Bridge with the Abbott

March 1, 2018 PavCo attorneys issue a cease-and-desist order to David Bird for stealing titles from this writer.

If that isn’t enough, his “Bridge with the Abbot” ripoff shows he can’t even spell. Anyone with half a brain knows that ‘Abbott’ is spelled with two tees, and some would claim three as in Hey…Abb-b-o-t-tt!

While surfing the Internet I came across a transcript of a 1938 radio broadcast from San Francisco — just a stone’s throw from Alcatraz, so pay attention, Mr. Bird, or you may bid its next Birdman. The half-hour comedy show featured Abbott and Costello. My interest was piqued by the following dialog, highlighting Bud Abbott’s penchant for bridge:

[Bud] Hey Lou! Let me give you a hand.

[Lou] No need to clap. We’re friends.

No, you nitwit. I mean a bridge hand.

Thanks, I get nervous on the Golden Gate.

You hold A-J-10 K-J-10 10-7-5-3 10-9-2.

That’s 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 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:

Radio show	?  ?  ?  ?
A J 10 K J 10 10 7 5 3 10 9 2		8 6 4 3 9 8 7 3 9 8 8 6 4
6 NT South	?  ?  ?  ?

Having occurred some 80 years ago, and with all parties long passed, hopes of 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… you can help restore this piece of history.

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

Multiple 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.

1. Before reading further, try to pick the winning South hand: 

A. K-Q-7 A-Q A-K-Q-J-6 A-K-J B. K-Q-7 A-Q A-K-J A-K-Q-J-7 C. K-9-7 A-Q A-K-Q A-K-Q-J-7 D. K-9 A-Q A-K-Q-J-6 A-K-Q-7

Quit

Top Bridge with the Abbott

Duncan Bell Wins

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

Rank	Name	Location	South HCP	South Sum
1	Duncan Bell	England	29	153
2	Martin Vodicka	Slovakia	29	153
3	Dan Gheorghiu	British Columbia	29	153
4	Konrad Majewski	Poland	29	153
5	Tim Broeken	Netherlands	29	153
6	Jean-Christophe Clement	France	29	153
7	Nigel Guthrie	Scotland	29	152
8	Gordon Ho	Scotland	29	152
9	Samuel Pahk	Massachusetts	29	152
10	Alessandro Rapuano	Italy	29	151
11	Tom Slater	England	28	151
12	Eric Lanier	Wisconsin	28	151
13	Jacco Hop	Netherlands	28	151
14	Gonzalo Goded	Spain	28	151
15	Walter Lee	Hong Kong	28	151
16	Foster Tom	British Columbia	28	151
17	Eddy Choi	Hong Kong	28	150
18	Leif-Erik Stabell	Zimbabwe	28	150
19	Sherman Yuen	Singapore	28	150
20	Radu Vasilescu	Pennsylvania	28	150
21	Franco Baseggio	New York	28	150
22	Carl Heinz Rosenthal	Austria	28	150
23	Charles Blair	Illinois	28	148
24	Gareth Birdsall	England	28	148
25	Nicholas Greer	England	27	150
26	Ryou Niji	Michigan	27	150
27	Ufuk Cotuk	England	27	149
28	Richard Stein	Washington	26	149
29	Nao Tabata	France	26	144
30	Joe Lafortune	Connecticut	26	144
31	Joe Clark	England	26	143
32	Harrison Luba	Massachusetts	25	147
33	Meelis Tiitson	Estonia	25	143
34	Bob Ondo	Michigan	25	143
35	Levi Katriel	California	25	143
36	Mark Raphaelson	Florida	25	142
37	Thijs Engberink	Netherlands	25	140

Puzzle 8N79 Main

Top Bridge with the Abbott

Solution

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

6 NT South	Q 7 5 2		Trick	Lead	2nd	3rd	4th
	6 5 4 2		1. W	10	3	4	Q
	4 2		2. S	9	10	Q	3
	J 5 3		3. N	J	6	7	2
A J 10		8 6 4 3	4. N	5	8	K	9
K J 10		9 8 7 3	5. S	A	10	2	4
10 7 5 3		9 8	6. S	A	3	2	8
10 9 2		8 6 4	7. S	K	5	4	9
	K 9		8. S	Q	7	4	6
	A Q		9. S	J	10	5	3
	A K Q J 6		10. S	6	J	2	7
	A K Q 7		11. S	K	A	5	8
				West is endplayed

Carl Heinz Rosenthal: On a minor-suit lead, win in hand and run the 9 through West, North covering with the queen; then run both minors to strip-squeeze West. If instead West rises with the A over South’s nine, simply preserve the J as an entry to win the 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 South	Q 5 2		Trick	Lead	2nd	3rd	4th
	6 5 4 2		1. W	10	3	4	J
	J 6 4 2		2. S	A	3	2	8
	5 3		3. S	K	5	4	9
A J 10		8 6 4 3	4. S	Q	7	6	3
K J 10		9 8 7 3	5. S	7	10	Q	4
10 7 5 3		9 8	6. N	J	6	9	10
10 9 2		8 6 4	7. N	5	6	Q	2
	K 9 7		8. S	A	9	2	8
	A Q		9. S	K	10	4	7
	A K Q		10. S	7	J	5	8
	A K Q J 7		11. S	K	A	5	8
				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 South	9 5 2		Trick	Lead	2nd	3rd	4th
	6 5 4 2		1. W	10	3	4	K
	4 2		2. S	K	A	2	3
	Q 7 5 3		3. W	9	5	6	A
A J 10		8 6 4 3	4. S	A	3	2	8
K J 10		9 8 7 3	5. S	K	5	4	9
10 7 5 3		9 8	6. S	Q	7	2	4
10 9 2		8 6 4	7. S	J	10	4	7
	K Q 7		8. S	6	10	5	8
	A Q		9. S	Q	10	5	6
	A K Q J 6		10. S	J	2	Q	8
	A K J		11. N	7	8	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 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 South	9 5 2		Trick	Lead	2nd	3rd	4th
	6 5 4 2		1. W	3	2	8	K
	Q 6 4 2		2. S	K	A	2	3
	5 3		3. W	5	4	9	A
A J 10		8 6 4 3	4. S	A	2	3	4
K J 10		9 8 7 3	5. S	K	9	5	6
10 7 5 3		9 8	6. S	Q	10	2	8
10 9 2		8 6 4	7. S	J	10	4	3
	K Q 7		8. S	Q	10	5	4
	A Q			continued below…
	A K J
	A K Q J 7

Dan Gheorghiu: Declarer succeeds with a repeating triple squeeze.

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

NT win 5	9		Trick	Lead	2nd	3rd	4th
	6 5		9. S	7	J	5	7
	Q 6		10. S	A	K	6	8
	—		11. S	Q	?
J		8 6		West is squeezed
K J		9 8 7
10 7		—
—		—
	7
	A Q
	J
South leads	7

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

Top Bridge with the Abbott

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