How many?

[qwery_hi]

In this puzzle assume S is always declarer. Take any random hand.

 

Let:

 

X = maximum tricks south can take in a non NT contract.

 

Now exchange 1 face card from S to E or W, so that net hcp doesn't change for S.

 

Let Y = maximum tricks south can take in a non NT contract in this new hand.

 

The goal is to maximize the difference between X and Y. For example, can it ever be 4?

[Flameous]

N

 

♠AKQJ

♥AK

♦AKQJ

♣AKQ

 

S

 

♠xxx

♥xxxx

♦xxx

♣xxx

 

W

 

♠xxxxx

♥QJxxx

♦xxx

♣-

 

E

 

♠x

♥xx

♦xxx

♣Jxxxxxx

 

On this layout 7♦ is making on any lead. Swap south's trump and east's spade and defence crossruffs first 7 tricks. Was this something you were thinking of?

 

Sry I can't use those hand diagrams :unsure:

[kenrexford]

N

 

♠AKQJ

♥AK

♦AKQJ

♣AKQ

 

S

 

♠xxx

♥xxxx

♦xxx

♣xxx

 

W

 

♠xxxxx

♥QJxxx

♦xxx

♣-

 

E

 

♠x

♥xx

♦xxx

♣Jxxxxxx

 

On this layout 7♦ is making on any lead. Swap south's trump and east's spade and defence crossruffs first 7 tricks. Was this something you were thinking of?

 

Sry I can't use those hand diagrams  :(

The principle is there, but it must be a "face card." Just make two Jacks into relevant pip-equivelants and you'd have it.

 

Of course, this misses the spirit of the thing.

 

It is easy to make a huge difference at notrump.

 

Give one opponent AKQJxxxxxxxxx in a suit. Give Declarer 13 tricks in two suits, with A-Kx in the side suits. 12 tricks for the taking if the opponent has KQJxxxxxxxxx-A. Switch the Aces, and 13 tricks for the defense.

 

So, the question is whether a specific honor breaks down a trump fit play wildly without some cross-ruff trick, I suppose. For, any number of cross-ruffs can live or die based on whether they can or cannot start with the stiff catching an Ace.