PCB Hand #7

[straube]

2 AT82 AQJ93 643

 

AQ7 KJ K852 AJT8

 

South is captain. Let's say North shows pattern and 7 QPs at 4C.

 

1) With rules of short to long (and stop with odd for singleton or doubleton), king parity last

 

4D-4S small spade, even club

4N-5H odd heart, even diamond

Cards placed

 

2) With rules of short to long (and stop with odd for singleton or doubleton), king parity first

 

4D-4N even king, small spade, even club

5C-5H odd heart, even diamond

Cards placed

 

3) With rules of long to short (and stop for singleton only), K parity last

 

4D-4H even diamond

4S-5C odd heart, even club (K AT82 J93xx KQx possible if stiff K=1)

5D-5S small spade, no DK

Cards placed

 

4) With rules of long to short (and stop for singleton only), K parity first

 

4D-4S even K, even diamond

4N-5D odd heart, even club (K AT82 J93xx KQx possible if stiff K=1)

5H-5N small spade, no DK

Cards placed

[foobar]

2 AT82 AQJ93 643

 

AQ7 KJ K852 AJT8

 

South is captain. Let's say North shows pattern and 7 QPs at 4C.

 

4C...4D (relay; even K parity....at this point we can have two black kings + rounded queens)

4H...4N (relay; nothing in S, even D, so both black kings go out of the window. Must hold A(D+Q) perforce and by induction A(H) (since KQ of clubs is impossible)

 

To be frank, I don't know if I can actually make this deduction at the table, but it seems promising.

[straube]

4C...4D (relay; even K parity....at this point we can have two black kings + rounded queens)

4H...4N (relay; nothing in S, even D, so both black kings go out of the window. Must hold A(D+Q) perforce and by induction A(H) (since KQ of clubs is impossible)

 

To be frank, I don't know if I can actually make this deduction at the table, but it seems promising.

 

To have the same starting place, please amend to have 4D be the relay ask. Also even K parity for 8 is skip (I'm using the parity recommendations by David and Mark I posted on the other thread); for now you might consider using those so you can compare outcomes.

[awm]

We would have:

 

4♦ - 4♥ (even diamonds)

4♠ - 5♣ (odd hearts, even clubs)

 

At this point it is ♠K, ♥Q, ♦AQ or the actual cards. Either one is actually fine (pitch the clubs on the spades or double hook), so we do not actually NEED another ask. But of course:

 

5♦ - 5♥ (small spade)

 

It seems like your case 2 still permits x AQxx Axxxx xxx?

[straube]

We would have:

 

4♦ - 4♥ (even diamonds)

4♠ - 5♣ (odd hearts, even clubs)

 

At this point it is ♠K, ♥Q, ♦AQ or the actual cards. Either one is actually fine (pitch the clubs on the spades or double hook), so we do not actually NEED another ask. But of course:

 

5♦ - 5♥ (small spade)

 

It seems like your case 2 still permits x AQxx Axxxx xxx?

 

Thank you. Corrected. Great point about how your method gives you useful combinations early. What's your substitute for the king parity question at the end? Atul and I have been discussing ways of solving the marriage problem or occasional AQ problem as well as other disambiguities and you appear to have already done so.

[awm]

Thank you. Corrected. Great point about how your method gives you useful combinations early. What's your substitute for the king parity question at the end? Atul and I have been discussing ways of solving the marriage problem or occasional AQ problem as well as other disambiguities and you appear to have already done so.

 

Here's what's in our current notes:

 

1. If there is a pair of suits where one has KQ and the other has none, the first such pairing (length order, followed by high to low) is picked. Stop if first suit has KQ, skip if second suit has KQ.

2. Otherwise if there is a pair of suits where one has K and the other has Q, the first such pairing (length order, followed by high to low) is picked. Stop if first suit has K, skip if second suit has K.

3. Otherwise this step does not exist.

 

This appeared slightly better than king parity in my (long ago, not going to be repeated) simulation.

[yunling]

4♦-4♠(ask 1&2, 0/3/6)

now there are many possible holdings but 6 is where we belong to

 

then

4NT-5NT(ask 2&3, 2/5 no stiff honor even K)

and cards placed

[sieong]

EPCB.

4♦ - 4♥ (even ♦)

4♠ - 4n (all odds, or ♥ is of different parity) [1]

5♣ - 5♠ (even ♠ odd ♥ even ♣)

 

Honor structure resolved.

 

[1] possible parities are ooeo oeeo eoee, which are ♠K ♥A ♣K, ♠K ♥AQ ♣Q, ♠K ♦AQ ♣Q, ♥A ♦AQ.Seems like EPCB is worse than classic PCB for determining slam. In particular, opposite the second combination, we may be forced to play 5n, which is inferior.

[nullve]

2 AT82 AQJ93 643

 

AQ7 KJ K852 AJT8

With some ambiguity left out:

 

(...)

4♣-4♦ (11-13 hcp (say); relay)

4♥-4♠ (even # of non-singleton Ks; relay)

5♥-? (odd # of non-singleton Qs, ♦Q, ♦J, no ♠J; ?).

Edited November 13, 2017 by nullve