7-5-1

[gnasher]

That all sounds about right and maybe I overstated it somewhat, my admittedly poor math suggests that there is a 33% chance that he will have 2 of any 3 of the 2 aces and q♣. The rest of the possible hands that he could have where it would still make or make on a finesse only increase the chances, maybe not all the way to >50% though. Someone should do the math, I'd bet its closer then you think and honestly if my gut is only off by 5-8%, kudos to my gut :)

How much would you like to bet?

 

The top honours that we can't see are ♠A, ♥A, ♥K, ♥Q, ♦K, ♦Q, ♣A and ♣Q. For his 1NT response partner will typically have two or three of these, but let's be optimistic and give him three. Suppose that these cards occur with equal likelihood (again, an incorrect but favourable assumption). The number of ways you can give him three of these cards is 8C5 = 56. The number of ways he can have two of the black honours is 15, and the number of ways he can have all three is one. So he will have two or three 16/56 of the time. That is, by making two favourable assumptions we get to a 30% chance that partner has the honour cards you want him to have. When he's got those cards, slam isn't necessarily making, of course.

[gwnn]

What is with this thread? 3♣ looks easy, but I don't object to 4♠ either. Anything else looks patently wrong.

4♥(if played as an autosplinter) is better than 4♠.

[ArtK78]

Thx all, I was afraid 3♣ wasn't forcing and as I can make 4♠ with almost no help, I wasn't sure it was the correct call. 4♠ also seemed to seriously undervalue how little help this hand needed to make slam and as such I was stuck for a bid.

 

Regardless, if you are curious, over 3♣ partner will bid 3nt over which you will bid 4♠ which is all you can make. Partner indeed had no help.

I thought 3♣ was highly invitational but nf

Hijack alert!

 

This is not a trivial issue. I had a good player come up to me after a regional pair game and ask questions about actions over responder's forcing 1NT response to one of a major. There is a lack of uniformity on how to handle rebids by opener on strong hands.

 

I found an article in The Bridge World some years back containing an interesting method for handling rebid issues. The main focus of the article was how to handle two suited strong but not game forcing hands. But it also contained methods for handling game forcing hands.

 

1♠-1NT:

 

3♣: Game forcing one or two suiter. Responder bids 3♦ to ask opener to clarify, and opener bids:

-----3♥ 5+ spades, 4+ hearts

-----3♠ 5+ spades, 4+ diamonds

-----3NT 5+ spades 4+ clubs not forcing

-----4♣ 5+ spades 4+ clubs forcing

-----4♠ one suited spade hand

 

3♦: 5-5 or better in spades and diamonds, invitational values, not forcing

3♥: 5-5 or better in spades and hearts, invitational values, not forcing

 

If the auction starts 1♥-1NT, the methods are:

 

2♠: Various hands, invitational or better. Responder puppets with 2NT, and opener rebids:

-----3♣ 5+ hearts, 4+ clubs, game forcing

-----3♦ 5+ hearts, 4+ diamonds, game forcing

-----3♥ 5 hearts, 4 spades, invitational values, not forcing

-----3♠ 6+ hearts, 5 spades, forcing

-----3NT 5+ hearts, 4 spades, forcing

-----4♥ one suited heart hand

 

3♣ 5-5 or better in hearts and clubs, invitational values, not forcing

3♦ 5-5 or better in hearts and diamonds, invitational values, not forcing

 

This method takes care of many hands and removes any ambiguity that partnerships may have in how they handle rebids after a forcing 1NT.

[Zelandakh]

1♠-1NT:

It seems to me you could add to this

 

1♠ - 1NT; 4♣ = club auto-splinter (4♦ asks, 4♥ = void, 4♠ = singleton)

1♠ - 1NT; 4♦ = void auto-splinter

1♠ - 1NT; 4♥ = void auto-splinter

 

and

 

1♠ - 1NT; 3♣ - 3♦; 4♦ = singleton auto-splinter

1♠ - 1NT; 3♣ - 3♦; 4♥ = singleton auto-splinter

 

 

 

If the auction starts 1♥-1NT, the methods are:

 

2♠: Various hands, invitational or better. Responder puppets with 2NT, and opener rebids:

-----3NT 5+ hearts, 4 spades, forcing

Are you sure about this one? It seems to me that this should be non-forcing and 4♣ (instead of 3NT) shows the forcing 5♥4♠ hand.

 

As with a 1♠ opening it seems logical to add

 

1♥ - 1NT; 3♠ = spade auto-splinter (3NT asks, 4♣ = void, 4♦ = singleton)

1♥ - 1NT; 4♣ = void auto-splinter

1♥ - 1NT; 4♦ = void auto-splinter

 

and

 

1♥ - 1NT; 2♠ - 2NT; 4♣ = singleton auto-splinter

1♥ - 1NT; 2♠ - 2NT; 4♦ = singleton auto-splinter

 

One thing that interests me here. Over 1♠ there are 2 ways of showing the black suits, one commits above 3NT and the other does not. Over 1♥ there is only one way of showing the majors with 4 spades and this commits the partnership above 3NT. Given that Responder has denied support for both majors by this stage this seems something of a strange design decision. Any idea what the logic is here? Note also that even if you decided to play the 3NT rebid as non-forcing with a forcing 4♣ rebid showing the same hand pattern but stronger, there would still be enough space to differentiate between singleton and void auto-splinters below 4♥ if desired (by incorporating a third splinter type into a direct 3♠ rebid).

 

An interesting question might be whether a codified scheme such as this works out better than an off-the-shelf and (comparatively) simple method such as transfers.

[ArtK78]

One thing that interests me here. Over 1♠ there are 2 ways of showing the black suits, one commits above 3NT and the other does not. Over 1♥ there is only one way of showing the majors with 4 spades and this commits the partnership above 3NT. Given that Responder has denied support for both majors by this stage this seems something of a strange design decision. Any idea what the logic is here? Note also that even if you decided to play the 3NT rebid as non-forcing with a forcing 4♣ rebid showing the same hand pattern but stronger, there would still be enough space to differentiate between singleton and void auto-splinters below 4♥ if desired (by incorporating a third splinter type into a direct 3♠ rebid).

Over 1♥ there are three ways of showing the majors, and none of them commits the partnership past 3NT:

 

1♥ - 1NT - 2♠ - 2NT - 3♥: 5 hearts, 4 spades, invitational values, not forcing.

1♥ - 1NT - 2♠ - 2NT - 3♠: 6 hearts, 5 spades, game forcing.

1♥ - 1NT - 2♠ - 2NT - 3NT: 5 hearts, 4 spades, game forcing.

 

I apologize for the misprint in my earlier post. Over the last action - 3NT - responder is allowed to pass. He knows that opener has about 19-20 HCP and 4-5-x-y distribution.

 

Over 3♠, showing 5-6-x-y distribution, 3NT is still a possible (albeit unlikely) final contract.

 

You can certainly play auto-splinters if you like, but I would guess that showing the suits that you have rather than the suits that you don't have would be better.

[Zelandakh]

Over 1♥ there are three ways of showing the majors, and none of them commits the partnership past 3NT:

 

1♥ - 1NT - 2♠ - 2NT - 3♥: 5 hearts, 4 spades, invitational values, not forcing.

1♥ - 1NT - 2♠ - 2NT - 3♠: 6 hearts, 5 spades, game forcing.

1♥ - 1NT - 2♠ - 2NT - 3NT: 5 hearts, 4 spades, game forcing.

 

I apologize for the misprint in my earlier post. Over the last action - 3NT - responder is allowed to pass. He knows that opener has about 19-20 HCP and 4-5-x-y distribution.

 

Over 3♠, showing 5-6-x-y distribution, 3NT is still a possible (albeit unlikely) final contract.

 

You can certainly play auto-splinters if you like, but I would guess that showing the suits that you have rather than the suits that you don't have would be better.

Aha, now I understand. So there is no forcing rebid with 5♥4♠. I suppose if you wanted this you could designate 4m rebids here as either fragment or shortage-showing. Or just use 4♣ for that as over the 1♠ opening. Perhaps there is no hand that would want this given the apparent absence of major suit fit though.

 

 

One of the points of an auto-splinter is that you do not necessarily have another suit, certainly not one that you want to show. It is a strong one-suited hand with a side shortage. I was suggesting that one way of using the additional space inherent in the system might be to differentiate between singletons and voids. Of course that is not the only way. You might, for example, decide to play instead:

1♥ - 1NT; 2♠ - 2NT:

 

3♠ = 4 spades, GF

3NT = 5611

4♣ = 5620 or 5710

4♦ = 5602 or 5701

 

Whatever, it is surely wasteful to leave these bids undefined when there are useful possibilities available.