[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

[kenrexford]

Flameous, on Sep 2 2009, 06:27 AM, said:

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.