25 May 2012

The loser that vanishes

Normally in suit contracts we count losers and in NT contracts, winners. From there agonize about how to decrease or increase that amount to our satisfaction. But it is not always so:

T 6 4 3
A K J 9 3
8 7 3
J T 7 6
Q 9 2
K T 9 6 4
K 8 4 3
J 8 7 5
Q 7 5
Q 5
A 9 5 2
T 8 6 4
A J 2

A small spade lead sees you longing for 3NT, until the diamonds don't break. 3NT is doomed but how are you going to make 5? It would appear that there are 3 losers; two in clubs and that pesky, didn't drop when it was supposed to, trump queen.

Count your winners, the two black aces, two top trumps, the top hearts and 5 ruffs is eleven!

Play it out. Win the lead, draw two trumps, cash everything that will cash, ruff everything that won't and you'll be left with two small clubs opposite Jx and 11 tricks.

This sort of thing is called a trick compression, the poor defense make their three defensive tricks in the space of two defensive tricks. Effectively you lost one trick twice!

On to the bidding challenge. You're playing matchpoints, everyone is vulnerable, partner opens 1, RHO passes and you are playing SAYC:
A Q J 3
K T 9 7 5 2
8 2


  1. MP is still bridge, I still have a spade suit, partner still hasn't rebid 2C; even then, partner is still not barred from holding fitting diamond cards... :)

  2. One spade.

    The hand is not strong enough to respond two diamonds and then game force with two spades, with the singleton heart being a huge negative to overbidding. It is better not to lose the spades.

    Over two clubs I'll bid two notrumps and hope they lead the unbid suit.

  3. I hate hands where the possibility of missing a fit looms large. Two or three diamonds may well be your best spot if partner holds three so the need to bid 2D seems sensible. Similarly 2S might also be the optimum contract with an uncomfortable 4-3 fit , but with the 1 spade response the diamond contract has gone forever.
    Anyway , as you always want to be in a major if you can then 1S it has to be.