# BBO Discussion Forums: PCB Hand #4 - BBO Discussion Forums

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## PCB Hand #4

### #1straube

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Posted 2017-October-18, 13:53

52 6 AQJ832 QJ65
KJ9 AJT K7 AK987

South is captain. Let's say North shows pattern and 5 QPs at 4C.

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

4D-5C small heart, even spade, odd club, even diamond
Cards placed (if not AQ x Jxxxxx QJxx)
5D-5H no club K

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

4D-5D even kings, small heart, even spade, odd club, even diamond
Cards placed (if not AQ x Jxxxxx QJxx)

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

4D-4H even diamond
Cards placed (if not AQ x Jxxxxx QJxx)
5D-5S small heart, no club K

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

4D-4S even kings, even diamond
Cards placed (if not AQ x Jxxxxx QJxx)
5H-5S small heart (I suppose step one since we don't zoom to jacks)
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### #2straube

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Posted 2017-October-18, 13:54

I've introduced a bit of unfairness here because I'm skipping with xx when I go from short to long but I'm stopping with xx when long to short.
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### #3straube

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Posted 2017-October-18, 14:37

The marriage (KQ) problem or occasional AQ problem can only occur with two evens. How about a question that asks whether the honor count in the longer or higher ranking even is equal or greater than the honor count in the shorter or lower ranking suit? Only consider the first two even suits (if three or more). So this would be the last question and only apply with two or more evens?
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### #4foobar

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Posted 2017-October-18, 22:44

straube, on 2017-October-18, 13:53, said:

52 6 AQJ832 QJ65
KJ9 AJT K7 AK987

South is captain. Let's say North shows pattern and 5 QPs at 4C.

Here's a Byzantine method with the following rules:

1) K-parity first (with K=2 QPs always)
2) Singletons and doubleton suits are scanned first, with ties broken in rank order. Skip with nothing or stop with A/K
3) 4+ cards suits are scanned in length order with ties broken in rank order. Stop with even and continue with odd

4 - 4 (even K-parity]
4 - 5 (nothing in , , even )

AQ of , Q perforce
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### #5yunling

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Posted 2017-October-18, 23:21

So either xx x AQxxxx Qxxx or AQ K xxxxxx xxxx

My method also has an ambiguity here, but it is a little better and you can usually bet p not holding all honors in short suits.
The ambiguity can be solved higher but I can't stop in a reasonable contract anyway(can stop on 5NT).
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### #6sieong

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Posted 2017-October-21, 23:29

EPCB.
4 - 4 (even )
4 - 4n (odd others or parity is different)
5 - 5 (even even odd )

Unresolved between AQ in vs . Same is true for classic PCB.

Note that classic PCB would have resolved at 5, so EPCB is worse by a step. Had there been a pair of relevant suits for the final ask, EPCB would have gained a step.
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### #7nullve

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Posted 2017-November-01, 10:54

straube, on 2017-October-18, 13:53, said:

52 6 AQJ832 QJ65
KJ9 AJT K7 AK987

Auction, with some ambiguity omitted:

(...)
4-4 (10-12 hcp (say); relay)
4-4 (even # of non-singleton Ks, no singleton K; relay)
4N-5 (even # of non-singleton Qs; relay)
6-? (Q, Q, J, J, no J; ?).

This post has been edited by nullve: 2017-November-13, 10:20

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