Puzzle 8N79   Main

Bridge with the Abbott

  by Richard Pavlicek

In my continued efforts to stop David Bird from stealing my titles, I remind him that spelling counts. One ‘t’ doesn’t cut it; anyone with half a brain knows there are two, and I could claim three as in “Hey Abb-b-o-t-tt.” And don’t think for a minute that adding ‘Monsignors’ to your stories changes anything! My attorneys have issued a cease-and-desist order, so be warned. Keep your beak off my puzzles! If these thefts continue, you may be the next Birdman of Alcatraz.

While surfing the Internet I came across the transcript of a 1938 radio broadcast from San Francisco — just a stone’s throw from Alcatraz, so pay attention, Mr. Bird. It was a half-hour comedy show featuring Abbott and Costello. About midway through my interest was piqued by the following dialog, highlighting Bud Abbott’s penchant for bridge:

[Bud] Lou! Let me give you a hand.

[Lou] No need to clap. We’re friends.

No, you nitwit. I mean a bridge hand.

Thank you. I get nervous on the Golden Gate.

You hold S A-J-10 H K-J-10 D 10-7-5-3 C 10-9-2.

That’s too heavy. I can’t even hold my breath.

What do you lead against six notrump?

First a right cross, then an uppercut.

No, no… Which card do you lead?

Diners Club is all that I have.

Never mind. It was cold.

I knew that!

Scribbled in Abbott’s handwriting in the margin was a Yarborough, presumably partner of the above. Beneath it was the note “no long suit in dummy, no accidents,” but that was it. No other bridge information existed, though one could surmise that whatever the missing hands were, Lou would be the perfect dummy. A partial construction shows:

6 NT South  ?
S A J 10
H K J 10
D 10 7 5 3
C 10 9 2
Table S 8 6 4 3
H 9 8 7 3
D 9 8
C 8 6 4
West leads  ?

Having occurred some 80 years ago, and with all parties long passed, hopes of further recovery are slim. Aside from Abbott’s allusion to 6 NT being cold and his curious note, which I’ll interpret as “no 5+ card suit in dummy, and no singleton or void anywhere,” the actual deal may be lost for eternity. Unless… unless… you can help restore this piece of history.

Construct a South hand with which 6 NT can be made against any defense.

Many solutions exist. Because Abbott loved high cards (and the dummy Costello couldn’t tell an ace from a deuce) tiebreaking goal is to make the South hand as strong as possible, first by HCP, and secondarily by the sum of all card ranks.

6 NT South North will
get what

S A J 10
H K J 10
D 10 7 5 3
C 10 9 2
Table S 8 6 4 3
H 9 8 7 3
D 9 8
C 8 6 4
Available cards:
West leads S   
S K Q 9 7 5 2
H A Q 6 5 4 2
D A K Q J 6 4 2
C A K Q J 7 5 3

To see if your solution is successful click

Duncan Bell Wins

This puzzle contest, designated “March 2018” for reference, was open for over a year. Participants were limited to one try, unlike my usual contests allowing entries to be revised with only the last one counting. Participation was abundant, of which 37 solutions were correct. Tiebreakers were (1) most South HCP, (2) greatest South rank sum, and (3) earliest date-time of submission, in that order of priority.

Congratulations to Duncan Bell, who was first of six solvers to submit the optimal solution (South with 29 HCP and 153 rank sum). Duncan’s victory is no surprise, having previously won The Twelve of Spades, Just Another Zero and High Stakes Rubber, besides regular high placings when he didn’t win.

Winner List
RankNameLocationSouth HCPSouth Sum
1Duncan BellEngland29153
2Martin VodickaSlovakia29153
3Dan GheorghiuBritish Columbia29153
4Konrad MajewskiPoland29153
5Tim BroekenNetherlands29153
6Jean-Christophe ClementFrance29153
7Nigel GuthrieScotland29152
8Gordon HoScotland29152
9Samuel PahkMassachusetts29152
10Alessandro RapuanoItaly29151
11Tom SlaterEngland28151
12Eric LanierWisconsin28151
13Jacco HopNetherlands28151
14Gonzalo GodedSpain28151
15Walter LeeHong Kong28151
16Foster TomBritish Columbia28151
17Eddy ChoiHong Kong28150
18Leif-Erik StabellZimbabwe28150
19Sherman YuenSingapore28150
20Radu VasilescuPennsylvania28150
21Franco BaseggioNew York28150
22Carl Heinz RosenthalAustria28150
23Charles BlairIllinois28148
24Gareth BirdsallEngland28148
25Nicholas GreerEngland27150
26Ryou NijiMichigan27150
27Ufuk CotukEngland27149
28Richard SteinWashington26149
29Nao TabataFrance26144
30Joe LafortuneConnecticut26144
31Joe ClarkEngland26143
32Harrison LubaMassachusetts25147
33Meelis TiitsonEstonia25143
34Bob OndoMichigan25143
35Levi KatrielCalifornia25143
36Mark RaphaelsonFlorida25142
37Thijs EngberinkNetherlands25140

Puzzle 8N79   MainTop   Bridge with the Abbott


This construction (28 HCP, 150 sum) was the most common theme of successful solvers:

6 NT SouthS Q 7 5 2TrickLead2nd3rd4th
H 6 5 4 21 WC 1034Q
D 4 22 SS 910Q3
C J 5 33 NC J672
S A J 10TableS 8 6 4 34 NC 58K9
H K J 10H 9 8 7 35 SC AH 10H 2S 4
D 10 7 5 3D 9 86 SD A328
C 10 9 2C 8 6 47 SD K549
S K 98 SD Q7H 4S 6
H A Q9 SD J10H 5H 3
D A K Q J 610 SD 6S JS 2H 7
Lead: C 10C A K Q 711 SS KA58
West is endplayed

Carl Heinz Rosenthal: On a minor-suit lead, win in hand and run the S 9 through West, North covering with the queen; then run both minors to strip-squeeze West. If instead West rises with the S A over South’s nine, simply preserve the C J as an entry to win the S Q.

Curious that Carl includes his middle name… a solution with 57 varieties?

A slight improvement (28 HCP, 151 sum) was submitted by Tom Slater and seven others:

6 NT SouthS Q 5 2TrickLead2nd3rd4th
H 6 5 4 21 WC 1034J
D J 6 4 22 SD A328
C 5 33 SD K549
S A J 10TableS 8 6 4 34 SD Q76S 3
H K J 10H 9 8 7 35 SS 710Q4
D 10 7 5 3D 9 86 ND JS 6S 910
C 10 9 2C 8 6 47 NC 56Q2
S K 9 78 SC A9H 28
H A Q9 SC KH 10H 4H 7
D A K Q10 SC 7S JH 5H 8
Lead: C 10C A K Q J 711 SS KA58
West is endplayed

Unblocking diamonds, followed by a spade to the queen, leads to the same strip-squeeze against West.

A greater improvement (29 HCP, 152 sum) was submitted by Nigel Guthrie and two others:

6 NT SouthS 9 5 2TrickLead2nd3rd4th
H 6 5 4 21 WC 1034K
D 4 22 SS KA23
C Q 7 5 33 WC 956A
S A J 10TableS 8 6 4 34 SD A328
H K J 10H 9 8 7 35 SD K549
D 10 7 5 3D 9 86 SD Q7H 2S 4
C 10 9 2C 8 6 47 SD J10H 4H 7
S K Q 78 SD 6H 10H 5H 8
H A Q9 SS Q1056
D A K Q J 610 SC J2Q8
Lead: C 10C A K J11 NC 7S 8S 7?
West is squeezed

In this case West can win the first spade to avoid the throw-in; but with clubs 3-3 declarer has 11 tricks, and West is eventually squeezed in the major suits. Note the Vienna coup at Trick 9.

Gordon Ho: The S 9 must be in dummy to execute the squeeze.

The optimal solution (29 HCP, 153 sum) was found by our winner Duncan Bell and five others:

6 NT SouthS 9 5 2TrickLead2nd3rd4th
H 6 5 4 21 WD 328K
D Q 6 4 22 SS KA23
C 5 33 WD 549A
S A J 10TableS 8 6 4 34 SC A234
H K J 10H 9 8 7 35 SC K956
D 10 7 5 3D 9 86 SC Q10H 28
C 10 9 2C 8 6 47 SC JH 10H 4H 3
S K Q 78 SS Q1054
H A Qcontinued below…
Lead: D 3C A K Q J 7

Dan Gheorghiu: Declarer succeeds with a repeating triple squeeze.

After cashing four clubs and the S Q (Vienna coup) the following ending is reached:

NT win allS 9TrickLead2nd3rd4th
H 6 59 SC 7H JH 5H 7
D Q 610 SH AK68
C11 SH Q?
S JTableS 8 6West is squeezed
H K JH 9 8 7
D 10 7D
S 7
South leadsC 7

At Trick 9 West is triple-squeezed. Suppose he lets go a heart as shown; the H A then drops the king, and the H Q squeezes him in the pointed suits. If West instead unguarded diamonds, declarer would overtake the D J with the queen; then the D 6 squeezes West in the majors. Similar if West unguarded spades.

Monsignor Moments

The Abbot: What a pathetic attempt to capitalize on my fame! Do you really expect anyone to believe this, when you can’t even spell Abbot correctly?

Brother Xavier: How dare you plagiarize my partner! Mr. Bird will see you in court!

Mark Raphaelson: Who’s on first?

Puzzle 8N79   MainTop   Bridge with the Abbott

Apologies to David Bird, and acknowledgments to
Lou Costello (1906-59) and Bud Abbott (1897-1974)
© 2021 Richard Pavlicek