Deals From Our Readers


On this page, we will publish interesting deals, sent to us by readers, together with our analyses. If you have an interesting or troublesome deal to submit, send it to us in an e-mail.

We can't respond personally to everybody, but if we find your deal interesting we will post it here for everybody to view it, together with our analysis.  If you don't want your name published, just include a signature.

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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:  When North-South play in with 9-card fit, they take only 8 tricks if the defenders start with 3 rounds of spades (promoting two trump tricks).  But when they play in with 8-card fit, they take 10 tricks with the aid of a finesse in .   One less trump, but two more tricks.

     If East-West play in with 10-card fit, a double-dummy defense will hold declarer to 8 tricks: a low club to North, two heart tricks, a second club trick and a third club ruffed with Q.   If East-West prefer their 7 spades, they take 9 easy tricks.   Three less trumps, but one more trick.

     So, when both sides persist in their longest fits, we see that 19 trumps produce 16 total tricks (–3).  But when 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 K is useless, East-West have 20 WP and 4 SST.  Nine tricks.

Hearts: K is doubleton onside, so it looks like North-South have 17 WP and an (adjusted) SST of 2, or 10 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 K offside, it looks like they have 17 WP (10 in diamonds, 7 in spades), or 9 tricks.   But since the defenders can arrange a trump promotion, another "bad thing" happens for the declaring side, which reduces their tricks to 8.

Clubs: With K doubleton onside, North-South have 20 WP and an SST of 3.  That is 10 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 provides more tricks than needed.  The only role North's singleton played was to stop the defenders from cashing 2 quick tricks in .


Deal 2.

I have an interesting hand that I'd like you to look at.  I think your formula 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 4.
WP = 6+7+7 = 20.  In fact, it may be higher because AK 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 overvalues the hand is that the clubs must be ruffed with high trumps.  This is a case where having a 9th trump is extremely 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 high trumps, thus incurring additional trump losers.  Therefore, he should downgrade 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."   It helps me try and visualize the distributions in the 4 hands, rather than just 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 the North-South cards, they have the potential for 10 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 1 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, 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 AK.  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 4 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

SouthWestNorthEast
1 pass
1 pass 2 pass
pass DBL pass 2
3 pass pass pass

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

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

Regards
Neil Hayward


Answer:  We agree with 3.  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 J ), 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 9 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 3 losers in dummy, or (if the defenders give up one of their trump tricks) take two ruffs in dummy but lose only one trump trick.

We also agree with West's pass over 3.  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 3 on a 4-3 fit isn't likely to produce a good result.

East was also correct to pass out 3.  He has decent values, but a mostly defensive hand, and hopes of defeating 3.  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
Priebe's main point is that if West leads K against 5 on the auction,

SouthWestNorthEast
1 2
4 5 pass pass
pass
East can give a suit preference signal for hearts, thus allowing E-W to score the first 4 tricks.

     From the "Law of Total Tricks" standpoint, one would expect 11 + 10 = 21 tricks, and yet best defense allows both N-S and E-W to score 9 + 9 tricks = 18 total tricks, a 3-trick over-estimation by the Law.  
     Now, Switch K for A, and N-S score 2 extra tricks for a total of 20, because K represents a control and allows N-S to pitch the remaining heart loser on diamonds.
     Switch A and K, and N-S will score 3 extra tricks for a total of 21.

Curiously, the SST + WP formula works exceptionally well with the E-W 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 A in the N hand, or K with A onside (modification #1) will be accounted for on this auction.

Sincerely,

Henry Sun
Benicia, CA


Answer:  The SST + WP formula works fine for N-S, too.  On the actual deal, K and KQ are wasted, so they have only 17 WP and an SST of 3.  That is 4 losers, or 9 tricks.

     If K is onside, N-S gain 5 WP, since both K and the Q become useful.  In that scenario, they have 22 WP.   Deduct one from your SST, and we have 2 losers, or 11 tricks.

     If North has A, then N-S gain 9 WP, because now J 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 4-card club support and fair values, most people would bid 5 over 4.   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 lies outside clubs.  For that reason, you could easily sell us a card-showing double over 4 with that hand. But on the actual deal, it probably couldn't matter, since South is likely to take out to 5.  And if N-S choose to defend 4, who are we to say it can't make ?   Give West one of East's diamonds in exchange for his club, and they will 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
Playing this hand, partner and I set the opponents' 4 2 times, while we had a cold 4 contract.

My partner complained about me not bidding 4, 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 4.

I asked quite a few people, almost everyone said I needed to bid 4.   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 2 or bid 3 as an extended preempt, but never consider 4 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 9 tricks, no more, no less.  Opposite KQJ sixth, out, you won't even make 3.   Those who said you should have bid 4 were wrong.

But we do NOT like your partner's bidding.   Her hand is simply too strong for a weak 2 — not because of her high-card points but because of her shape.   A hand with a 6-4 distribution and a strong 4-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 3 over 3, we think your partner could have made up for her previous underbid by going on to 4 all by her own. The extra distribution is the key, not some number of high-card points or trumps.

Since your side can win 11 tricks in hearts, it means 4 is on, even if we move one diamond to spades (or clubs).  Why ?   The explanation is that with K onside, North's 9 HCP are eqivalent tp 12 WP.  So, in effect you have 23 HCP and an SST of 4 in that scenario, which equals 10 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 1, not 2.


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
Edgar's write-up reads, in part, "When the Vu-Graph audience saw the Closed Room result, +650 in 4, they anticipated another swing to PRECISION, for 6 looked both cold and biddable.  But the North-South competition in the Open Room turned the swing around... Declarer lost only two clubs, one spade, and one heart — +530 [in 3X]."

From a Law perspective, there are 8 spades plus 9 hearts = 17 total trumps, but 11 + 9 = 20 total tricks.  If one declares the E-W hand in clubs, there are 9 clubs + 9 hearts = 18 total trumps, but 12+ 9 = 21 total tricks.

From an 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 11 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, E-W can make both 6 and 6, but N-S can take only 7 tricks in hearts (if East leads his singleton 10, he gets two ruffs).

So, if E-W and N-S play in their longest fits, 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?

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

N-S have 16 WP and an SST of 3.   That means they have the potential for 9 tricks.  The fact that they take only 7 tricks is due to the factors (adverse ruffs) over which the method has no control.  When that happens, even our formula comes up with the wrong answer.


Copyright c 2007, Mike Lawrence & Anders Wirgren
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