Puzzle 8F23 Main

Yarborough Fair

by Richard Pavlicek

Could a Yarborough win all 13 tricks in notrump? Yes, with cooperation.

Imagine you hold the South hand, playing in a fantasy world:

NT win 13 in	—		Trick	Lead	2nd	3rd	4th
South hand	A K Q J 10 5 4 3 2		1. W	2	A	A	6
	—		2. S	9	3	K	K
	K 10 7 3		3. S	8	4	Q	Q
A K Q J 10 5 4 3 2		—	4. S	7	5	J	J
—		—	5. S	5	2	3	4
—		A K Q J 10 5 4 3 2	6. S	9	10	2	10
Q 9 6 2		A J 8 4	7. S	9	J	10	2
	9 8 7 6			Claim the rest
	9 8 7 6
	9 8 7 6
West leads	5

South wins four spade tricks as West obligingly ducks, while North and East toss red honors. Similarly, South wins four heart tricks as North ducks, and four diamond tricks as East ducks. Somewhere in the process South leads the 5 to fetch the two, three and four! Unreal, but the issue is not what would happen in practice; only what is possible via legal plays.

A puzzle for you:

What is the weakest hand that could win all 13 tricks in notrump?

Assume South is declarer (though it doesn’t matter) and must win all 13 tricks in the South hand. You can dictate the play of all four hands to achieve the goal. Weakness is judged as the sum of South’s card ranks. The above sum is 95. How low can you go?

Tim Broeken Wins

Except for a few token Americans, it was a European sweep, mainly England and Netherlands, and I’ll tell you why. The Yarborough was named after the Earl of Yarborough, a British nobleman. How the Dutch got in the picture I’m not sure, but I vaguely remember an Earl of Holland. Maybe they got together for lunch with the Earl of Sandwich. Or was he a Duke? No, I seem to have landed in a 1960s time warp with Duke of Earl blaring from the jukebox, but I digress.

Winner List

Rank	Name	Location	Sum
1	Tim Broeken	Netherlands	54
2	Hendrik Nigul	Estonia	54
3	Pavel Striz	Czech Republic	54
4	Charles Blair	Illinois	54
5	Audrey Kueh	England	54
6	Howard Liu	Illinois	54
7	Dan Baker	Texas	54
8	Ufuk Cotuk	England	54
9	Jim Munday	California	54
10	Jonathan Mestel	England	54
11	Julian Wightwick	England	54
12	Mike Wenble	England	55
13	Reint Ostendorf	Netherlands	55
14	James Lawrence	England	55
15	Perry Groot	Netherlands	55
16	Jan Kuipers	Netherlands	55
17	John Auld	England	55

Puzzle 8F23 Main

Top Yarborough Fair

Solution

How low can you go? It was easy to improve on my example Yarborough (sum 95) at the puzzle introduction, but most reduced the sum only to the low 60s or high 50s. Any solver with experience at the game of Hearts, or the French game Misere, would have an edge here. Or as one of my regulars noted:

Charles Blair: “Anyone can raise a cheap cheer by chucking an ace. The true
master shows his worth in the manipulation of twos and threes.” -Victor Mollo

Charles was one of our token American solvers, not quick enough to stem the European tide. Perhaps if he changed his name to Tony?

Before showing the perfect solution, let’s look at a few that came within a hair. From Mike Wenble, England:

NT win 13 in	—		Trick	Lead	2nd	3rd	4th
South hand	J 10 9 6		1. W	6	J	A	7
	10 9 8 7		2. S	5	4	10	K
	10 9 8 7 3		3. S	3	2	9	Q
A K Q J 10 9 8 6 4 2		—	4. S	8	6	6	7
—		A K Q 7	5. S	5	6	10	A
6		A K Q J	6. S	4	5	9	K
6 5		A K Q J 2	7. S	3	A	8	Q
	7 5 3		8. S	2	K	7	J
	8 5 4 3 2		9. S	4	Q	3	2
	5 4 3 2			Claim the rest
West leads	4

After nine tricks, South’s 5-4-3-2 are high. Note that the West, North and East hands are interchangeable. For instance, if you swap North and West, the 6 lead would just reorder the first four tricks.

Balanced diet

The 55 sum can also be attained with a balanced South hand, and the highest card a seven. From Reint Ostendorf, Netherlands:

NT win 13 in	—		Trick	Lead	2nd	3rd	4th
South hand	10 9 8 7 6		1. W	2	9	A	3
	9 8 7		2. S	5	4	8	K
	10 9 8 6 3		3. S	7	6	7	10
A K Q J 10 9 8 6 4 2		—	4. S	7	Q	6	5
5		A K Q J	5. S	4	J	3	2
Q J		A K 10	6. S	6	5	10	A
—		A K Q J 5 2	7. S	5	A	9	K
	7 5 3		8. S	4	K	8	Q
	4 3 2		9. S	3	Q	7	J
	6 5 4 3 2		10. S	2	J	6	A
West leads	7 4			Claim the rest

After 10 tricks, South’s 4-3-2 are high. Here also, the West, North and East hands are interchangeable.

Hand 54, where are you?

The absolute weakest hand to win 13 tricks sums to 54 and is unique except for suit identity. South must hold suits of 7-5-4-3-2, 5-4-3-2, 7-5-3 and a singleton four. Note that this is the same as the first example (sum 55) except the eight is reduced to a seven.

The following solution from our winner plays out like a broeken record:

NT win 13 in	—		Trick	Lead	2nd	3rd	4th
South hand	A K Q J 10 3		1. W	2	K	Q	3
	K 9 8 6		2. S	5	4	9	J
	8 7 6		3. S	7	6	8	10
A K Q J 10 9 8 6 4 2		—	4. S	4	A	3	2
—		9 8 7 6 5 2	5. S	7	10	6	A
A		Q J 10	6. S	5	9	8	K
10 9		A K Q J	7. S	4	A	7	Q
	7 5 3		8. S	3	K	6	J
	4			Claim the rest
	7 5 4 3 2
West leads	5 4 3 2

After eight tricks, South’s 2 5-4-3-2 are high. Once again the other three hands are interchangeable.

Earning style points were Howard Liu, Dan Baker and Julian Wightwick. Each not only produced a perfect solution (sum 54) but also made North as weak as possible. In the above layout, swapping the North-East hearts and the K-10 effects their solution. Dan and Julian also correctly noted that a second Yarborough is impossible if South has a sum of 54, but close with North requiring one 10.

A new bridge Law?

Considering that the worst bridge hand, 4-3-2 4-3-2 4-3-2 5-4-3-2, has a sum of 41, it is curious that adding 13 allows all the tricks to be won — or one trick per pip. Move over, Jean Rene Vernes! The Law of Total Tricks is now superseded by Pavlicek’s:

Law of Total Pips

The sum of the pips of any bridge hand, minus 41, equals the
number of tricks that can be won playing against lunatics.

Of course you ask: Are there enough lunatics in the bridge world for the Law to be practical?
Well, if you keep reading you may find convincing evidence…

Looney Tunes

Howard Liu: Next time I will open 7 NT with the South hand.

Jim Munday: This nullo defense is fun. The way I bid, I need my opponents to master it.

And if you’re still not convinced, this should clinch it:

Jonathan Mestel: I wrote a tune and will sing it for you…

Tell her to make me a notrump grand;
Parsley, sage, rosemary and thyme.
With barely a seven held in her hand,
She will ever be a true love of mine.

Puzzle 8F23 Main

Top Yarborough Fair

© 2011 Richard Pavlicek