[straube]

My first "marriage" problem. Per suggestion, I've started to eliminate less "interesting" hands but I've included this as someone may have a fix. The key to this is knowing whether North has the KQ of diamonds vs clubs. Statistically you would bet on the 4-cd suit but...

AT632 6 KQT9 JT6
KQ5 AJT3 AJ4 A98

South is captain. Let's say North shows pattern and 6 QPs at 3N.

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

4C-4H small stiff (no Q), even club
4S-4N even diamond
5C-5H odd spade, first odd (spades) doesn't have a king
Either Axxxx x KQxx xxx or Axxxx x xxxx KQxx

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

4C-4S odd kings (with 6 we skip), small stiff (no Q) but even club
Either Axxxx x KQxx xxx or Axxxx x xxxx KQx

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

4C-4H odd spade, even diamond
4S-4N even club
5C-5H small stiff (no Q), first odd (spades) doesn't have a king
Either Axxxx x KQxx xxx or Axxxx x xxxx KQx

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

4C-4S odd kings (with 6 we skip), odd spade but even diamond
4N-5C even club
Either Axxxx x KQxx xxx or Axxxx x xxxx KQx

[straube]

A few observations so far...

1) king parity first seems to be a big winner. I'm considering abandoning king parity last. I'm also wondering whether a simple rule (skip with odd/stay with even) would work almost as well.

2) If I count the Q as 0 (and I think I'm the only one doing that), I'm perhaps unfairly advantaged by scanning the stiff first because if asker doesn't hold the Q (but has the AK) he could never know until the singleton is scanned whether pard has it or not.

3) OTOH, counting stiff honors as full value (A3,K2,Q1) means that scanning the singleton has more importance than counting (A2,K1,Q0)

4) It may be worth considering the marriage problem. Say you hold a KQ combination. Should captain always be able to tell that you do? If so, you could trigger an ask that asks whether it's in a particular suit or not. But often you could hold the KQ in one of three suits (or even four suits) and the more suits, the more expensive the information.

[foobar]

Couple of comments:

For K-parity, my recollection is that Imprecision defines it as the K in the first suit with odd parity. This method appears to use the actual number of Ks as parity.

Edit: "After showing all parities, we next shows the number of RP in the rst suit with only one honor, if such suit exist, and stop if odd (A or Q) or skip a step if even (K). We may zoom into answering this, but we never zoom beyond this ask."

Isn't treating Q=0 equivalent to the classic DCB that only scans A/K on the first round?

[straube]

 foobar, on 2017-October-18, 10:05, said:

For K-parity, my recollection is that Imprecision defines it as the K in the first suit with odd parity. This method appears to use the actual number of Ks as parity.

Yes, it's the parity of actual kings (stiff kings aren't included).

 foobar, on 2017-October-18, 10:05, said:

Isn't treating Q=0 equivalent to the classic DCB that only scans A/K on the first round?

I don't understand. What I think is I could count stiff Qs as 0 or 1 or 2 or 3 or whatever but the important thing is whether we scan (stop) for them or whether we give up on finding them. I.e. I can pick up a stiff Q even if I count it as 0. It clearly takes some space to find stiff Qs, especially if singletons are scanned last (and sometimes even when the singleton is scanned first)

I think downgrading stiff honors just makes sense because any book I've ever read on hand evaluation downgrades stiff honors considerably and for good reason. I still want to scan for stiff kings and have some ambivalence about stiff queens.

[foobar]

Your original post just mentioned counting Qs as 0...stiff-Qs as 0 is probably the norm?

[foobar]

Your original post just mentioned counting Qs as 0...stiff-Qs as 0 is probably the norm?

[straube]

I'd guess so and I think most would count stiff kings as 1. But awm gives full value to each honor and Zel seemed surprised I wasn't scanning for a stiff queen (why I changed to looking for it now).

There is additional uncertainty and then cost for being able to scan for stiff queens. If you're only searching for a or king and have the other honor you know right away. If you have only A or K and include queen search then you don't. Big win though if you have both A and K.

[foobar]

 straube, on 2017-October-18, 08:20, said:

My first "marriage" problem. Per suggestion, I've started to eliminate less "interesting" hands but I've included this as someone may have a fix. The key to this is knowing whether North has the KQ of diamonds vs clubs. Statistically you would bet on the 4-cd suit but...

AT632 6 KQT9 JT6
KQ5 AJT3 AJ4 A98

South is captain. Let's say North shows pattern and 6 QPs at 3N.

With IMP (long to short, skip with odd-parity unless singleton, K-parity in suit with a single honour, K=2 always):
4C...4H (relay; odd ♠, even ♦, S(A) perforce)
4S...4N (relay; even clubs)
5C...5H (relay; no stiff H(K))

Basically we are down to S(A)+D(KQ) || S(A)+C(KQ).

[foobar]

 straube, on 2017-October-18, 08:20, said:

My first "marriage" problem. Per suggestion, I've started to eliminate less "interesting" hands but I've included this as someone may have a fix. The key to this is knowing whether North has the KQ of diamonds vs clubs. Statistically you would bet on the 4-cd suit but...

AT632 6 KQT9 JT6
KQ5 AJT3 AJ4 A98

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♣ - 5♦ (relay; odd K-parity, nothing in ♥, nothing in ♣, odd ♠, even ♦)

Perforce A♠, KQ of ♦

[yunling]

4♣-4♥(ask 1&2, 0/3/6)
cards placed

[awm]

Latest version of IMPrecision does not use king parity ask. We would get:

4♣ - 4♥ (odd spades, even diamonds)
4♠ - 4NT (even clubs)
5♣ - 5♦ (even hearts)
5♥ - 5♠ (KQ in first scan suit, so diamonds)

At this point the hand is completely resolved.

Of course, trick twelve on this hand is a little tough to come by -- you can extract it from the club spots but it's hard to know about these in a relay sequence. Without this you can play for a dummy reversal by ruffing three diamonds; this requires spades 3-2, and also might run into trouble if diamonds are 5-3 with shortness behind the long spade hand. You also need a bunch of entries for this which carries some risk. It seems possible to sign off A LOT earlier in a real auction, figuring that the actual hand is pretty much the best case and even so slam would be mediocre absent the club Jack-Ten (which you can't find out).

[sieong]

EPCB.
4♣ - 4n (odd ♠, even all others)
5♣ - 5♦ (♦KQ)

Honor structure resolved.

[nullve]

 straube, on 2017-October-18, 08:20, said:

AT632 6 KQT9 JT6
KQ5 AJT3 AJ4 A98

Same method: Teller shows, in order, 3-point range, king parity, queen parity, king location, queen location, jack location.

(...)
3N-4♣ (10-12 hcp (say); relay)
4N-5♣ (odd # of non-singleton Ks, no singleton K, odd # of non-singleton Qs, no ♠K; relay)
5♥-5♠ (♦K, no ♠Q; relay)
6♣-6♠ (♦Q, no ♠J; contract)
P

The last two relays seem pointless. Maybe passing 4N is best.

This post has been edited by nullve: 2017-November-13, 10:15