Horrendous

	A 8 7 3
	5 4 3
	8 6 5
	8 7 4
K Q		J 10 5 2
A K J 9 6		Q 10 8 7
A K Q 2		J 7 4
3 2		A K
	9 6 4
	2
	10 9 3	Vul: None
	Q J 10 9 6 5	Opening lead: K of spades

North	East	South	West
Pass	Pass	3	4
5	5	Pass	6
Pass	Pass	7	Pass
Pass	Dbl	All	pass

What on earth! (Again). This hand would qualify for Counting, which would mark the first time I suggested a declarer would need mathematics higher than 13, as well as the first time I'd introduced counting into Bidding. Well, lemme see. Assuming E-W can make their non-vul slam, how many tricks could N-S afford to go down? And how many could they justly count on? Remember the rule of 2 & 3?

Well, the rule of 2 & 3 holds that not-vul, you might justly overbid your hand by 3 tricks, which would cost 500 doubled (as would a vulnerable overbid by two tricks doubled). So it would seem that South has four tricks coming, and North, with an ace and otherwise undistinguished hand (on 4-3-3-3 distribution) has one. Which adds up to 5, I believe. And strangely, that's just what South wound up with.

And now we come to a different type of counting: how many tricks do N-S need for a good sac, or in other words, how many winners do they need? That shouldn't be difficult. If E-W can make their slam, their non-vul slam, N-S can afford to go down only 4 tricks (minus 800). Which is to say, they need 9 winners for that 7 club bid. For it's a serious mistake to think that you can afford almost anything if you're not vul and they've got slam. For you can't.

This pair wound up with the 5 winners, the predictable 5 winners, which in a seven bid means down 8! Which comes out to minus 2000 when those guys can't even make 1000! Not a very wise idea.