## Question 25

Congratulations to a fine book, but something is missing. Look at the following deal, from Bergen: Declarer Play etc., p 144.

 A 10 8 A 6 4 A 9 7 5 4 3 6 J 9 6 Q 7 K Q J 9 10 8 7 3 2 Q 10 K J 8 2 Q 10 9 4 K J K 5 4 3 2 5 6 A 8 7 5 3 2

North-South have a fit in spades, a SST of zero (three singletons) and WP 19. So they should win 13 tricks. But it is impossible to win more than 10 tricks.
What is wrong in the calculation? Are there adjustments that are not shown in your book?

Regards

Stig Bryde Andersen
Brovst, Denmark

If you click on the link 'Errors' (under 'What's important?') in the left frame, you can read about some factors which may cause our formula to come up with a false result.
The first thing is that even though North-South have 19 HCP, they have less than 19 WP, because they have too many aces. That may surprise you, since we have all been taught that aces and kings are undervalued in suit play.
Working points (WP) is the sum of our trick-taking honors, so in this example North-South have five trick-taking honors: the four aces and the trump king. Usually, 19-21 WP will be the same as six trick-taking honors, but it might be seven or five. When you have seven trick-taking honors, you will get one extra trick, but if you only have five, you will take one too few. That is the case here.
The second missing trick comes from the very low SST. When one of the sides has an SST of 0 or 1, they will sometimes take less tricks than estimated if they (a) don't have enough tricks to ruff all their losers (like here), and (b) no strong side-suit taking tricks (they don't have that either).
Another thing is that there is no way for estimating a lucky split. If West gets one of East's hearts in exchange for a club, North-South take 12 tricks by ruffing out clubs. But when clubs are 4-2 (with two trumps in the short hand), there are only 10 tricks.