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Fewest HCP To Make Notrump

by Richard Pavlicek

This page shows the fewest HCP needed to make a notrump contract against any defense (1) with favorable distribution and (2) against any distribution. Each contract can be declared by North or South, unless stated by South only, then declarer is crucial. Most examples are not unique but show just one possible layout.

7 NT

6 NT

5 NT

4 NT

3 NT

2 NT

1 NT

Besides the fewest HCP, each example has the lowest possible rank sum for North-South. This is evaluated as the sum of all card ranks: Ace = 14, king = 13, queen = 12, jack = 11, etc.

Seven Notrump

with 11 HCP by South only

1. 7 NT	— — 8 7 6 5 4 3 2 7 6 5 4 3 2
K K Q J 10 9 8 7 6 5 4 3 2 — —		— — A K Q J 10 9 A K Q J 10 9 8
Sum 167	A Q J 10 9 8 7 6 5 4 3 2 A — —

On a humorous note, this gives a whole new meaning to the principle of fast arrival. Not only does South make 7 NT, but so does West. Better bid it first!

with 17 HCP

2. 7 NT	2 5 4 3 2 A 4 3 2 A 4 3 2
K K Q J 10 9 8 7 6 9 8 7 6		Q 9 8 7 6 K Q J 10 K Q J 10
Sum 163	A J 10 9 8 7 6 5 4 3 A 5 5

against any distribution with 19 HCP

3. 7 NT	2 A 4 3 2 A 4 3 2 A 4 3 2
Sum 165	A K 10 9 8 7 6 5 4 3 5 5 5

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Six Notrump

with 9 HCP by South only

4. 6 NT	— — 8 7 6 5 4 3 2 8 7 6 5 4 3
K K Q J 10 9 8 7 6 5 4 3 2 — —		Q — A K Q J 10 9 A K Q J 10 9
Sum 163	A J 10 9 8 7 6 5 4 3 2 A — 2

with 13 HCP

5. 6 NT	2 A 4 3 A 5 4 3 7 6 5 4 3
K K Q J 10 9 8 10 9 8 7 6 A		Q 7 6 5 K Q J K Q J 10 9 8
Sum 157	A J 10 9 8 7 6 5 4 3 2 2 2

Thanks to Derek Wang, Taiwan, for the above construction.

against any distribution with 18 HCP by South only

6. 6 NT	3 2 4 3 2 A 4 3 2 A 4 3 2
Sum 169	A Q J 10 9 8 7 6 5 4 K 5 5 —

against any distribution with 19 HCP

7. 6 NT	4 3 2 4 3 2 A 4 3 2 A 3 2
Sum 170	A Q J 10 9 8 7 6 5 A 5 5 4

Six notrump is the only contract for which the fewest HCP to make against any distribution is affected by which hand declares. (Compare Examples 6 and 7.)

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Five Notrump

with 9 HCP

8. 5 NT	2 A 3 7 6 5 4 3 7 6 5 4 3
K K Q J 10 9 8 7 6 5 4 A A		Q — K Q J 10 9 8 K Q J 10 9 8
Sum 152	A J 10 9 8 7 6 5 4 3 2 2 2

Thanks to Derek Wang, Taiwan, for the above construction.

against any distribution with 17 HCP

9. 5 NT	K 2 Q 10 9 8 7 6 A 2 A 3 2
Sum 182	A 10 9 8 7 6 5 4 3 5 4 3 2 — —

Note that opponents cannot take more than two hearts with A-K-J alone.

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Four Notrump

with 8 HCP

10. 4 NT	2 K 3 7 6 5 4 3 7 6 5 4 3
K A Q J 10 9 8 7 6 5 4 A A		Q — K Q J 10 9 8 K Q J 10 9 8
Sum 151	A J 10 9 8 7 6 5 4 3 2 2 2

against any distribution with 16 HCP

11. 4 NT	5 4 3 2 3 2 A 3 2 A 4 3 2
Sum 176	A Q J 10 9 8 7 6 J 10 9 4 4 —

South’s 9 is essential only if North is declarer to prevent two lead-throughs if East has honor-third. With South declarer, J-10-5-4 is sufficient.

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Three Notrump

with 6 HCP by South only

12. 3 NT	— — 8 7 6 5 4 3 2 7 6 5 4 3 2
A K 6 5 4 3 A K 7 5 4 3 2 — —		— — A K Q J 10 9 A K Q J 10 9 8
Sum 177	Q J 10 9 8 7 2 Q J 10 9 8 6 — —

Note the necessity of South’s 6; i.e., swapping the 6 and 5 allows West to defeat 3 NT. Playing it out to understand this is a good exercise in two-hand technique.

with 7 HCP

13. 3 NT	3 2 Q 3 2 7 6 5 4 7 6 5 4
K A K J 10 9 8 7 6 5 4 A A		Q — K Q J 10 9 8 K Q J 10 9 8
Sum 150	A J 10 9 8 7 6 5 4 — 3 2 3 2

against any distribution with 15 HCP

14. 3 NT	6 5 4 3 2 10 9 8 3 2 2 A 2
Sum 174	A Q J 10 9 8 7 — A 4 3 5 4 3

North’s 8 is essential only if South is declarer to prevent two lead-throughs if West has honor-third. With North declarer, 10-9-4-3-2 is sufficient.

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Two Notrump

with 4 HCP by South only

15. 2 NT	— — 8 7 6 5 4 3 2 7 6 5 4 3 2
A K Q 5 4 3 A K 7 5 4 3 2 — —		— — A K Q J 10 9 A K Q J 10 9 8
Sum 171	J 10 9 8 7 6 2 Q J 10 9 8 6 — —

with 5 HCP

16. 2 NT	A J 10 9 10 9 8 7 6 5 4 7 8
— Q J K Q J 10 9 8 K Q J 10 9		K Q 8 7 6 5 4 3 2 A K A A
Sum 160	— 3 2 6 5 4 3 2 7 6 5 4 3 2

No matter how the play begins, East must lead spades at least twice as the A-K are driven out. Thanks to Tom Slater, winner of Jolly Old Saint Nicholas, for this construction.

against any distribution with 14 HCP

17. 2 NT	K 4 3 2 J 10 9 2 Q 10 9 8 7 —
Sum 193	A 10 9 8 7 6 5 — 6 5 4 3 2 A

North’s red suits are immune to damaging lead-throughs, because the defense has limited communication with only A-K-J between them.

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One Notrump

with 3 HCP by South only

18. 1 NT	— — 8 7 6 5 4 3 2 7 6 5 4 3 2
A K Q 5 4 3 A K J 7 5 3 2 — —		— — A K Q J 10 9 A K Q J 10 9 8
Sum 164	J 10 9 8 7 6 2 Q 10 9 8 6 4 — —

South’s hearts cannot be jack-high (like spades) else West could establish and enjoy his long heart. The actual holding nets four tricks if West continually attacks hearts (not wise); else three tricks combined with four spade tricks. While far from obvious, South’s 6 and 6-4 are essential, as reducing any would allow West to prevail.

Bridge can be a crazy game. West makes 7 NT as declarer but cannot beat 1 NT on-lead.

with 4 HCP

19. 1 NT	8 7 10 9 8 7 6 5 4 K J 9 2
K Q J 10 9 K Q J 10 9 8 Q J —		A A A K A Q 10 8 7 6 5 4 3
Sum 151	7 6 5 4 3 2 6 5 4 3 2 3 2 —

This problem was the topic of my October 2016 puzzle contest, Fewest HCP Notrump. Congratulations to the winner, Tina Denlee, Quebec, whose construction is shown above.

against any distribution with 13 HCP

20. 1 NT	K 5 4 3 2 3 2 J 10 9 8 A 3
Sum 186	A 10 9 8 7 6 J 10 9 8 3 2 2

Both red eights are essential, else a triple lead-through would be possible.

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© 2008 Richard Pavlicek