I have two questions for you.

In your book you have a few examples where you add three or more WP for long cards in a suit (see Pages 140 and 143 for examples). Are you recommending that this be done as part of hand evaluation whenever you can anticipate that a suit will be a source of tricks? If so, would the rule be something like "count a jack as three points, and add three additional points for each card beyond four in a running side suit"?

I believe I have a simpler method of counting tricks that in each case I tried came out to the same number as with your method:

- Assume that with half (19-21) of the working points and a good trump fit, you'll take seven tricks.
- Add or subtract one trick for every three working points your side holds above or below half.
- Add one short suit trick for each (nonduplicated) side-suit doubleton, two for each singleton, three for each void.
- The total will equal your expected tricks on the deal.

Assuming that works, I find it an easier calculation although others might prefer the approach in the book.

A good hand to test these methods came up the other night at our local club:

Q J | x x x | |

J x x x x x | A K Q x | |

x x | A Q 10 x | |

Q x x | J x |

West |
North |
East |
South |

1 | |||

pass | 1NT | d'ble | 2 |

3 | pass | ? |

West's 3 bid is clearly dubious with only 3 WP, but he also has 4 SST. (Perhaps, however, he shouldn't give himself credit for the spade doubleton since it probably duplicates shortness in partner's hand. What do you think?) If he gives his partner 14-15 WP he gets 13 - 4 SST - 1 trick in WP = 8 tricks. (I'd count 7+2 (doubleton diamond, doubleton spade) - 1 trick in WP = 8). Therefore his 3 bid seems reasonable since he'd prefer not to sell out to 2. (Note that the LOTT would certainly find 3 – maybe even 4 – acceptable.)

Now turn to East. He's heard his partner come in with 3 and he's looking at 16 HCP. Should he go on to 4? He has 13-15 WP (the club jack is worthless, the diamond queen may be), and he can be fairly sure his partner has no more than 5 or 6 WP. The partnership might have 19 WP, but it might not. If he assumes West has a doubleton spade but no other useful doubleton, East's calculation would be 13 - 4 SST = 9 (or my way, 7+2 for the doubletons = 9) with perhaps - 1 for fewer than 19 WP. Clearly, 3 is high enough.

Even though some of both East's and West's assumptions were slightly wrong, the errors pretty much cancelled out. As the cards lie, the diamond king was onside so 3 just makes while 4 has no play.

Ed Herstein

**Answer**

When you have a long suit and know you can use it for discarding losers from the other hand, you should of course take that into account in valuing your cards. If these long cards take tricks, they should be valued as tricks, i.e. 3 points each. Therefore, we agree with what you say about long side-suits.

Regarding your proposed method for counting your tricks, we refer to our answer to question 6. Your method is excellent. It is just as good as ours or the one suggested by Ben Hooyer. But we still prefer our own "If we have half the deck in working points, we will lose as many tricks as our SST. For every full 3 WP more than average, we deduct one loser; for every full 3 WP less than average, we add one loser".

In your example, we might not have doubled 1NT for take-out with the East hand (we hate doubling with wrong shape unless lots of extra values, especially when both opponents are bidding and haven't found a fit), but once we have, going on to 4 is too much. East has no reason to expect more than their share in working points and an SST of 4. It might even be worse. Pass is indicated. As it was, East had 15 WP but West only had 1 WP. The black queens and jacks were all useless, the diamond queen might have been also. That they still took nine tricks with an SST of 4 is because of their third (useful) doubleton, which adjusts their SST to 3.

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