Suit Establishment - how many entries?
Take this side-suit in a trump contract:
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:
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 |
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| ♠ |
5 |
| ♥ |
10 8 7 6 |
| ♦ |
J 10 6 3 |
| ♣ |
Q 10 9 4 |
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| |
| ♠ |
A Q 10 9 8 7 2 |
| ♥ |
A 3 |
| ♦ |
A K 4 |
| ♣ |
6 |
|
| West |
North |
East |
South |
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1 ♠ |
| Pass |
2 ♣ |
Pass |
3 ♠1 |
| Pass |
4 ♠ |
Pass |
6 ♠ |
| Pass |
Pass |
Pass |
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- Forcing to game facing a two-over-one.
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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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