## Question 27

I'm now on my second reading of I Fought The Law and I ask your comments on the following hand from the Master Solvers' Club in The Bridge World (May 2004, problem A):

K Q 9 8 7 3
A J 10
Q 9 8 6

The auction proceeds:

 RHO You LHO Part 1NT 2 3 4 5 ?

Per the conditions of the problem, 1NT appears to be strong, 2 = natural, 3 = both minors/short spades.
How can one go about using the SST + WP formula here? I have a void in clubs and partner probably has a singleton diamond, so my SST = 1. I have no idea whether my partner's points are in spades, hearts, or clubs but I do know that if they are in clubs they don't help me on offense.
I suppose that one could argue that if partner's points are in hearts they are valuable for offense or defense while if they are in clubs they are valuable for defense only. But give parter a yarborough like

x x x x
x x x
x
x x x x x

and as long as spades are 2-1 and the heart honors are split, it looks like I can take 5 spades, 2 hearts, and 2 ruffs = 9 tricks, whereas the opponents will lose only 2 hearts = 11 tricks. Give partner

x x x x
K x x
x
x x x x x

and assuming that the Q can be located then I'll take 5 spades, 3 hearts, and 2 ruffs = 10 tricks whereas the opps may lose 3 hearts = 10 tricks.
But obviously if partner has

x x x x
x x x
x
Q J 10 x x

then we do far better to defend than declare.

Since partner did, apparently, have something along the lines of that hand, it would have worked better for partner to make a non-fit bid of 4, but that would also have given away the show to the opponents.
Bottom line is that while I can easily see how to estimate SST during the auction, the estimation of WP seems far more nebulous (except as it relates to holding like Qx in their suit or AQJ when partner has denied length in the suit) and would function best as a post-mortem analytical tool.

Sincerely,

Henry Sun
Benicia, CA

The hand you present is difficult because it is possible both that neither side can make a 5-level contract or that both sides can. It is further complicated by the fact that 5 may be a good save against 5, or that 5 is a save against 4. As you write, it largely depends on where partner has his values (if any).
You ask 'How can we use the WP + SST formula to decide what to do?'. The answer is that we can use it to estimate what partner needs for us to take nine, ten or 11 tricks. Assuming our SST to be 1, we need roughly 16 WP for 11 tricks, 13 WP for ten tricks and 10 WP for nine tricks. If we assume South has 10 WP, it suggests nine tricks opposite a yarborough, which is a correct estimation if partner has your first example hand.
It is also possible that North has a diamond void. If so, our SST is 0, which gives us the potential for another trick (but no guarantees: on a trump lead, North's 4-3-0-6 yarborough won't take more tricks than when he is 4-3-1-5). In the Errors section, you can see that a very low SST often overvalues your tricks if you have no strong side suit.
Therefore, we can conclude that bidding 5S is likely to lead to a minus result.
If we do the same estimation for the opponents, we know that their SST is likely 4 (3-3-4-3 opposite 1-3-4-5, e.g.) or 3 (3-2-5-3 opposite 1-3-4-5, e.g.). In the first case they need 26 WP, in the second they need 23 WP – and all this assumes that the breaks are normal. Here, we can see that both minors are breaking badly and that we have a good defensive holding in hearts behind the stronger opponent (if partner has the heart queen, declarer's king of hearts isn't the 3 WP he thought it was). Therefore, it is likely that they will lose (at least) one extra trick due to the bad breaks, so that they will need 29 or 26 WP, respectively, to make their game. Can we expect them to have that much?
We think this 'bad breaks argument' is so strong that we vote for double. After all, it is possible that South can defeat 5 without help from North. And with a little luck (like North's having the ten or the jack singleton in diamonds plus one stopper in clubs), we might defeat them two or more tricks.