	NORTH AT64 AQ T86 AT74
QJ53 K7653 Q3 32		8 JT92 KJ7 QJ965
	K972 84 A9542 K8

Dealer: North Vul: Both MP Scoring

West

Pass
Pass

North
1
2
Pass

East
Pass
Pass
Pass

South
1
4

Trump Promotion

West led what looked like a 4th best heart.
Dummy's

Q won and there appeared to be just two diamond losers if trump divided.
A 'safety play' looked possible: Win a top trump and play toward the other hand's 2nd highest trump.
How does one choose after just trick one which of the opponents is more likely to have the spade length?
North South held 4 hearts combined, the outstanding hearts being:

KJT976532
One of the defenders would have at least 5 hearts and the other at most 4.
Which hand was more likely to have 5 or more?
If West had led the

2 playing 4th best leads, East would have at least 5 hearts.
Then West on average would have more "vacant spaces" or non-hearts in which to hold spade length.
If leading the

3, West might or might not have a 5th heart which would of necessity include the

2.
What about the

5 lead? There would be lower the

3 and

2 as possible 5th and maybe even 6th hearts! Then again, from

532 the

5 might be led (6 hearts then with East that didn't preempt.)
Unfortunately declarer rightly pegged West with longer hearts than East and at trick two won the

A.
West showed on the following spade led then towards the

K.
West was entitled to the

Q and

J. Two diamond tricks seemed unavoidable.

There was a winning line try on a different layout featuring an obscure small heart lead from an original

K5.

	NORTH T6 A T86 AT74
QJ K J73 Q963		JT976 KQ J52
	97 8 A9542 K8

After winning the

K, cash

K, then

A, ruff a club,

A and ruff a club.
Cash

A and exit a diamond to an East that must conced a trump promotion to dummy. Any of the top two of the outstanding three diamond honours with East would suffice.