No Game with How Many HCP?
Author: Larry Cohen
Date of publish: 01/01/2012
Level: Intermediate to Advanced
When Richard Pavlicek posed this puzzle (see below), I was astounded by the answer. He asked:
"What is the most HCP North-South can have, yet not be able to make any game contract?"
Certainly, if they have all 40 HCP (the AKQJ in every suit) they must be able to make at least 3NT (if not 7NT).
So, how about, say 30 HCP? How about 31 HCP?
Shocking as it is, he showed me the following construction with 39 HCP (yes, all but one jack) where North-South can not make any game!
Vul:
Dir: |
 --
 --
 AKQJ32
 AKQJ432
|
|
J109876
1098765
5
--
|
|
5
--
1098764
1098765
|
| |
AKQ432
AKQJ432
--
--
|
|
There you have it. Every HCP in the deck other than one jack!
He didn't analyze it for me, so here goes my attempt:
Let's try 3NT by South (first).
West leads the J.
The best declarer can do is to win and start working on his hearts. The 6-0 break spells doom. He has to let West in with the fifth round of hearts and West continues spades. Declarer wins and knocks out West's sixth heart. West plays another spade. Declarer has established his 7th-round heart winner, but that gets him up to only 8 tricks. Or, looked at another way, he has to lose his three low spades and two of his low hearts for down one.
How about other games? Let's debunk them all, starting with:
4: West leads the J and the play goes as it did in 3NT--declarer is in fact down two.
4: Similar to the other two contracts we've examined. Declarer has to lose lots of major-suit tricks.
5: The defense leads a diamond. The North hand must play on minors, and East will get lots of club/diamond tricks.
5: Similar to 5. Diamonds are led, and North will have to lose to East's long cards in the minors.
Now, can anyone figure out if North-South can get that J into their collection (to have all 40 HCP) and still have no game?
Updated: Oct 2022
