Author Topic: inferences and such  (Read 1462 times)

kenberg

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inferences and such
« on: October 21, 2021, 01:17:25 PM »
I had pretty much decided that the sleight of hand thread was dead but Blu indicated some interest so here is something that I have been thinking about.

I don't recall the vul or the form of scoring and I don't recall the hands exactly but I think I can come close enough.

Pard opens a spade, Rho doubles, you hold

S: A8x
H: xx
D: Jxxxx
C: AQx

An 11 count with a bit of shape support for pard.

XX seems right. It was suggested that a conventional 2NT might be right but my styke is that 2NT is invit plus with four card support, when I have three card support I XX and then show my support on the next round. I believe this to be reasonably standard.

So:

1S    X   XX-  Pass
Pass 1NT ?


I don't regard my hand as minimal for the XX but how good is it? I just bid 2S, some thought I should bid 3S, I think they might be right. I'll get back to that, but I titled this post inferences. And so?

I have 2 hearts. My Lho did not bid 2H over my XX. That at least suggests that he does not have four hearts. So it's reasonable to think that Rho has four hearts, Khi does not have four hearts, but then pard has four hearts.  So pard is 5-4-?-?. Since I have more Ds than Cs, perhaos pard has more Cs than Ds, let's see where that line of thinking goes. Lho also did not bid 2m over my XX. That could be because he does not have four of either, that would give him at least four spades. And that would give Rho at most one spade. But he did bid 1NT. Well, perhaps he is 1=4=4=4 and trusting that he will not really end up playng in NT, but also maybe he is 3=4=3=3. As of yet he only knows that I have a XX, he does not know that it is based on support, he could ell thing 1NT is as good as anything for the moment. Giving Lho 3=4=3=3 and Lho 2=3=4=4 makes pard 5=4=3=1, and this all seems very consistent with the auction as it has gone so far. And it turns out that is how it is.

Now I could pass 1NT. The usual agreement is that afer the XX the opponents cannot lay at a low level undoubled, so the pass would be forcing. But if I pass or if I double, the auction will probably move on to 2 of something, and I will never convince pard that I have decent three card support unless I show it now. So I am going to show it now.  2S or 3S? I am open to opinions. Maybe 3S is right. Now as it happens, pard is 5=3=3=1 as noted, and she has an 11 count. Perfectly fine opening with the stiff, but I doubt she is accepting a 3S invit. Clearly 3S is passable, if I want to be in game on this I bid 4S.

Now to the two hands, putting declare at the bottom of the diagram:

S: A8x
H: xx
D: Jxxxx
C: AQx


S: KJ9xx
H: Axxx
D: x
C: Kxx
As mentioned I am not completely positive about this but I think it is right. I think the 9 and 8 of spades are important cards.
Probably the three clubs cash without a ruff, the hear ace is a rick.
The X of 1S was on the declarer's left, and the opening lead is the heart K, ducked, the continuation is the heart Q, taken. Now what? That's four tricks. Suppose we are in 4S, where do we find 10 tricks? Possibly we could get one ruff and set up a long diamond by roughing in hnd but that seems a little tough. Can we get 6 trump tricks? Well maybe. And maybe not.

If we go with the inferences above then Lho has 3 spades, Rho has 2. Rho's spade holding could be Qx, or Tx, or QT, or xx. Of course if we get one ruff and bring in the spades w/o loss we have out six spade tricks. but also we might be able to ruff tice and then lose one sade, that's also 6 tricks. Suppose Rho started with the 2=3=4=4 shape and his spades are Qx or QT, or xx.

Ruff a H, club to hand, ruff a heart with the 8. It holds if Rho has xx in spades, it drives out the Q if Rho holds Qx, and while Rho can win with the T if he holds QT his Q now drops. Unfortunately, he holds Tx so he wins with the T and Lho eventually score the Q. The winning play is t ruff with the A and eventually drop the T makinng the 9 in hand high.

We can probably agree that it is delicate. Perhaps we reason that for the 1NT call Lho must have the spade Q. Maybe so. But of sohis other teo spades might be xx or Tx. If QTx, we should uff that last H with the 8, and if Qxx we should ruff the last H with the A.

The odds about balance. The outstanding spades are the Q, the T, and three non-descript cards x, y, and z.
So Lho , if he holds the Q, might have begun with
QTx or QTy or QTz.
Or with Qxy or Qxz or Qyz.

All in all, I think we do not want to be in 4S. And I think we probably would not be in 4S even if I had bid 3S instead of 2S.

But it can be amusing to think of what might have been.

 



Ken

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Re: inferences and such
« Reply #1 on: October 21, 2021, 08:49:43 PM »
It was sad thatit appears I  (the t/o doubler) couldn't be doubled and loose the first 9 tricks  -- our 'stopper was Q32 facing the 10-5.   What do you think about passing 1NT around hoping Pat could double it. but with her scattered 11  5=4=1=3  why would she.
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kenberg

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Re: inferences and such
« Reply #2 on: October 21, 2021, 11:56:27 PM »
Indeed, althiugh I am not sure it would have stayed in 1NTX. My Lho can pull, although where to? probably almost anything is better for EW than 1NTX.

I took a simple view. I XXed with the idea that I would then show my spades. So I did. But I agree that it would have been more profitable to follow a different path! A lot more profitable!
Ken

kenberg

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Re: inferences and such
« Reply #3 on: October 26, 2021, 09:03:16 PM »
This is just a note. I said above that I regard 1S-(X)-2NT as showing four card support. I was playing against the robots today and Lhrobot opened 1S, partner robot doubled, Rhorobot bid 2NT holding Axx in support. And a diamond void so maybe that merits it. Either four card support or a decent reason for doing on three might be the right way of thinking about it. I still think 2NT to show four, redoubling and then supporting to show three is where to start even if we get off on our own on occasion.

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