Is it the same?

[gnasher]

[hv=d=e&v=n&n=s73haj42dkqj75cat&s=sak8hkt9da643cq95]133|200|Scoring: IMP[/hv]

There were several interesting things about this hand. Here is the first:

 

You reach 6♦ after East has doubled a club bid for the lead. West leads ♣2 (3rd/low). You win the ace and draw trumps. East has a singleton, and throws two clubs. You play ♠AK and ruff, West playing 5-10-Q and East playing 2-4-9.

 

Now you play a club to East's king, and East returns a club to your queen, West following.

 

At one table, East had doubled a Stayman 2♣. At the other, the auction had begun 1NT-2NT (transfer), and East later doubled a 4♣ cue-bid.

 

Are the two declarers in the same position?

[JLOGIC]

Did we deny a major over stayman? If so I guess one difference is that our the stayman opp can falsecard the Q with QTxx in spades, whereas the non stayman opp cannot.

 

I think they would be equally likely to lead a club rather than a spade from QJTx(x) over a stayman double or a double of a cuebid.

 

Some might say RHO should not double stayman with KJxxx and out, so that would point to him having the HQ, but really I think when you're that weak the HQ is irrelevant and doubling stayman is better without it (since you are more likely to need a club lead), so I wouldn't put much weight into that. Personally I would double stayman with neither hand, but I doubt many people exist that would X with the HQ and not without or vice versa.

[JLOGIC]

RHO has done well to pitch 2 clubs, this might have been wrong if we had 5 of them, and this would be less likely to be possible if partner had bid stayman since we might have redoubled, I guess that is also a slight difference but probably RHO would pitch clubs either way so w/e.

[ceeb]

Dissimilar. The Stayman doubler appears to be 5215 (♠QJ10x on lead is unlikely, double with only 4 ♣s nutty), whereas the other RHO can be 4414.

[JLOGIC]

Dissimilar. The Stayman doubler appears to be 5215 (♠QJ10x on lead is unlikely, double with only 4 ♣s nutty), whereas the other RHO can be 4414.

LHO led the 2 of clubs playing 3rd and 5th, and RHO has already shown up with 5 clubs (he followed to the ace, pitched 2, won the king and exited one).

[gnasher]

Did we deny a major over stayman? If so I guess one difference is that our the stayman opp can falsecard the Q with QTxx in spades, whereas the non stayman opp cannot.

Yes, in the Stayman sequence responder will have denied a major.

 

However, in the non-Stayman sequence opener's shape will be known: he must have three hearts, because otherwise he's gin, and presumably East will have managed to signal his club length during the first four tricks.

[JLOGIC]

Did we deny a major over stayman? If so I guess one difference is that our the stayman opp can falsecard the Q with QTxx in spades, whereas the non stayman opp cannot.

Yes, in the Stayman sequence responder will have denied a major.

 

However, in the non-Stayman sequence opener's shape will be known: he must have three hearts, because otherwise he's gin, and presumably East will have managed to signal his club length during the first four tricks.

Will he have signalled that he doesn't have KQ of clubs also?

 

Whatever, I think its much easier to falsecard the Q from QTxx when you know they don't have 4 spades. It's not like you'll have that much time to work out they are cold if they have AKJx by ruffing 2 hearts immediately so you can falsecard the Q etc etc.

[pooltuna]

I would argue they are not as the X of 4♣ is more likely to be weaker than the X of Stayman where with the spots the doubler is missing should be worrysome if either opp found a XX

[ceeb]

Dissimilar. The Stayman doubler appears to be 5215 (♠QJ10x on lead is unlikely, double with only 4 ♣s nutty), whereas the other RHO can be 4414.

LHO led the 2 of clubs playing 3rd and 5th, and RHO has already shown up with 5 clubs (he followed to the ace, pitched 2, won the king and exited one).

I see. Watching the discards. Subtle clue.

[WellSpyder]

I was a little surprised looking at the results to see what a high proportion of declarers made this contract. If I've read them correctly, 7 out of 8 pairs in the top division bid to 6♦ and 5 out of 7 declarers made it. In the second division 4 out of 8 pairs bid the slam and all of them made it (including my partner, after a double of Stayman). Looks like a lot better than a guess, somehow...

[gnasher]

This is what I was thinking:

 

- In the double-of-cue-bid sequence, the information that the clubs are 3-5 doesn't tell us anything. East had to throw something on the trumps, and he wasn't going to throw hearts. If he had thrown two spades, we'd have known that spades were 3-5; if he had thrown a spade and a club, we'd have known that he had at least four of each black suit. All his discards tell us is that he was dealt at least eight black cards, but we knew that anyway (barring hearts 1=5). Thus, in this situation the only useful information we have comes from the trump suit, and East is 6:5 to hold ♥Q.

 

- In the double-of-Stayman sequence, the information about the club layout was known before the play started. That makes it useful information, so the distributional information we have says that it's evens whether ♥Q is on the left or on the right. (I realise that the other inferences people have suggested may change that - I'm just talking about what we can infer in terms of vacant spaces.)

 

Am I right?

[Simplicity]

Am I right?

Essentially your vacant spaces argument looks sound.

 

However I think it's interesting as East might not practically have freedom of choice over his discards:

 

If East is 3415 with the ♥Q then he won't discard a spade - We know he has at most 5 clubs so he cant afford to show out on the 3rd spade. Else when we find out he is 3415 we'll surely finesse against him. Consequently restricted choice would apply and P(3415):(4315):(5215) is 6:2:1, or East having 4 hearts is 9:2.

 

Theoretically this all means East should discard 2 clubs with all these shapes so that we have no idea whats going on, but in reality he probably won't.

 

Of course if East did discard a spade from 3415 we'd surely just play for East to have made a mistake? Consequently i think i depends a bit on our read of East.

 

All in all i think its probably right to play East for the ♥Q both times. Or is this all gobbledygook?

[MFA]

When he doubles the cuebid we can assume it shows ♣KJ. Those two cards we can count in. But I agree not the small ones we see during the discards.

In other words, before the play starts, LHO will be 13:11 favourite to hold any other specific card. So knowing about ♣KJ is significant for vacant spaces calculations.

[MFA]

When he doubles the cuebid we can assume it shows ♣KJ. Those two cards we can count in. But I agree not the small ones we see during the discards.

In other words, before the play starts, LHO will be 13:11 favourite to hold any other specific card. So knowing about ♣KJ is significant for vacant spaces calculations.

If we are not sure that the double shows ♣J as well as ♣K, then we should just leave out that card and count one space for ♣K only.

[dake50]

Expect 4C doubler to say "cannot help any other lead" here no HQ. (wheels within wheels may be: he hopes to win HQ by demanding C-lead as if no H-help)

 

Whereas 2C doubler has CK and enough length to not lose set if C-lead against 4M/3N.

[JLOGIC]

When he doubles the cuebid we can assume it shows ♣KJ.

You only double cuebids with KJ and not the king?