Lowest recorded point count for 7NT bid and made

[SimonFa]

I'm reading some back issues of the EBU magazine (kudos to them for putting them online) and in the August 2010 issue Heather Dhondy reports on a 25HCP 7NT contract that was bid and made at Ealing Bridge Club:

 

♠AK ♥2 ♦KT876543 ♣AK

 

♠T9 ♥A98543 ♦A2 ♣842

 

The bidding started with an Acol strong 2♦and reached 7NT via Blackwood. It made when Diamonds behaved.

 

She discusses some theoretical hands, including the famous James Bond Moonraker 7♣ on 8HCP and a 7NT on 19HCP.

 

She than asks if the low of 25HCP for 7NT bid and made has been beaten in the wild assuming best defence. Does anyone know?

[whereagles]

moonraker 7♣..? I don't remember that lol

[ggwhiz]

moonraker 7♣..? I don't remember that lol

[hv=pc=n&s=s5432h5432dc75432&w=sakqjhakqjdakckj9&n=shdq8765432caqt86&e=st9876ht9876djt9c]399|300[/hv]

 

It's in the book but not the movie.

 

Drax is cheating and Bond sneaks this (approximate) hand into the game. 7♣ with extra side bets on top of the redouble rolls home and the accurate hands are in one of the Bergen books too. Points Schmoints?

[PrecisionL]

[hv=pc=n&s=sat7653h8765432dc&w=skhakqjdakqjcakqj&n=sj9842hd8765432c2&e=sqht9dt9ct9876543]399|300|7[/hv]

7♠ by south with 5 hcp.

Reference: The Italian Blue Team Bridge Book, Garozzo & Forquet, 1969.

[Cyberyeti]

What's the lowest theoretical for 7N ? I can see an 11 point layout.

[manudude03]

Nonsense deleted for now, it had 11HCP :)

[daro1950]

It would seem that this type of layout would be needed.

 

This appears to be the minimum possible. 3 Aces and 2 Jacks = 14HCP. I wonder what the odds are? Lol

Of course South must be the Declarer.

 

 

[hv=pc=n&s=s32h3d32cat875432&w=shkqjt9876542dkck&n=shadajt987654cj96&e=sakqjt987654hdqcq]399|300[/hv]

[kenrexford]

I had a weird related experience. With 4045 pattern and a 0 hcp hand, I got our contract to 3Dx opposite 0454 pattern, AK in both minors, and both minors splitting 2-2.

 

The lead determines whether 12 or 13 tricks are available.

 

Partner, howewr, panicked and played for down one. It was still a 1 imp gain against 3M their way.

[WellSpyder]

This appears to be the minimum possible. 3 Aces and 2 Jacks = 14HCP. I wonder what the odds are? Lol

Of course South must be the Declarer.

Give West the remaining small ♥s in exchange for ♣K, and N the ♦s in the S and E hands in exchange for his ♣s and you are down to 11 HCP since S no longer needs ♣A

[inquiry]

Without requiring a void in opening leaders hand, declarer holds SAT98765432 HA DA CA and partner has singleton spade Jack..... spades divide 10-1-1-1

 

That is 17 total points....

 

 

 

[PhilKing]

Just to clarify, one example of an 11 pointer is:

 

W: ♠K♥KQJT98765432♦♣ and North ♠AQJT98765432 ♥A♦♣

 

The other two hands are irrelevant except that East has all the points. You can obviously go lower with inferior defence - a zero count is clearly possible with enthusiastic unblocking.

[Cyberyeti]

Just to clarify, one example of an 11 pointer is:

 

W: ♠K♥KQJT98765432♦♣ and North ♠AQJT98765432 ♥A♦♣

 

The other two hands are irrelevant except that East has all the points. You can obviously go lower with inferior defence - a zero count is clearly possible with enthusiastic unblocking.

 

That was exactly the hand I had in mind.

[AyunuS]

But the topic says 7NT bid and made. With a singleton K and all but the K in another suit, you wouldn't ever want to bid 7NT. You'd bid 7 in your suit for fear that they'd lead one of the suits you didn't have. It wouldn't be worth the risk to go for 7NT.

 

Lowest HCP I'd think you'd bid 7NT with is if you have - A A AK1098765432, where one opponent bids 7 of the suit you have none of, and possibly redoubles if you double that instead of overcalling, so you can be pretty confident that they have all 13 of them. Then bid 7NT where their other hand has to lead so you can be sure they have to lead one of your suits. Then you can bid and make 7NT with 15 HCP.