[ralph23]

THis isn't necessarily a b/i question but anyway .... it seems like a good exercise.

Suppose you are South, the declarer.

What is the minimum number of high-card points that NS can have, and make 7NT against any defense (in fact, you will claim upon seeing the opening lead)?

Oh yea... one little detail ... you can arrange all four hands any way you like!!

And give the hands, of course.

Adv & exp pls hide your answers.

NB to b/i's -- I gave this problem to 3 or 4 very advanced players at a post-sectional-game gathering at the hotel bar, and they had a hard time with it... so be proud of yourself for getting the right answer!! (No one was drunk btw ).

[FrancesHinden]

just as a starting point, I can do it easily with 11 between the NS hands. But I'm sure that was too easy...

[goobers]

The best I can do is 17... obviously you can't be off an ace, and the hand I construct requires a jack

Edit: Read Frances' post, I forgot they can be void in the suit where their partner holds an ace

[ralph23]

deleted

[ralph23]

ralph23, on Aug 6 2007, 11:22 AM, said:

FrancesHinden, on Aug 6 2007, 11:09 AM, said:

just as a starting point, I can do it easily with 11 between the NS hands.  But I'm sure that was too easy...

Can it be done with less?

E.g., Can it be done if NS hold only a single Ace between them?

Try a reductio -style proof on the single-Ace (not to be confused with the single bullet) hypothesis....

PS -- I will add (though I think it was clear before), that you must succeed against ANY defense, and you will claim on the opening lead.

Hey, don't complain ....I did let you arrange all four hands after all !!!

This post has been edited by ralph23: 2007-August-06, 10:38

[FrancesHinden]

Not with someone else on lead.

I just thought I got to 11 too quickly if that is the answer.

There's another problem which I've seen before (though I can't remember the answer) which is what is the minimum you need if all four players co-operate during the play.

[Simplicity]

I can't be done with one ace:

either i) LHO holds two suits in which case the opposition have a cashing ace
or ii) LHO has 13 top tricks and 7NT isn't the place to be

personally i can't beat 11 points

[TimG]

Reminds me of the time I held eleven hearts to the Jack...

[jtfanclub]

Let's see, if I have:

AQJT98765432
A
--
--

And LHO has

K
KQJT98765432

then I make 7NT.

I must have the ace in any suit LHO has. If LHO has only one suit, then he has all 13 tricks. Therefore, he must have at least two suits, and we must therefore have at least 2 aces.

Hmmm...if I have

AJT98765432
A3
--
--

and partner has
--
JT9876542
32
32

and the two majors break 1-1, I make it, but now we're into the usual problem where opener has to have at least one minor suit card. So I think 11 is the minimum.

[ralph23]

jtfanclub, on Aug 6 2007, 12:06 PM, said:

Hmmm...if I have

AJT98765432
A3
--
--

and partner has
--
JT9876542
32
32

and the two majors break 1-1, I make it, but now we're into the usual problem where opener has to have at least one minor suit card. So I think 11 is the minimum.

Well, in the hypo example, EW would share 22 minor suit cards, so it looks like West will have one to lead ...

[david_c]

FrancesHinden, on Aug 6 2007, 05:37 PM, said:

There's another problem which I've seen before (though I can't remember the answer) which is what is the minimum you need if all four players co-operate during the play.

These hands are pretty bad:

---

In 7NT, South can win the first five tricks in the majors, and E/W discard all their diamonds.

To make 7NT with two Yarboroughs you need at least three nines: for the first five tricks at least one of the opponents must follow suit, so between them they must have at least five cards below a nine. That leaves only 23 cards below a nine for you and your partner, so you must have three nines. So you can't beat the example above by much, if at all.

[Quantumcat]

Can you give us the answer, Ralph? Surely it has to be impossible with only one ace, LHO always has another card besides his long suit, with one ace there is always the other one they can cash...it'd be easier if it was 6NT, then we could have blocked suits etc.

However, if it was 7 of a trump suit, you'd only need 5:

---

[ralph23]

Quantumcat, on Aug 6 2007, 09:02 PM, said:

Can you give us the answer, Ralph? Surely it has to be impossible with only one ace, LHO always has another card besides his long suit, with one ace there is always the other one they can cash...it'd be easier if it was 6NT, then we could have blocked suits etc.

However, if it was 7 of a trump suit, you'd only need 5:

---

Nice hands!

Yes, jt has the answer defined well in his post. 11 is it.

Simplicity has a nice proof that one Ace is impossible as well.

Just take a complete suit for South, and remove the King and give it to West. Then take any Ace of another suit for South's 13th card, and give the remainder of that suit to West. West now has the inverse of the South hand.

[kenrexford]

See! This is why I always start cuebidding sequences even when we have at least 5 HCP's opposite 6 HCP's.