S/—		♠	K 6 2
		♥	A 2
		♦	8 3 2
		♣	A K 10 9 8
		♠	A 7 5 4
		♥	K 9
		♦	K Q J 5
		♣	J 5 4

West leads the ♥5.

Declarer counts six top tricks: the ♠AK, the ♥AK and the ♣AK. So he has to develop three extra tricks. Is he going to set op the clubs or the diamonds?
Suppose he finesses for the ♣Q. If he succeeds, nine tricks are there. But if East wins with the ♣Q, declarer is almost certainly defeated:

		♠	K 6 2
		♥	A 2
		♦	8 3 2
		♣	A K 10 9 8
♠	J 9 8			♠	Q 10 3
♥	Q 10 7 6 5			♥	J 8 4 3
♦	9 7 4			♦	A 10 6
♣	7 6			♣	Q 3 2
		♠	A 7 5 4
		♥	K 9
		♦	K Q J 5
		♣	J 5 4

...since EW set up their heart suit. As soon as declarer plays a diamond — in order to set up his ninth trick; after all he has only eight tricks to take — the defence wins with the ♦A and cashes at least three more hearts (in this case exactly three).
So by finessing for the ♣Q declarer puts all his eggs in one basket: a 50% chance of success.

Declarer can do much better by combining chances: he wins the lead with the ♥A, cashes the ♣A (you never know...) and plays a diamond from dummy. If East has the ♦A (a 50% chance), the contract is made:
- If East wins with the ♦A, South has three master diamonds, enough for his contract.
- If East ducks, South scores the ♦K. With a diamond trick in the bag and still a heart guard, declarer now changes tack by finessing for the ♣Q, not caring too much whether the finesse wins or fails.
If it wins, he cashes the clubs and switches back to diamonds, making eleven (diamonds not 3-3) or twelve tricks.
If the club finesse fails (the actual layout), he safely makes nine tricks.

If, on a different layout, West wins with the ♦A and clears the hearts, declarer first tests the diamonds. If this suit is 3-3 (a 35.5% chance), he has nine tricks. If not, he tries his last chance: the club finesse (50%).

Following this line of play, declarer will only be defeated if West has the ♦A and the diamond suit is not 3-3 and East has the ♣Q. The total chance of defeat is 16.8%. If declarer suffers that fate, he has reason to complain...