I'll look at A. It's a recurring theme. I decided to amuse myself by making up a hand for each of the other three players and then seeing what the bots would do:
K95
KQ76
AJ43
82
QT7 8643
T8432 A95
Q62 K9
AJ KT93
AJ2
J
T875
Q7654
The uncontested bot auction went 1
- 1NT- Pass
Off two, -200.
Yes, the South bot self-alerted that he held either 2 or 3 cards in each major. Of course he does not have a good bid.
Me, I think I would bid 1
over 1
. Partner would rebid 1NT and then, I think, I would go back to 2
. It looks as if 2
would be off 1. Can the opponents make 2
, assuming they can somehow bid it? Maybe, but I sort of doubt it. We can certainly take 2 spades, 2 hearts, and 1 diamond, but just how does declarer plausibly get the other 8 tricks?
I chose the three non-S hands w/o trying for any particular result, I just chose some cards, but of course there are many possibilities. With this deal, I don't want to be in 3
. My 1
call, followed by the correction to 2
, seems to work out reasonably well.
But what should we really do? The choices seem to be:
1
on
AJ2
1NT holding a stiff
3
on a hand that does not much appeal to me.
As I say, this is a recurring theme.
I am pretty sure that at the table I would bid 1
. But I am hardly prepared to strongly defend that call.
We could vary the hands a little, trading the
6 for the
6:
K965
KQ7
AJ43
82
QT7 843
T8432 A965
Q62 K9
AJ KT93
AJ2
J
T875
Q7654
The bot auction would go the same, since the S hand has not changed and N has no reason to pull 1NT. The defense against 1NT could get interesting. Of course after my 1
call N would raise. Well, I have one top diamond, two hearttricks to be developed. Can I manage five spade tricks? Maybe. Without then first taking six tricks? Maybe. The opponents now have a 9 card
fit but still, can they find it? And even if they do, can they take 9 tricks? So 2
making would be great, 2
off 1 might well be acceptable.
Still. What should be done? Beats me. I doubt that it is hard to deal the cards so that 1
works very badly.