[Valardent]

Hello,

We play a modified Heeman set of responses on a 1NT 15-17 opening.

For all invitationnal hands with 4 or 5 M, we go trough a 2♦/♥ transfer and rebid 2NT.
(In the actual version of Heeman, an unbalanced invitationnal hand 5M-4m or 4M-5m rebids 3m at a certain point which I think is not optimal as you might end on a misfit at the 3 level with both hands being minimal)

If opener is minimum, he passes except with 4M when he couldn't break the transfer (true that if responder is 4-6/5-5 it's ppbly better to play in the minor suit, but in my experience the misfit is more frequent than the minor sidefit making 3m superior to 2NT)

If opener is max, he responses following this set which I think can be optimalised :

3♣: 2cardsM or flat hand with 3M; then: 3♦ shows 5M-4+m (3♥asks), 3♥/♠ is 4M-5+m, 3NT bal 4/5M

3♦: 3cardsM or 4M333 NT orientated; then: 3♥/♠ is 4M-5+m, 3NT bal 4M, 4x 5M

3M: 4(bad)cardsM mini

3OM: undefined

3NT: to play

The main problem is when you unveil a decent/great minor fit but you have run out of space to investigate wheather 3NT or 5m(6m) is best. For example, if opener has xxx and KQ10 in the remaining suits, he might face a singleton towards xxx which makes 5m great or towards KQ10 when 3NT is surely better.

Instead of responder showing the side minor, one could work with showing the short suit if any.

Might this be better?

[Zelandakh]

After a 2♦ transfer, try reversing your 2♠ and 2NT rebids - the extra step on invitational hands makes all the difference. I think packing all of these 4+ spade hands into a 2♥ transfer is misguided and would look to change the system somewhere else to unload this response. The idea of showing shortage rather than a second suit has been discussed on these forums before. Iirc, the result of the discussion was that it works ok but it was a fair while ago.

[Valardent]

Zelandakh, on 2013-May-02, 02:23, said:

After a 2♦ transfer, try reversing your 2♠ and 2NT rebids - the extra step on invitational hands makes all the difference. I think packing all of these 4+ spade hands into a 2♥ transfer is misguided and would look to change the system somewhere else to unload this response. The idea of showing shortage rather than a second suit has been discussed on these forums before. Iirc, the result of the discussion was that it works ok but it was a fair while ago.

Tx for showing some interest Zel. If winning a step was on option I would have tried it, but it's defenitely not.

So based on shortness description method, this is what I brainstormed so far based on a ♥transfer start:
(Responder denies 4 in ♠)

1 NT (-) 2 ♦(-)
2 ♥ (-) 2 NT (-)
??
passe minimum
3 ♣ maximum, fit ♥3e or bad fit ♥ 4e flat ; GF
3 ♦ maximum, misfit ♥ or fit ♥ 3e flat ; GF
3 ♥ minimum, bad fit ♥4e
3 ♠ maximum, xx ♠, decent fit ♥3e ; GF
3 NT to play
4 ♥ maximum, very bad fit ♥4e


1 NT (-) 2 ♦ (-)
2 ♥ (-) 2 NT (-)
3 ♣ (-) ??
3 ♦ 4 ♥, x♠; then 3 ♥ relay : 3 ♠ 4 ♥-5+♣ 3 NT 4 ♥-5+♦ 4 m 1444, best m
.......................or 3 ♠ relay : 3 NT 4 ♥-4/5 m
........................................... 4 m 4 ♥-6 m

3 ♥ 4/5♥ no shortness, if 5♥ 5332 ;
then 3 ♠ relay :3 NT 4 ♥
...................... 4 ♥5♥ 5332

3 ♠ 4 ♥, x♣; type 3451
3 NT 4 ♥, x♦; type 3415
4 m 6m-5♥
4 ♥ 5 ♥ not 5332

1 NT (-) 2 ♦ (-)
2 ♥ (-) 2 NT (-)
3 ♦ (-) ??
3 ♥ 4/5 ♥, x♣; type 3(5)(4)1
3 ♠ 4/5 ♥, x♦; type 3(5)1(4)
3 NT 4/5 ♥, x♠; type 15(4)(3), 1444, 14(5)(3), 14(6)(2)
4 m 6m-5♥

Laying this out on the bbf editor is a nightmare, hope it's readable.

Any remark wlc.

[Zelandakh]

I found an online write-up of Heemann (thanks Free) and it seems to me that you are using it quite differently to how it was intended. In particular, the transfer followed by 2NT sequences specifically show a 4 card suit. It seems to me that the answer to the problem of getting too high on a misfit is to treat 4M5m hands as balanced and only go to the 3 level with 4M6+m. If this is your modification then the obvious question is how you are using the 2♣ response, which should normally be handling the INV+ 5M hands. In other words, before fixing this in isolation, I would first like to see how it fits into the 1NT structure as a whole. Very often a NT structure issue is easier to fix elsewhere than at the point where the problem appears to be.

[Valardent]

Zelandakh, on 2013-May-03, 01:36, said:

I found an online write-up of Heemann (thanks Free) and it seems to me that you are using it quite differently to how it was intended. In particular, the transfer followed by 2NT sequences specifically show a 4 card suit. It seems to me that the answer to the problem of getting too high on a misfit is to treat 4M5m hands as balanced and only go to the 3 level with 4M6+m. If this is your modification then the obvious question is how you are using the 2♣ response, which should normally be handling the INV+ 5M hands. In other words, before fixing this in isolation, I would first like to see how it fits into the 1NT structure as a whole. Very often a NT structure issue is easier to fix elsewhere than at the point where the problem appears to be.

Heeman uses 2♣ as a puppet to 2♦ (then 2♥-2♠ for 5♠) followed by 2NT/3m showing an invitationnal hand 5M332 possibly 5M4m22/5M5m. The 5M4m31 is the problem hand.

Situation is similar after the direct M transfer where the 4M5m31 is the problem (possible misfit at the 3 level).

As the 5M-5m and the 4M-6m invit. handtypes have a low frequency, I prefer to keep those bids for GF (shapely-)hands and put all the invitationnal hands in 2NT after a direct M transfer. I'm convinced this treatment is overall better.

Of course the 2NT 'invite' is quite crowded but I think the scheme above is quite superior the the former one except for the last part where ...2NT 3♦ 3♥ needs to be 2-way so that responder can show a hand without shortness.

So the focus (if any) should be on that scheme as I'm really not interested in changing the structure around.

[Zelandakh]

What I am suggesting is that you can avoid the issue by treating 5431 hands as balanced for the purposes of inviting, then allowing Opener to check back for shortage if they want to accept. So

1NT - 2♣; 2♦ - 2♥; 2♠
==
2NT = invitational with 5 spades, possible shortage (51(43), 53(14))
... - 3♣ = GF, asks for shortage

--
1NT - 2♣; 2♦
==
2NT = invitational with 5 hearts, possible shortage (15(43), 35(14))
... - 3♣ = GF, asks for shortage

--
1NT - 2♦; 2♥
==
2NT = invitational with 4 hearts, possible shortage (14(53), 1444, 34(15))
... - 3♣ = GF, asks for shortage

--
1NT - 2♥; 2♠
==
2NT = invitational with 4 spades, possible shortage (41(53), 4144, 43(15))
... - 3♣ = GF, asks for shortage

The only catch is that you have to play 2NT with an unbalanced hand opposite a minimum Opener without a fit for the major. But so does the rest of the room so it is not really a problem. This seems a clearly better solution than mixing the 5M4m hands into the transfers.