Main     Puzzle 8K41 by Richard Pavlicek    

Wrong-Sided Notrump

Timothy Tenace was visibly agitated as he entered the Puzzlers Anonymous meeting. “I know we’re here to be cured, but I have one last puzzle for you: Last night on Board 7, Grover used that fourth-suit crapola to make me play 3 NT, and I couldn’t make it with the defense I got. Just as I suspected, the hand record showed it was makable by North only. Funny thing, the East-West hands were as flat as a pancake — not even a doubleton. The hand record also showed at least one other makable game.” Timothy then wrote on the chalkboard:

Timothy Tenace
3 NT S A 10 4
H A J 8 7 6 5
D K 10
C 8 3
Both Vul


West

Pass
Pass
Pass


Grover
North

1 H
2 S
3 NT



East

Pass
Pass
All Pass


Timothy
South
1 D
2 C
2 NT



 
Table


 
S J 8 6
H 3
D A Q 8 6 4
C A Q 6 5

Assign the missing cards to West and East to create a deal consistent with the dialogue. Multiple solutions exist. Contest tiebreakers were (1) the nearest-to-equal East-West card sums, and (2) the fewest makable games for North-South.

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Tim Broeken Wins Again!

In May-June 2011 this puzzle was presented as a contest, with 57 participants from 22 locations. Thanks to all who entered, and congratulations to the 16 who produced a layout where 3 NT makes by North but not by South. All but three equalized the West and East card sums, so the significant tiebreaker was the fewest makable games for North-South. (Remaining ties are resolved by date and time of entry.) Only three found the optimal solution to limit N-S to three games.

Another record broeken, as Tim Broeken wins for the second month in a row, and for the third time in this contest series. Hmm… Timothy Tenace, Timothy Broeken… I think I see it all now. Or maybe his secret to success is living in the lowlands — I said to the levee — so I now have a plan to cure the nether regions of puzzle mania. I won’t divulge the details, but it has to do with sea kelp. Man the dikes!

RankNameLocationW-E SumsGames
1Tim BroekenNetherlands100-1003
2Pavel StrizCzech Republic100-1003
3Jonathan MestelEngland100-1003
4Bozidar PutCroatia100-1004
5Manuel PauloPortugal100-1005
6Charles BlairIllinois100-1005
7Simon CreaseyEngland100-1005
8John ReardonEngland100-1006
9Dean PokornyCroatia100-1006
10Jim MundayCalifornia100-1006
11Edouard BonnetFrance100-1007
12Brian WeikleMinnesota100-1007
13David BrooksAustralia100-1007
14Jonathan FergusonTexas99-1017
15Richard SteinCalifornia89-1116
16Cenk TuncokMassachusetts87-1136

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Solution

Superficially it would seem that 3 NT is better played by South to shield the C A-Q from the opening lead. (The C K must be offside, else declarer would have nine top tricks given a 3-3 diamond split.) But Timothy says otherwise; and who are we to argue with such a handsome nerd? If only North can make 3 NT, what condition would be favorable to North but unfavorable to South? Aha! Both spade honors with East, as 12 of the 16 successful solvers prescribed. A typical construction:

3 NT S A 10 6
H A J 8 7 6 5
D K 2
C 8 4
Trick
1. W
2. E
3. E
W 1 L 2
Lead
S 3
C J
H 2
2nd
6
5
4
3rd
Q
3
K
4th
5
4
A
S 9 7 4 3
H K 10 9
D 9 7 6
C K 7 3
Table S K Q 2
H Q 3 2
D J 8 5
C J 10 9 2
John Reardon
England

S J 8 5
H 4
D A Q 10 4 3
C A Q 6 5

To defeat 3 NT by South, a triple attack is required: West must lead a spade (ducked); East must shift to a high club (shrewdly ducked) and then a low heart is led to the king. Note that if East leads a second club, declarer can duck again (or ace then low) to establish the queen. With the three-prong attack, declarer cannot develop a ninth trick without the defense winning five.

If North declares 3 NT, a spade cannot be led (easy ninth trick). If East starts the C J, declarer ducks to leave the defense stymied; a heart shift now nets the defense only four tricks as declarer sets up the C Q. East does best to start a diamond to attack declarer’s communication; but the 10 is won in South, and a heart is led to the jack and queen; East cannot continue diamonds (hearts set up) so shifts to the C J (ducked); then any lead allows declarer to develop a ninth trick.

Hyper-Moysian fit, anyone? A game in spades must always make on this puzzle regardless of the honor distribution, declarer and opening lead, given that both West and East are 4-3-3-3. Declarer has six top tricks (one spade, one heart, three diamonds, one club) and can ruff twice in each hand for 10. Note that North’s second club is discarded on the third diamond to prepare a crossruff of clubs and hearts.

Tim Broeken: So if you don’t like partner bidding the fourth suit to let you play 3 NT, just raise the fourth suit to game!

Charles Blair: Hurray for 3-3 fits! Is Timothy’s partner Grover Grosvenor?

How about other games? Four hearts by North makes; C J lead (best) is won by the ace, and a heart is led to the jack and queen; C 10 ducked; S K to the ace; H A; diamond to 10; club ruff; D K to ace; then D Q and C Q allow declarer to pitch both spade losers as only West can ruff with the high trump. Five diamonds by North or South also makes, since the defense cannot dislodge North’s S A (any spade lead is ducked) so hearts are easily established with two ruffs.

Altogether, the above solution allows six makable games: 3 NT North, 4 H North, 4 S North and South, 5 D North and South. The next solver improved on this, eliminating the heart game, by moving the D J to West, which also required a subtle shift of the C 7 to let North make 3 NT, as shown below:

Timothy Tenace
3 NT (N) S A 10 6
H A J 8 7 6 5
D K 2
C 8 4
Trick
1. E
2. N
3. S
4. S
5. S
W 5 L 0
Lead
D 8
D 2
D Q
D 10
D 4
2nd
3
5
J
S 3
S 4
3rd
7
A
H 5
H 6
C 4
4th
K
9
6
C 7
H 2
S 9 7 4 3
H K 10 9
D J 9 7
C K 3 2
Table S K Q 2
H Q 3 2
D 8 6 5
C J 10 9 7
Manuel Paulo
Portugal

S J 8 5
H 4
D A Q 10 4 3
C A Q 6 5

After a diamond lead, declarer cannot win the 10 (as on previous deal) but must run diamonds to reach the ending shown. Next a heart is led to the ace (East cannot gain by unblocking) then a heart goes to the queen. The C J shift is covered by the queen and king; then West cannot cash the H K but must shift to a spade, ducked to the queen. East’s only safe exit is a club, but ace and a club endplays him in spades. If West held the C 7, East could escape the endplay by unblocking in clubs. South
leads
S A 10 6
H A J 8 7
D
C 8
S 9 7
H K 10 9
D
C K 3 2
Table S K Q 2
H Q 3
D
C J 10 9
S J 8 5
H 4
D
C A Q 6 5

If North plays 4 H, the mislocated D J prevents the entry-gaining finesse of the D 10 previously described, leaving no successful route to 10 tricks. Trust me, or you’ll soon be joining the ranks of Timothy with the prognosis incurable.

“Five diamonds can be defeated!” –Timothy Tenace

So far we’ve determined that three games cannot be defeated per the puzzle conditions: 3 NT by North and 4 S by North or South. Manuel’s solution neatly defeats 4 H; but what about the pesky game in diamonds? The only way to defeat 5 D is for the spade honors to be split and for West to lead his honor, or for East to lead a low spade if North declares. While costing a spade trick, this kills North’s entry, preventing declarer from using the long hearts, after which two clubs may eventually be lost.

Alas, split spade honors wreaks havoc with the puzzle’s basic requirement. West would have no attack suit against 3 NT, so it seems incredulous that South must fail while North succeeds. But that’s the stuff puzzles are made of. Only three solvers found the delicate layout where all games fail, except 3 NT by North and the impregnable 4 S. Here it is, with the play shown for North declaring 3 NT:

3 NT (N) S A 10 6
H A J 8 7 6 5
D K 2
C 8 4
Trick
1. E
2. E
3. E
4. S
5. N
6. S
7. S
8. S
W 6 L 2
Lead
C J
C 10
C 3
D 3
D 2
D Q
D 10
D 4
2nd
5
6
A
5
9
7
H 2
H 9
3rd
2
7
9
K
A
H 6
H 7
S 6
4th
4
8
H 5
8
6
J
S 2
S 3
S Q 9 7
H Q 9 2
D 7 6 5
C K 9 7 2
Table S K 4 3 2
H K 10 3
D J 9 8
C J 10 3
Tim Broeken
Netherlands

S J 8 5
H 4
D A Q 10 4 3
C A Q 6 5

Declarer must duck two club leads, then run diamonds. Note that West cannot pitch a spade (else a spade can be established) so must come down to a blank H Q. East must keep three hearts (else a heart can be established) so must come down to a doubleton spade. Declarer next wins both major aces and exits with a spade. If East wins the S K, he is endplayed and must give North the last trick with the H J. If East unblocks the S K to avoid the endplay, West must give South the last trick with the S J.

So if North can make 3 NT after a vicious club attack, why can’t South with C A-Q protected? Curiously, more vicious than a club attack by East is a heart attack by West — better known as “the big one” to Lou Costello. West leads the H 2, which East wins as cheaply as possible (assuming declarer ducks) and shifts to a club honor. Whether declarer wins the C A or holds up, he cannot establish hearts without losing five tricks (note the defenders control who wins their second heart trick). If declarer tries to exert pressure by running diamonds, the lack of communication in hearts proves fatal.

Pavel Striz: An interesting and funny deal! In addition to honors, S 9-7 uniquely protect spades, H 9-10 protect hearts, and C J-10-3 (or J-10-2) helps a lot… The only other makable game is 4 S from both sides — no wonder Timothy was so agitated!

The optimal solution found by Tim Broeken, Pavel Striz and Jonathan Mestel is unique. Strategically, S K-Q, H 3-2, C 3-2 and every diamond can be swapped, but the highest options must go to East to achieve equal card sums. If East is given H K-10-9, the status of 3 NT is unchanged; but 4 H by South, however unlikely, is now makable — exactly how is left to anyone whose puzzle mania is still not cured. Now in intensive care:

Jonathan Mestel: I’ve looked at game from both sides now, in red suits too; and still, somehow, it’s cards’ illusions I recall. I really don’t know bridge at all.

On that note I declare this P.A. meeting adjourned. Good night, Timothy.

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© 2011 Richard Pavlicek