and hearts, forcing to 3♣

(Here, we assume opener will accept the transfer when he has 12-14 balanced with 2-3 spades, so that 1NT can be the starting point of transfer rebids, as in 1♦ opening. Note that we are not using 2♣ here for clubs and diamonds, since hearts are more valuable.)

With these transfer rebids, we don't have a need for reverse Flannery - they are problematic anyway when opener has 1-suited minor.

So, we are OK (meaning that we can always locate 4-4 major-suit fits) when opener has a (43)(51)and responder bids a major, using transfer rebids above. The remaining issue is when responder does not have a major. When 1♣ (balanced club)is opened, if responder has 4+ clubs, he can bid 1NT (see the structure I describe in another comment below), we will be also OK (meaning that we can find 3NT or 5♣).

The only two sequences that I have trouble with are:

1. 1♦ - 1NT (4+ clubs); and

2. 1♣ - 1♠ (4+ diamonds)

I don't have satisfactory answers to either.

Kai-Ching Lin

1♣-2♣

2X-2NT : GF balanced hands with 4-5 diamonds

and

1♣-2♣

2X-3♦ : GF 1-suited long diamonds

In fact, this 2♣ transfer can (and should, imo) cover 5♦-4M GF hands as well (responder just bids his major after the transfer.). You did not explicitly say how you would bid these hands. I assume you would start with 1♦ or 1♥, which may not be best. Starting with the 2♣ can get your relative suit-lengths sorted out correctly.

The structure I play reverses your 1♠ and 1NT. That is,

1♣ - 1♠ = 4+ diamonds (could be 5♦-4M GF hands)

1♣ - 1NT = 4-5 clubs, non-forcing

1♣ - 2♣ = 4+ clubs, forcing to 2NT or 3♣

1♣ - 3♣ = 6+ clubs, non-forcing

In terms of effectiveness, we are probably about the same. But the structure above somehow is more natural to me.

Neither of our structures could solve Yuan's Death Hand ((41)=3-5, 15-17) completely. Over my 1NT or your 1♠, which shows 4+ clubs, opener with the Death hand can explore above 2♣ (say 2M), knowing that we are at least safe at 3♣. So, 3NT or 5♣ can still be found. However, if responder shows diamonds (my 1♠ or your 1NT), opener still has a problem. A possible partial solution is to extract hands with 8-10 HCP and 4-5 diamonds from the diamond bid, and add it to the club response. So, in my structure, 1♣-1NT is now either 4-5 clubs, 6-10 or 8-10 balanced. Opener with the Death Hand can still bid above 2♣ safely (stopping at 2NT or 3♣) when responder bids 1NT. And when responder bids 1♠ (diamonds), he will downgrade his hand and rebid 2♣, knowing that responder does not have 8-10 balanced. The trouble with this structure, in addition to its complexity, is that when the opponent enters the auction with 2M, our side may not be able to compete effectively, if opener has the unbalanced club hand.

Kai-Ching Lin

Kai-Ching Lin

Kai-Ching Lin

Therefore, with AKx AQJxxx Kx AQ or AK AQJxxx xx AKQ, opener can bid 3NT.

Kai-Ching Lin

♠K3 ♥AKJ ♦T98763 ♣A9

=

AAJ

KK339

109876 (straight flush)

Dummy has the worst hand.

Kai-Ching Lin

Kai-Ching Lin

Kai-Ching Lin

He also recommended, over 1M-(2M/2NT), using X = negative X hands and pass = penalty-oriented hands.

Kai-Ching Lin

Kai-Ching Lin

I was also wrong to say the simulation should be run with 15-17. After South's acceptance, it can almost only be 16-17.

Kai-Ching Lin

Let me try to understand the numbers better. Are these two numbers, EVns(pass): 69.92 ± 334.5 and EVns(double): -1.20 ± 511.36, the expected values of all the 10,000 deals you ran, or the expected values of the 4183 deals when a small diamond is led?

I suppose if the simulation is run with 1NT = 15-17, instead of 16-17, the results would be even better. I am a bit surprised by the fact that the ♦A lead is only slightly inferior to the small diamond lead (by less than 1%).

Thanks!