North dealer, East-West vulnerable
|A Q J 10 3 2|
|Q 10 9 2|
|Q 4 2||K J 10 9|
|K 8||7 4|
|6 5 4||J 7 3|
|A K Q J 10||8 5 4 3|
|A 8 7 5|
|9 6 5|
|A K 8|
|9 6 2|
Playing the hand above, partner and I set the opponents' 4 2 times, while we had a cold 4 contract.
My partner complained about me not bidding 4, and I said I did not have the distribution to do so, square hand, no ruffing power, needed too much luck (which most sorrowfully was there).
My calculation was that even if partner had a singleton and all our points were working, we had an SST of 4, and since the WP we had could at the best be 20, we should not be in 4.
I asked quite a few people, almost everyone said I needed to bid 4. So, where did I go wrong?
By the way, even before I read you book, when I used to rely more on the LAW, I would probably have either passed 2 or bid 3 as an extended preempt, but never consider 4 because of my shape.
We have nothing to complain about your bidding. And your calculation is correct. You expect an SST of 4, and if your partner has a maximum weak two-bid (9-10 WP and 6331/6322), you will take nine tricks, no more, no less. Opposite KQJ sixth of hearts, out, you won't even make 3. Those who said you should have bid 4 were wrong.
But we do NOT like your partner's bidding. Her hand is simply too strong for a weak 2 – not because of her high-card points but because of her shape. A hand with 6-4 distribution and a strong four-card side suit is almost one trick better than the ordinary 6331 or 6322 distribution, which is the norm for a weak two-bid. Had you known your side's SST was 3, you would have at least invited game. But you shall expect an SST of 4, and proceed on that basis.
When you competed with 3 over 3, we think your partner could have made up for her previous underbid by going on to 4 all by her own. The extra distribution is the key, not some number of high-card points or trumps.
Since your side can win eleven tricks in hearts, it means 4 is on even if we move one diamond to spades (or clubs). Why? The explanation is that with the K onside, North's 9 HCP are the eqivalent of 12 WP. So, in effect you have 23 HCP and an SST of 4 in that scenario, which equals ten tricks. Sometimes things are like that, but there is no way for us to know when a suit like AQJ10xx is just as good as AKQJxx. Had North known that, she would have opened the bidding with 1, not 2.
[ 1 | 2 | 3 | 4 | 5 | 6 ]