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Batting at the Bermuda Bowl - 2
(Page of 5)

In the first part of this article, we assessed this problem

Hans
10984
A82
AJ832
K
Nunn
AK7
J7
9
AQJ8764
W
N
E
S
1
2
X
2
4
P
4
P
4
P
6
P
P
P

My partner Tony Nunn declared 6 on the lead of the K. A similar situation existed at the other table and the tournament bulletin complimented the Miguel Villas Boas's successful line of play.

Edgtton
QJ652
KQ1096
106
10
Campos
10984
A82
AJ832
K
Gosney
3
543
KQ754
9532
Villas-Boas
AK7
J7
9
AQJ8764
W
N
E
S
 
1
2
2
P
4
P
4
P
4
P
4N
P
5N
P
6
P
P
P
D
6 South
NS: 0 EW: 0
 

On the K lead, declarer ducked the heart rectifying the count to catch West in a major suit squeeze. Say the defense continues with the Q, declarer runs all of his trumps and crosses to the diamond ace in this end-position. 

West
QJ6
10
North
10
8
AJ
East
3
3
KQ
South
AK7
9
W
N
E
S
 
1
2
2
P
4
P
4
P
4
P
4N
P
5N
P
6
P
P
P
D
6 South
NS: 0 EW: 0

When South plays a diamond, West cannot find a discard.

A similar ending operates if West switches to a diamond at trick two, with  the heart ace now serving as the transportation to dummy. 

What did Tony do ?

 

 

Hans
10984
A82
AJ832
K
Nunn
AK7
J7
9
AQJ8764
W
N
E
S
1
2
X
2
4
P
4
P
4
P
6
P
P
P

Tony's Line of Play

1. Won the A

2. K

3. 8 to A

4+. Run clubs (discarding or unblocking the spade spots)

West discarded accurately and declarer was an easy one down.

Was this a sloppy line of play by a declarer who forgot to rectify the count or was there more to this hand ?

 

As pointed out in the comments to the previous post,declarer was playing for a strip-squeeze. He was hoping East held the singleton J or Q. If that were the layout, on the run of the trumps the four card ending would be.

West
Q6
Q10
North
4
82
J
East
43
KQ
South
A7
J
4
D

West is squeezed on the last trump. If he discards a heart, declarer can throw him in with the J to get the last two tricks in spades (note the spade spots unblock)

So which line is better ?

Line A (Australian line): Works if West has Qxxxx or Jxxxx of spades.

Line B (Brazilian line) : Works if West holds the KQ109 of hearts or any six card suit.

Sidetracks

If the spades are 6-0, both lines run the risk of a defensive ruff. 

If the clubs are 5-0, the contract has no chance on either line.

Playing for the vigorish of a doubleton KQ doubleton is a very risky proposition. 

We conclude we can ignore the side-tracks.

Is it time to bring out the calculator ?

Maths Time

It is rare for a bridge problem to come down to a mathematical exercise of determining the best odds. Here is a simplified attempt that ignores cases where neither line works.

Line A : East has the singleton Jack or Queen. Since East must hold a singleton, there are six total layouts (two spade honours and four spot cards). The chances of East's card being an honour is exactly two in six. Which is 33.33 percent.

Line B: West has all the hearts 

a. KQ109x opposite xxx : The KQ are known so the problem is effectively 109x opposite xxx. Again, this is straight-forward to compute. This is exactly four cases (depending on which exact x is held by East). The total number of layouts is 6C3 = 20.

4 winning cases out of total 20 comes to 20 percent. 

However, as pointed out by Henry Bethe in the comments, declarer will go down when West's shape is 5512 with 10x or 9x of clubs if West switches to a spade at trick two. Using Chris Baker's odds in the same section, 5512 vs 5521 is 40:60 and within that the 10x or 9x of clubs is 70 percent (three 10x cases, three 9x cases, one 109 case out of ten total cases for 5C2)

The real odds of success assuming West is 55 in the majors (assuming perfect defense) come down to only 6 percent !! (20* (1-.7))

Let us say that the defender will switch to a spade one time in three.

 So the real odds come down to something like 20 percent (2/3 times) and 6 percent (1/3 times) for a total effectiveness of about 15 percent

b. All six hearts : The 65 layouts will occur far less often than the 55 layouts. 6502, a losing layout, vs 6511 is 20:80 according to Chris Baker's calculations. Also with 65 shape West have chosen an alternate route or bid over 4

Generously, lets assume the 65 shape scenario (65 shape plus chose michaels plus did not bid again) occurs one time in five. And liberally lets assume the success rate is 80 percent.

So the total odds of Line B become

15 percent (4/5 of the time when West is 55) plus 80 percent (1/5 of the time when West is 65) = 12 +16 = 28 percent

Conclusion

Even a generous allocation to the possibility of the 65 shape struggles to bring the line up to scratch to the elegance of the strip-squeeze.

A real mathematical answer

Would probably involve notations for some of the approximations done above. And while my original intention was to aim to write it down, this article has already given me enough of a  headache that i'll duck out of this one.

Is that all there is or is there more to the truth ? 

Your choice about what would you like to believe 

A. The simplified mathematical analysis attempted on the previous page

B. A real mathematician's answer with formulas and uncertainty factored in

C. David Burn's comment : "What is clear, though, is that nothing about this hand is clear."

D. Tony Nunn is just an unlucky declarer versus Brazil

I'm kinda leaning towards D...

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