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Peter Fredin won this slam. Will you...?
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Master Point Press will soon release a book written by me about the Swedish bridge star Peter Fredin. The flamboyant Swede is in my biased opinion (we are friends) one of the most spectacular bridge players alive and arguably among the top three declarers in the world. The book is called "Master of Bridge Psychology" and is a compilation of Fredin's best, worst and funniest hands. The deal below unfortunately is so new that it didn't make it into the book. It is a thing of beauty. Try to see if you can make this slam on a lucky diamond lead (don't press "next" more than 3 times):

West
North
K1043
Q2
Q73
Q1065
East
South
75
AK53
AK106
AK4
W
N
E
S
 
2
P
2
P
2N
P
3
P
3
P
4N
P
6N
P
P
P
D
11
6NT South
NS: 0 EW: 0
2
3
5
10
3
1
0
6
8
Q
6
1
2
0
7
8
K
4
3
3
0
A
9
3
4
3
4
0
A
3
5
2
3
5
0
K
7
6
9
3
6
0
4
8
10
9
1
7
0
Q
J
5
J
1
8
0
4
A
7
2
2
8
1
10
3
7
Q
1
9
1
K
6
5
Q
1
10
1
2
9
K
J
3
11
1
A
J
10
8
3
12
1
N/S +990
13

Peter Fredin, sitting South, found himself in a questionable 6NT, but got the fortuitous lead of the 2 (3.5). East followed with the 5 and Fredin won in hand with the 10. When he led a diamond to the queen at trick 2, East discarded the 6. Fredin now continued with a small diamond from dummy to the ace in hand and watched in dismay as East discarded the 8.

Before you go the next page:

Who has the A, East or West? And if you decide that East has the ace, is there a plausible way of making the slam?

East's second spade discard is bad news for declarer because East now for sure must be a heavy favorite to have the A, since he would never discard a spade from Qxxx(x) or QJxx(x) missing the ace. The standard way to declare this contract was originally to play West for the A and try to get a good count in order to optimize the chances for finding the J, bringing home the slam by way of 1 spade trick, 3 heart tricks, 4 diamond tricks and 4 club tricks. But not anymore.

With the A in East, and with no possibility of rectifying the count for one of the many possible squeezes, the chances of making this contract on the surface looks very bleak indeed.

But not for Peter Fredin. He found a simple and brilliant way to win in this position:

West
North
K1043
Q2
Q1065
East
South
75
AK53
K
AK4
W
N
E
S
 
2
P
2
P
2N
P
3
P
3
P
4N
P
6N
P
P
P
D
11
6NT South
NS: 0 EW: 0
2
3
5
10
3
1
0
6
4
Q
6
1
2
0
7
8
A
8
3
3
0
A
2
5
3
3
4
0
K
9
3
4
3
5
0
K
7
6
9
3
6
0
4
8
10
9
1
7
0
7

Before you read on. Try if you can emulate the Swedish star. He won the contract with the lie above within a minute.

Here is the full deal: If you didn't find a winning plan single dummy, have a go at it double-dummy.

West
Q2
J7
J9842
J873
North
K1043
Q2
Q73
Q1065
East
AJ986
109864
5
92
South
75
AK53
AK106
AK4
W
N
E
S
 
2
P
2
P
2N
P
3
P
3
P
4N
P
6N
P
P
P
D
11
6NT South
NS: 0 EW: 0
2
3
5
10
3
1
0
6
8
Q
6
1
2
0
7
8
K
4
3
3
0
A
9
3
4
3
4
0
A
3
5
2
3
5
0
K
7
6
9
3
6
0
4
8
10
9
1
7
0
Q
J
5
J
1
8
0
4
A
7
2
2
8
1
10
3
7
Q
1
9
1
K
6
5
Q
1
10
1
2
9
K
J
3
11
1
A
J
10
8
3
12
1
N/S +990
13

Hint: think of end positions for the East hand.

The solution is simple. If you are convinced the A is off-side, you will have no trouble winning this contract as long as East has length in hearts!

West
Q2
J7
J9842
J873
North
K1043
Q2
Q73
Q1065
East
AJ986
109864
5
92
South
75
AK53
AK106
AK4
W
N
E
S
 
2
P
2
P
2N
P
3
P
3
P
4N
P
6N
P
P
P
D
11
6NT South
NS: 0 EW: 0
2
3
5
10
3
1
0
6
8
Q
6
1
2
0
7
8
K
4
3
3
0
A
9
3
4
3
4
0
A
3
5
2
3
5
0
K
7
6
9
3
6
0
4
8
10
9
1
7
0
Q
J
5
J
1
8
0
4
A
7
2
2
8
1
10
3
7
Q
1
9
1
K
6
5
Q
1
10
1
2
9
K
J
3
11
1
A
J
10
8
3
12
1
N/S +990
13

Fredin instantly realized that if he could find the J (see next page), East would be doomed as long as he had four or more heart, as he will be the victim of a "compound count-squeeze" in hearts and spades. Fredin made sure he was in dummy in this six-card position:

West
Q2
J7
J
J
North
K104
Q2
Q
East
AJ
10986
South
75
AK53
D

When Fredin cashed his Q, poor East was forced to stiff his A in order to hold on to four precious hearts, making it possible for the Swede to duck a spade into East, making his 12th trick on the K. 

As you have seen Fredin decided to finesse West for the J instead of playing for clubs breaking 3-3. Taking the club finesse is in Fredin's own words "obvious". Why isn't it a coin toss, since the slam also wins when West is 2353 and East 5413?

The answer to the question lies in West's opening lead of a small diamond. Ask yourself what you would lead from:

West
J9842
842

or

West
J9842
Jxx

or

West
J9842
J873

You would for sure lead a club on the first hand. On the second hand the choice is more even with diamonds still being the most likely, while a club lead on the last hand looks highly inferior to a diamond lead (in fact it wasn't). All in all this makes West a big favorite to have the J.

If you would like to see more wizardry (and some spectacular catastrophes) from Fredin's hand, the book about this truly brilliant bridge player will be available in a few months.

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