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Suit Establishment - how many entries?

Take this side-suit in a trump contract:

 
 6 5 4 3 2
K J 9
 
Dummy
W   E
Declarer
 
A Q 10 8
 
7
Winning a trick may look far-fetched but on the expected 4-3 split as above (62 per cent), you can set up and enjoy a fifth-round length winner. You’ll lose the first round; you’ll get over to dummy (that’s one entry) to ruff a second diamond; you’ll get over to dummy (that’s a second entry) to ruff a third diamond; you’ll get over to dummy (that’s a third entry) to ruff a fourth diamond. And you’ll need a fourth dummy entry to enjoy the fifth diamond.

The number of entries required to set up a suit is equal to the number of ruffs plus one (to get back at the end). Take this side-suit in a trump contract:
 
8 6 5 4 3 2
??
 
Dummy
W   E
Declarer
 
??
 
7
You don’t know how the six opposing clubs will split. You’ll naturally hope for a 3-3 split but a missing even number of cards does not usually split evenly (3-3 is just 36 per cent). You can hypothesise.

If you have four dummy entries, you can set up clubs if they split 4-2 (lose a club; three club ruffs; back at the end to enjoy two long cards). If you have three dummy entries, you’ll need clubs to split 3-3. If you have just two dummy entries, I’m afraid you can forget about setting up dummy’s clubs for the lack of entries. I hope there’s a Plan B.
South Deals
None Vul
K J 3
J 4
7 5 2
A 7 5 3 2
6 4
K Q 9 5 2
Q 9 8
K J 8
 
N
W   E
S
 
5
10 8 7 6
J 10 6 3
Q 10 9 4
 
A Q 10 9 8 7 2
A 3
A K 4
6
West North East South
      1 
Pass 2  Pass 3 1
Pass 4  Pass 6 
Pass Pass  Pass  
  1. Forcing to game facing a two-over-one.

As declarer, you win West’s  K lead and count 11 winners. The 12th can only come from dummy’s fifth club. This can be established if the suit splits 4-3, provided there are four dummy entries (three ruffs plus one).  A,  K and  J are three entries. The fourth is more elusive — it’s  3 but only if you retain  2.

At trick two, lead  6 to  A and ruff  2 with  7 (not with  2). Lead  8 (not  2) to  J (both following — a 3-0 spade split would have prevented  3 from being an entry) and ruff  3 with  9. Both opponents follow, revealing the 4-3 club split: your plan will work.

Lead  10 to  K, ruff  5 with  Q and at trick eight (fanfare), you lead  2 to  3. You cash  7 at that’s 12 tricks and slam made. Only an unlikely opening spade lead, removing a dummy entry prematurely, would have scuppered the slam.

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