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Negative double


Hanoi5

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You hold:

 

Qxx

8

A98xx

AT63

 

It goes:

 

1 1 ?

 

1 2 ?

 

1 3 ?

 

I bet 2 is a possibility in the first one, but what about the third? Would you use a negative double there?

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Over 1, I'd show a high-card raise in clubs. It's quite likely that LHO is about to bid 3, so showing my support is quite important.

 

Over 2, I'd still support clubs. In standard methods there's only one way to support clubs without going past 3, and this is a long way from a minimum for that. I'd upgrade it to a game-force.

 

Over 3, I'd double, for want of a good alternative.

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Over 1, double shows 4 spades*, and I have less than 4 spades. Therefore I will not double. I will bid 2.

 

Over 2 and 3 double might contain less than 4 spades sometimes, so double is an option. Over 3 it's hard to see anything else working out better so double it is.

 

Finally I think I'd bid 3 over 2, a calculated underbid.

 

*yes, if double denies 4 spades then it is an easy double, but I think OP would have told us if this were the case.

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X on all three. Double is very easy for me on the first one as my double denies a holding of 4 or more S.

 

I see some top pairs are going to "double=4+ in spades", and 1 to deny. This seems to work well in combination with support doubles/redoubles by opener and/or a 1S rebid (after the neg double) to show only 3 of them. Haven't changed to that, yet. But we probably will.

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X on all three. Double is very easy for me on the first one as my double denies a holding of 4 or more S.

 

Well, I play 1S as denying 4 spades (so it's effectively the same as your double), but I wouldn't do it, I'd show a good club raise like gnasher.

I'd raise clubs on the second as well.

ON the last I'd double like everyone else.

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In general, if p opens a minor, and I have 4 cards, I assume

we have a fit in the minor.

 

Simple and it works.

 

A neg. X will lead quite often to 4-3 fits, if p has

4 spades => 4-3 fit, if he has 4 diamonds, he has 5 clubs.

 

As a conseq., I would show the fit, and I would show inv.

strength, i.e.

 

#1 2H

#2 3C - best would be, if we play good - bad, but 3C wont be

dead min anyway

#3 X, at this high, p can not for sure assume, that I have 4

spades, so if we end up in 4S, so be it, at least, that will

be game.

 

With kind regards

Marlowe

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I'm intrigued by the fact that all of the responses so far have talked about shape, but the title of the thread is "What strength do you need?". How would any of the answers be different if the A were instead the J?

 

Than I would not have inv. strength, hence

 

#1 2C - a single raise

#2 3C - this time on the light side, still the distribution makes up for the

missing HCP

#3 Passe, at this high you need something like 10HCP

 

With kind regards

Marlowe

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Double on the first hand is really wrong, we have an invitational hand with a club fit, there is no need to lie about spade length (assuming double shows 4+ spades).

On the second, we again have a club raise, the only problem is that 3 is an underbid and 3 is an overbid. I still slightly prefer either of them over double. (Are we going to pass 2 and play a 3-3 fit? If we have clear agreements that partner can bid 2NT without a stopper, double becomes better, as we can then frequently bid 3 over that.)

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On the second, we again have a club raise, the only problem is that 3 is an underbid and 3 is an overbid. I still slightly prefer either of them over double. (Are we going to pass 2 and play a 3-3 fit? If we have clear agreements that partner can bid 2NT without a stopper, double becomes better, as we can then frequently bid 3 over that.)

 

Does

1
(2
) dbl

2NT       3

show extras, or just a different shape from a direct 3?

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