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PCB Hand #7

#1 User is offline   straube 

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Posted 2017-October-19, 20:27

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
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#2 User is offline   foobar 

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Posted 2017-October-19, 23:42

View Poststraube, on 2017-October-19, 20:27, said:

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.
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#3 User is offline   straube 

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Posted 2017-October-19, 23:57

View Postfoobar, on 2017-October-19, 23:42, said:

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.
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#4 User is offline   awm 

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Posted 2017-October-20, 06:54

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?
Adam W. Meyerson
a.k.a. Appeal Without Merit
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#5 User is offline   straube 

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Posted 2017-October-20, 07:09

View Postawm, on 2017-October-20, 06:54, said:

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.
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#6 User is offline   awm 

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Posted 2017-October-20, 07:14

View Poststraube, on 2017-October-20, 07:09, said:

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.
Adam W. Meyerson
a.k.a. Appeal Without Merit
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#7 User is offline   yunling 

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Posted 2017-October-20, 21:16

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
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#8 User is offline   sieong 

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Posted 2017-October-22, 00:42

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.
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#9 User is offline   nullve 

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Posted 2017-November-01, 11:30

View Poststraube, on 2017-October-19, 20:27, said:

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; ?).

This post has been edited by nullve: 2017-November-13, 10:24

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