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Last updated: August 29, 2007

Deal 1

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

Please, analyze this deal for me. It's from the final of the 2002 Rosenblum Cup. I find it odd that both sides will do better in their second longest suit than in their longest.

"Curious"

Answer

If North-South play in hearts, their nine-card fit, they can only take eight tricks if the defenders start with three rounds of spades (promoting two trump tricks). But if they play in clubs, their eight-card fit, they take ten tricks with the aid of a heart finesse. One less trump, but two more tricks.

If East-West play in diamonds, their ten-card fit, a double dummy defense holds declarer to eight tricks: a low club to North, two heart tricks, a second club trick and a third club ruffed with the queen of diamonds. If East-West prefer their seven spades, they take nine easy tricks. Three less trumps, but one more trick.

So if the sides prefer their longest fits, we see that 19 trumps produce 16 total tricks (-3). But if they play in their second longest fits, 15 trumps take 19 total tricks (+4). This is a good illustration to what you should know by now, namely that how many tricks you can take is not related to how many trumps you have.

What would our formula say? Let's see.

Spades: Since diamonds are 2-1 but the heart king useless, East-West have 20 WP and an SST of 4. Nine tricks.

Hearts: The heart king is doubleton onside, so it looks like North-South have 17 WP and an (adjusted) SST of 2, or ten tricks. But the 5-2 spade split and South's bad trump spots mean the defenders can promote two trump tricks. As we have said before, "the formula predicts how many tricks one side will take if nothing bad happens", and here spades 5-2 was such a "bad thing".

Diamonds: East-West have an (adjusted) SST of 3. Trumps are 2-1, so with the heart king offside it looks like they have 17 WP (10 in diamonds, 7 in spades), or nine tricks. But since the defenders can arrange a trump promotion, another "bad thing" happens for the declaring side, which reduces their tricks to eight.

Clubs: With the heart king doubleton onside, North-South have 20 WP and an SST of 3. That is ten tricks. Note that if the defenders lead trumps, South can't enjoy North's singleton diamond. But he doesn't need to, since North's heart suit provides more tricks than needed. The only effect North's singleton had was to stop the defenders from cashing two quick tricks in diamonds.

Deal 2

I have an interesting hand that I'd like you to look at. I think yourformula overvalues it. The hand is from The Complete Book on Balancing by Mike Lawrence, page 199, 4th hand.

I will relist it here:

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

West	North	East	South
		1	pass
1NT	pass	pass	2
3	3	pass	3
pass	pass	pass

I think your formula would suggest bidding 4S. WP = 6+7+7 = 20. In fact it may be higher because the heart suit may drop an honor, or even 2.

SST = 0+2 = 2, with -1 for the 2nd Doubleton = 1. 13-1 = 12 tricks, 20 WP = no adjustment

I think the reason it over values the hand is that the clubs must beruffed with high trumps. This is a case where having a 9th trump isextremely valuable.

The solution is to use judgment, and not just plug and chug a formula.The North player should realize he will have to ruff Clubs with hightrumps, thus incurring additional trump losers. Therefore he shoulddown grade his hand.

I find the formula helps me tie together what I've read in the 3 books"The Complete book on" Hand Evaluation, Overcalls, and Balancing. Ithelps me try and visualize the distributions in the 4 hands, rather thanjust use a generic formula.

Thank you,
Brandon Einhorn

Answer

As we have written elsewhere, our formula predicts how many tricks one side can take if nothing bad happens. And, indeed, if we only look at North-South's cards, they have the potential for ten or more tricks: on a non-club lead, hearts may be ruffed out, giving South four trumps, five hearts and two diamonds (11 tricks); and on a club lead, South can ruff three clubs in dummy to go with four red-suit tricks and three trump tricks in his own hand (10 tricks).

But if we view the situation from North, he knows his long side-suit won't produce lots of tricks after East's 1H opening bid. Furthermore, North doesn't have lots of trumps to ruff with. Therefore, he should downgrade his hands, just as you say. If you click on What's important->Errors in the left frame, you'll find a discussion on this very topic under the heading (2) Not enough potential. The deal above fits into this category (very low SST, not lots of trumps, no trick-taking side-suit).

It is also true that a fourth spade would be useful. And if North has 4-6-3-0 instead, the potential goes up so that the formula predicts accurately. On normal breaks, you can expect four trump tricks in hand, one or two club ruffs in dummy, five hearts and ace-king of diamonds. Once again, a bad split in trumps or hearts may mean North-South lose one (or two) of those tricks, but that doesn't mean the prediction is wrong.

Conversely, if North's distribution is 3-4-6-0, he can expect his long side-suit to be much more useful than when East has bid it. West's club bid also means that South isn't likely to have lots of wasted values opposite the void. Taking a shot at 4S in that scenario looks good to us.

Deal 3

I've been reading your book on the "new" Law. This hand came up this morning:

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

South	West	North	East
		1	pass
1	pass	2	pass
pass	DBL	pass	2
3	pass	pass	pass

Do you agree with the 3C bid? And should West have bid 3D over 3C?

I think it's a neat hand from the point of view of competitive bidding.

Regards
Neil Hayward

Answer

We agree with 3C. South doesn't have much, but he has trump support in combination with distribution. Therefore, this hand will usually take as many tricks in a club contract as a 4-3-3-3 nine count.

Let's do the math. North may have more or less than 12 WP, but 12 is a good approximation, so assume he has that much. South has 4 WP (not counting the jack of spades), and his two doubletons means his side has an SST of at most 4. Since North didn't raise spades, which he might have preferred with three spades, South can expect North to have at most a doubleton spade and adjust his side's SST to 3. If North really has 12 WP, their 16 WP and an SST of 3 should give them good play for nine tricks.

Here, North has a singleton spade, so North-South's SST is as low as 2, but since they run into a bad trump split (two unexpected losers) they will only take nine tricks. As long as North plays carefully, he will either be able to ruff three losers in dummy, or (if the defenders give up one of their trump tricks) take two ruffs in dummy but only lose one trump trick.

We also agree with West's pass over 3C. The main reason is that he has no idea what his partner has. It is possible that East's distribution is 4-3-3-3 or 3-3-3-4, when going on to 3D on a 4-3 fit isn't likely to produce a good result.

East was also correct to pass out 3C. He has decent values, but a mostly defensive hand, and hopes of defeating 3C. With the same values, no club honors, and better distribution, say a major suit doubleton, East might have considered taking the push. Not now.

Deal 4

Here is a hand from the book Matchpoint Defense, by Jim Priebe (Masterpoint Press, 2006), that you might be interested in (page 36):

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

From a Law of Total Tricks standpoint, one would expect 11 + 10 = 21 tricks, and yet best defense allows both NS and EW to score 9 tricks = 18 total tricks, a 3-trick over-estimation by the Law. Switch the KS for the AH, however, and NS score 2 extra tricks for a total of 20, because the KH represents a control and allows NS to pitch the remaining heart loser on diamonds. Swith the AH and the KH and NS score 3 extra tricks for a total of 21.

Curiously, the SST + WP formula works exceptionally well with the EW hands (EW have 20 effective HCPs and a SS count of 4 = 9 tricks), but I don't think, in real life, the value of the AH in the N hand, or the KH with the A onside (modification #1) will be accounted for on this auction.

Sincerely,

Henry Sun
Benicia, CA

Answer

The SST + WP formula works fine for NS too. On the actual deal, the heart king and king-queen of diamonds are wasted, so they have only 17 WP and an SST of 3. That is four losers, or nine tricks.

If the heart king is onside, NS gain 5 WP, since both the heart king and the diamond queen become useful. In that scenario, they have 22 WP. Deduct one from your SST, and we have two losers, or 11 tricks.

If North has the heart ace, NS gain 9 WP, because now the jack of diamonds is also useful (we value all trick-taking cards lower than the queen to 3 WP each, so the precious jack isn't 1 WP, but 3 WP). With 26 WP and an SST of 3, the formula says "12 tricks". And so it is.

The difficulty with the deal is for North to know the value of his red honors. With four-card club support and fair values, most people would bid 5C over 4S. But the doubleton spade is a warning sign (a duplication when you know that South can't have more than two spades), just like the fact that most of North's strength is outside clubs. For that reason, you could easily sell us a card-showing double over 4S with that hand. But on the actual deal, it probably wouldn't matter, since South is likely to take out to 5C. And if NS choose to defend 4S, who are we to say it can't make? Give West one of East's diamonds in exchange for his club, and they even score an overtrick!

Deal 5

North dealer, East-West vulnerable

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

South	West	North	East
		2	pass
pass	3	pass	pass
3	pass	pass	4
pass	pass	pass

My partner complained about me not bidding 4H, and I said I did not have the distribution to do so, square hand, no ruffing power, needed too much luck (which most sorrowfully was there).

My calculation was that even if partner had a singleton and all our points were working, we had an SST of 4, and since the WP we had could at the best be 20, we should not be in 4H.

I asked quite a few people, almost everyone said I needed to bid 4H. So, where did I go wrong?

By the way, even before I read you book, when I used to rely more on the LAW, I would probably have either passed 2H or bid 3H as an extended preempt, but never consider 4H because of my shape.

Thank you,
Sara Bilgoray

Answer

We have nothing to complain about your bidding. And your calculation is correct. You expect an SST of 4, and if your partner has a maximum weak two-bid (9-10 WP and 6331/6322), you will take nine tricks, no more, no less. Opposite KQJ sixth of hearts, out, you won't even make 3H. Those who said you should have bid 4H were wrong.

But we do NOT like your partner's bidding. Her hand is simply too strong for a weak 2H – not because of her high-card points but because of her shape. A hand with 6-4 distribution and a strong four-card side suit is almost one trick better than the ordinary 6331 or 6322 distribution, which is the norm for a weak two-bid. Had you known your side's SST was 3, you would have at least invited game. But you shall expect an SST of 4, and proceed on that basis.

When you competed with 3H over 3C, we think your partner could have made up for her previous underbid by going on to 4H all by her own. The extra distribution is the key, not some number of high-card points or trumps.

Since your side can win eleven tricks in hearts, it means 4H is on even if we move one diamond to spades (or clubs). Why? The explanation is that with the HK onside, North's 9 HCP are the eqivalent of 12 WP. So, in effect you have 23 HCP and an SST of 4 in that scenario, which equals ten tricks. Sometimes things are like that, but there is no way for us to know when a suit like AQJ10xx is just as good as AKQJxx. Had North known that, she would have opened the bidding with 1H, not 2H.

Deal 6

Here's a deal (June 1972 Bridge World page 13),where it appears that BOTH the law of total tricks AND the SST+WP analysis comes up short:

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

From a Law perspective, there are 8 spades plus 9 hearts = 17 total trumps, but 20 total tricks. If one declares the EW hand in clubs, there are 9 clubs + 9 hearts = 18 total trumps, but 21 total tricks.

From a SST + WP perspective, whether one declares in spades or clubs, the SST + WP count remains the same since the short suits are in hearts and diamonds and hence are not affected by which black suit is trumps, and yet played in spades there are 10 tricks while played in clubs there are 12.

Would either of you care to comment on this hand? The reality of the situation is that playing in clubs means both there are more ruffing tricks (due to the 9 card fit that breaks 2-2 versus the 8-card fit that breaks 4-1) plus a useful discard on the long spade, but I'm not certain how that is accounted for in either model.

All the best,

Henry Sun
Benicia, CA

Answer

Thank you for the interesting deal.

Double dummy, EW can make both 6C and 6S, but NS can only take seven tricks in hearts (if East leads his singleton, he gets two ruffs).

So if EW and NS play in their longest fit, there are 18 trumps and 19 tricks. A Law fan would have expected more tricks, since (a) the deal is pure, (b) there is a void, and (c) the deal is a double fit. ALL these factors point towards more tricks than trumps, so one more trick than trumps sounds too little. Still, the Law ain't that bad here.

What about SST + WP?

EW have 23 HCP and an SST of 1, but since the HA isn't useful (6C still makes if it's the H2), we could either think of the deal being 20 WP + 1 SST, or 23 WP + 2 SST (pretending West has a heart to follow to the HA). In both cases, the formula says there should be 12 tricks. So it is.

NS have 16 WP and an SST of 3. That means they have the potential for nine tricks. The fact that they only take seven tricks is due to factors the method has no control over (defensive ruffs). When that happens, even our formula comes up with the wrong answer.