The "Bridge Methods" Blog

In the previous blog entry, I described levels one to three of DIP’s adopted control showing method:  QPs, Ace Parity and King Parity.  It is amazing how often just knowing these levels unlocks the hand, either completely (i.e. you know exactly what partner has) or sufficiently (i.e. the possible options have narrowed such that you can know, at least on balance of probability, what to do).  This blog entry details the rest of the levels, just in case you need them.

The important concept in the remainder of this method is to assume that partner knows exactly what you’ve got from what you’ve already indicated.  So, if you’ve shown EVEN ace parity, and you have two aces, you assume that partner knows you have exactly two (rather than zero or four).  It won’t always be the case that he does know, but he usually will and often if not, that itself will be a sign to hold back.

Let’s consider responder’s ♠Qxx ♥xx ♦Axxx ♣KJxx in response to 1♣ to work through as an example.

The auction is likely to have gone

1♣ 1♦ 9+ hcp, ART FG
1♥ 2♣ BAL
2♦ 2♥ no M
2♠ 2NT both m’s
3♣ 3♥ 3=2=4=4

It might then continue

3♠ 4♣ 6 QP’s
4♦ …

Responder is going to bid 4NT or higher.  He will bypass 4♥ as he has expected (ODD) ace parity and will bypass 4♠ as he has expected (ODD) king parity.  The scan order is ♣♦♠♥ (lowest of even length first).  As it happens, he also skips 4NT as he has the king in the first scan suit (♣).  At this point, something interesting happens.  Responder gives no further information about kings (i.e. he exists the level four scan and he will never enter the level six scan).  This is because he assumes you know, from the level one to three responses, that he only has one king and you have already shown it.

Actually, the above is clear here as opposed to some higher QP scenarios.  To be odd aces and odd kings, you have to be QKA (KKKA, for example, has too many QP’s).  This wouldn’t be as clear with an 11 QP hand, where you might be KAAA or QQKKKA.  In this scenario, however, opener almost certainly has either 2+ aces or 2+ kings, so can usually eliminate one of the above options, and if not, he probably has 3+ queens, in which case he could once again eliminate an option.  The worst honour dispersion for clarity is JJJJQQKA with a few tens thrown in, in which case, he should probably be trying to sign-off rather than relay further anyway!

So, we move to level five scanning.  Here, we finally stop, as we don’t have specifically either the ♣A or ♣Q.

… 5♣ Odd A’s, Odd K’s, ♣K, neither or both of ♣AQ
5♦ …

If at this point opener still doesn’t have enough information (maybe, for the grand) he can relay again with 5♦ as shown above.  Responder continues level five scanning and bypasses 5♥ as he does have the ace or queen of diamonds.  This ends level five scanning and normally we would go to level six scanning (kings in short suits) but as mentioned earlier, opener “knows” responder hasn’t got any of those, so we skip level six scans.  We go straight to level seven scans and scan the longest of the short suits (♠).  As we do have the ace or queen of spades, we bypass 5♠ as well.

We have now shown all our QP’s, so we don’t do a level seven scan on hearts but instead go straight to level eight.  Lo and behold, we have the ♣J, so we bypass 5NT.  We don’t have the ♦J, so responder comes to rest in 6♣.  At this point, opener places the contract (possibly by passing).

DIP uses the agreement that the last possible scan is 5NT, so no further relays are possible.  You won’t get too much argument from me, however, if you don’t want to have this arbitrary limit in which case you could scan for the ♠J using 6♦ here.  Also, some relayers don’t allow you to automatically scan for jacks (in which case, you would bid 5NT not 6♣ in the above auction).  Whether you could bid 6♣ to then ask for jacks depends, of course, on whether you subscribe to the 5NT “last ask” agreement, or not.

There’s a fair bit to take in above (indeed, it took me some time to get my head around the inferential approach inherent and I have played relay methods for some time).  However, I hope it piques your interest, whether for DIP as a whole and/or for this type of control showing method in particular.

Regards, BM.

Notwithstanding the frustration expressed in my previous blog entry, for use in DIP I currently advocate a [modified] method of Scandinavian origin.  This may be somewhat an odd choice,  as I haven’t had first hand opportunity to discuss the formative thinking with its original authors.  Having said that, the more it is tested the probable original considerations become clearer.

Firstly, an overview of the levels (read “steps”) involved after you have shown your exact hand shape – I’ll expand on these later.

1. Show Queen Point controls (QPs) versus an expectation based on your minimum hcp count.
2. Show Ace parity versus an expectation based on the control count from level (1).
3. Show King parity versus an expectation based on control count from level (1) modified by Ace parity from level (2).
4. Show/deny Kings in long (4+) suits.
5. Show/deny Aces and/or Queens in long (4+) suits.
6. Show/deny Kings in short (3-) suits.
7. Show/deny Aces and/or Queens in short (3-) suits.
8. Show/deny Jacks, in normal suit order:  longest (else lowest) suit first.

Now for some elaboration.  In level one, show your QPs, i.e. A=3, K=2, Q=1 modified as follows

• A singleton King counts as 1 QP (and for the balance of the auction is considered Queen), and
• A singleton Queen counts as 0 QPs (and for the balance of the auction is considered a Jack).

In some literature, the modified QP scheme above is referred to as “ZZ points“.

The level of QPs expected is based on 60% of the  minimum hcp count (or any other 40 point deck method such as “Fifths Point Count“ as used in DIP), rounded down if needed..  For the common DIP  opening ranges, rounding is not necessary:  for 10 hcp it’s 6 QPs and 15 hcp it’s 9 QPs.  However, for the 1♦ positive response to 1♣ showing 9 hcp, the QP range is rounded down, from 5.4 to 5.

For levels two and three, you have to consider the most likely honour dispersion, a priori, for the QP count you have shown.  I can’t, yet at least, prove the pseudo-logic outlined below, but it intuitively feels right and is recommended for DIP.  If you have 1 QP it must be a Q, 2 it is assumed to be a K (rather than QQ), 3=A, 4=AQ, 5 = AK, 6 = AKQ.  7 to 12 controls assume the same pattern added to AKQ for 6, so 7=QQKA, 8=QKKA, 9=QKAA, 10=QQKAA, 11=QKKAA, 12=QQKKAA.  You can calculate 13 and above by repeating the pattern again.

From the above picture building exercise, you calculate “Expected Parity”.  So for 10 QPs, you expect an EVEN Ace parity (2) and an ODD King parity (1).  For level two, Ace parity, GO if you have the expected parity (you skip a step) and STOP if you don’t (you bid the next step).

For level three, King parity, you derive it as indicated above, but you reverse your expectation if the Ace parity in level two was not as expected.  So, imagine the same 10 QP example, if partner STOPPED, indicating unexpected Ace parity, you would reverse your King parity expectation from ODD to EVEN, and on the next scan, GO with EVEN kings and STOP with ODD.

Note, the Scandinavian examples I saw were not quite as scientific in this regard, they simply always stopped with EVEN (for both Aces and Kings).  This does not seem to approximate an optimal approach – that is, optimising the number of GO then GO and/or minimising the number of STOP then STOP parity auctions.

On another note, I originally considered putting the King parity level before the Ace parity one. Counter intuitive as this might initially appear, it is quite practical, even sensible, but only if at least one of the following two conditions hold true

• You are only going to incorporate one parity level, i.e. Ace or King parity, but not both, or
• You are not going to use modified expected parity (and instead use a more simplistic approach like the original Scandinavian one)

To explain, anecdotal evidence would suggest that King parity alone is better than Ace parity alone, which is why (along with its overall effectiveness) a small school of relayers have incorporated it as an additional standalone step into “standard” denial cue based methods.  However, it is often more difficult to know how to adjust expected Ace parity from unexpected King parity, than the other way around (where an additional or missing Ace can usually sensibly be assumed to replace/add a KQ combination (rather than the far less likely QQQ alternative).  As in DIP we intend (and expect) to show both parities, we show Ace parity first.

This continues to be a long blog entry, so it will be split and continued in the next iteration.

Regards, BM

Specifically in DIP, but more generally in most relay methods, I am mildly frustrated with which way to head on this function.  To be clear, I mean when you have shown your shape, and shown your (usually A=3 based) control count, how to facilitate the showing of your exact honour location.

This is not to say that DIP hasn’t adopted a decent method, which I may describe a little below.  It is just that I have no means (other than very anecdotal) of telling how good it is with respect to other options, AND, when it doesn’t work well it tends to work horribly.  In this sense, it is the same as almost every other method I have come across.

For a long time, I played a form of parity control showing.  In this, you scan suits, longest first, and STOP with EVEN and GO with ODD.  When you scan a second time, if originally EVEN, you STOP with 0/4 QPs and GO with 2/6 QPs.  If originally ODD, you STOP with 1 honour (i.e. A or Q) and GO with 2 honours (i.e. AK or KQ).  Then, you assume partner knows which you’ve got, and start scanning for jacks, STOP without and GO with.

In the original version I used, you didn’t at any stage scan the shortest suit, so with, say 3=5=3=2, you scanned HDS,HDS,HDSC.  Which order you scan equal lengths is arbitrary – I do lower first, in keeping with the order DIP shows suit lengths.  You don’t scan the shortest suit, clubs here, for AKQ:  you assume partner can work it out from the others.  You do scan the shortest suit for jacks, however, unless you have no places left to hold one (e.g. you have ♣AQ or similar and partner now already “knows” this).

In recent versions I have seen of this method, you ignore the second shortest suit instead.  In the above case, it would be spades, so you would scan HDC, HDC, HDSC.  I suppose the theory is that the shortest suit is quicker/easier to scan (definitively) than the second shortest, maybe making it more efficient?

I certainly don’t mind the above method still and it mostly works well.  Also worth considering is the more common (possibly standard) version, where on the first pass, you STOP with AKQ or none and GO with 1 or 2 thereof.  On the second pass, if you stopped last time you ignore it and assume partner knows which you’ve got.  If you went last time, on the second pass you STOP with 1 and GO with 2.  On the third pass, you scan for jacks as before.  There is a variant where on the “AKQ or none”, you explicitly show which on the second scan.  There is another variant (possibly the original one) where A=2 & K=1 which can be used.

One improvement to the above method (and for the jacks component of other ones) is that if you only have 1 card left in the suit, you reverse the order:  i.e. STOP with the card and GO without.  The odds certainly favour this approach being more efficient and I recommend adopting it where relevant.  So, let’s say you have the 3=5=3=2 above and you “went” in clubs on the first pass, showing 1 or 2 of the ♣AKQ.  Using this approach, on the second pass you would GO with 1 and STOP with 2 honours.

I started describing the Scandinavian-based method that DIP has currently adopted, but it was taking too long.  I’ll split this out into a second post.

I have described two of the main families of honour showing relays above and alluded to another.  However, the problem remains – each of them are not always ideally suited and there appears no easy way to choose which one to prefer or how yet to sensibly splice more than one of them into a system.

Regards, BM.

If you want the thinking behind this, see the previous blog entry.  However, I think I’ve convinced myself to add 5m4m22’s into the balanced schema in all scenarios in DIP.  Also, I believe I’ll go with the 1NT+ juggling instead of the 1♠ & 1NT juggling for BAL resolution, when relevant.

So, as a reprise, I’ll outline the net effects of these below.

1NT 2♣ (Stayman) 2♦ 2♥ R in 1st/2nd seat

• 2♠:  both minors, then 2NT R
  • 3♣:  2=3=4=4 (most common shape first, high shortage first)
  • 3♦:  3=2=4=4
  • 3♥:  2=2=4=5 (low length first)
  • 3♠:  2=2=5=4, B or B+1 (exceptionally B-1) QPs
  • 3NT:  2=2=5=4, B+2 or B+3 QPs
  • 4♣:  2=2=5=4, B+4 QPs
• 2NT:  5♣332, then 3♣ R
  • 3♦:  2=3=3=5
  • 3♥:  3=2=3=5
  • 3♠+:  3=3=2=5, QP’s as above
• 3♣:  4m333, then 3♦ R
  • 3♥:  3=3=3=4
  • 3♠+:  3=3=4=3, QP’s as above
• 3♦:  2=3=5=3
• 3♥:  3=2=5=3
• 3♠+:  3=3=5=2, QP’s as above

1♣ 1♦ [1♥] 2♣ 2♦ R in 1st/2nd seat

• 2♥:  no M, then 2♠ R
  • Resolves as for 1NT 2♣ 2♦ 2♥ R above, except up one step, so the B to B+3 QPs collapse into a single 3NT response where relevant
• 2♠:  4♥432, then 2NT R
  • 3♣:  4♠4♥32, then 3♦ R
    • 3♥:  4=4=2=3
    • 3♠:  4=4=3=2, B to B+1 QPs
    • 3NT:  4=4=3=2, B+2 to B+3 QPs
    • 4♣+:  4=4=3=2, B+4 QPs etc
  • 3♦+:  Resolves analogously 4♠4m32 below
• 2NT:  4♠4m32, then 3♣ R
  • 3♦:  4=2=3=4
  • 3♥:  4=3=2=4
  • 3♠:  4=2=4=3
  • 3NT:  4=3=4=2, B to B+3 QPs
  • 4♣+:  4=3=4=2, B+4 QPs etc
• 3♣:  5♥332, then 3♦ R
  • Resolves as for 3♥+ with 5♠332 below
• 3♦:  4M333, then 3♥ R
  • 3♠:  3=4=3=3
  • 3NT:  4=3=3=3, B to B+3 QPs
  • 4♣+:  4=3=3=3, B+4 QP’s etc
• 3♥:  5=2=3=3
• 3♠:  5=3=2=3
• 3NT:  5=3=3=2, B to B+3 QPs
• 4♣+:  5=3=3=2, B+4 QP’s etc

1♣ 2♦ (BAL, no M) 2♥ R in 3rd/4th seat

Exactly as for 1♣ 1♦ 2♣ 2♦ 2♥ 2♠ R (and the other two comparable auctions above) in 1st/2nd seat, but down one step.  No need to switch anything around as there is no major and both minors have already been bid.

1♣ 2♣ (BAL, at least 1M) 2♦ R in 3rd/4th seat

Similar to 1♣ 1♦ 2♣ 2♦ 2♠+ (and the other two comparable auctions above) in 1st/2nd seat, but down one step.  However, this time, we do sometimes need to switch the majors around to avoid bidding one we have, so …

• 2♥:  4♠432, then 2♠ R
  • 2NT:  4♠4♥32, then 3♣ R
    • 3♦:  4=4=2=3
    • 3♥:  4=4=3=2, >= B+4 QPs
    • 3♠:  4=4=3=2, B to B+1 QPs
    • 3NT:  4=4=3=2, B+2 to B+3 QPs
  • 3♣+:  Resolves as for 4♥4m32 below
• 2♠:  4♥4m32, then 2NT R
  • 3♣:  2=4=3=4
  • 3♦:  3=4=2=4
  • 3♥:  2=4=4=3
  • 3♠:  3=4=4=2, B to B+1 QPs
  • 3NT:  3=4=4=2, B+2 to B+3 QPs
  • 4♣+:  3=4=4=2, B+4 QPs etc
• 2NT:  5♥332, then 3♣ R
  • 3♦+:  Resolves as for 5♠332 below
• 3♣:  4M333, then 3♦ R
  • 3♥:  4=3=3=3
  • 3♠:  3=4=3=3, B to B+1 QPs
  • 3NT:  3=4=3=3, B+2 to B+3 QPs
  • 4♣+:  3=4=3=3, B+4 QP’s etc
• 3♦:  5=2=3=3
• 3♥:  5=3=2=3
• 3♠:  5=3=3=2, B to B+1 QPs
• 3NT:  5=3=3=2, B+2 to B+3 QPs
• 4♣+:  5=3=3=2, B+4 QP’s etc

Note, with the 5M332’s, we don’t need to switch the order around (lower still first) as there are no problems.  However, with the 4M432’s and the 4M333’s, we do.

Hope this helps clarify the potential outcomes from the previous blog entry.

Regards, BM

Alas, referring to the title of this blog entry, I’m not sure one exists.  However, I’ll explore it a little today and discuss what comes to light.  Warning:  this is a fairly long, theoretical discourse.

THE 1NT OPENING

The first case to consider is DIP’s 1NT opening.  I can’t recall expounding on it in detail previously, so here is the version currently proposed

1NT 2♣ (i.e. Stayman)

• 2♦:  No major
• 2♥:  3=4=3=3
• 2♠+: 4=3=3=3 (2NT+ shows the same shape with various types of invitational acceptance)

After 1NT 2♣ 2♦ 2♥

• 2♠:  Both minors, then after 2♠ R
  • 3♣:  2=3=4=4
  • 3♦:  3=2=4=4
  • 3♥:  2=2=4=5
  • 3♠+:  2=2=5=4, showing controls
• 2NT:  5♣332
• 3♣:  4m333
• 3♦:  2=3=5=3
• 3♥:  3=2=5=3
• 3♠+:  3=3=5=2, showing controls

I’ll assume you can figure how the single suiters resolve.

1ST/2ND SEAT 1♣ OPENING

OK, now let’s switch to the first and second seat [1/2] 1♣ openings with a semi-POS 1NT response. Currently, 5m4m22 hands are not included, but maybe they should be? 5M332 are included, for the record.  So, after 1♣ 1NT 2♦ (remember, we are now almost certainly going to keep 1♣ 1NT 2♣ as some sort of Staymanic device) then continuations are

• 2♥:  minor 1/S, then over 2♠ R
  • 2NT:  5♣332
  • 3♣:  4m333
  • 3♦+:  5♦332, resolving shape then controls
• 2♠:  4+♥, <4♠, then over 2NT R
  • 3♣:  5♥332
  • 3♦:  4♥4♣32
  • 3♥:  3=4=3=3
  • 3♠:  2=4=4=3
  • 3NT:  3=4=4=2
• 2NT:  4♠, <4♥
  • Then after 3♣ R, resolves analogous to 1♣ 1NT 2♠ 2NT 3♦+
• 3♣:  5♠332
• 3♦:  4♠4♥32
• 3♥:  2=3=4=4
• 3♠+:  3=2=4=4, showing controls

Once again, I hope you can figure out how to resolve the not fully detailed bits (remembering DIP’s low length & high shortage first precepts).  The 3♦+ bids are arranged as they are to avoid bidding the majors if you have them.

Clearly, since the hands without at least one major resolve up to 3♠ (whereas, the hands with at least one major may resolve up to 3NT) there is scope to handle the 5m4m22 hands through 1♣ 1NT (rather than 1♣ 2♦ where they currently reside).  If so, the structure could be switched to something more akin to the 1NT 2♣ 2♦ 2♥ continuations.  Nevertheless, this is not yet part of DIP.

Now, on to the [1/2] 1♣ 1♦ 2♣ 2♦ or 1♣ 1♦ 1♥ 2♣ 2♦ auctions.  Noting 1♣ 1NT 2♣ being reserved as Staymanic and 1♣ 1NT 2♦ instead as the relay, we can use analogous relay continuations for these two sequences.  Though there is some merit in considering bidding the majors “naturally” after 1♣ 1♦ 1♥ 2♣ 2♦ as on balance, opener is likely to be stronger than responder, this is more than offset IMO by keeping the known hand concealed.  By “natural”, I mean playing 1♣ 1♦ 1♥ 2♣ 2♦ 2♠ as showing spades rather than hearts.

So, for the present, we can assume that 1♣ 1NT 2♦ 2♥+ and 1♣ 1♦ 2♣ 2♦ 2♥+ and 1♣ 1♦ 1♥ 2♣ 2♦ 2♥+ all resolve identically.  What, then, if we bring 5m4m22’s into the 1♣ 1NT structure?  It would probably then also, to ease memory strain if nothing else, make sense to bring them into the 1♣ 1♦ 2♣ and 1♣ 1♦ 1♥ 2♣ structures as well.

3RD/4TH SEAT 1♣ OPENING

Finally, we consider 3rd/4th seat [3/4] 1♣ continuations (which you got a taste of in the previous blog).  Simplistically, the default is to switch to a fairly classic style of Symmetric Relay, with1♦ as the NEG (i.e. < 7 hcps in context, as [3/4] 1♣ shows 17+ hcp), and 1♥+ POS according to the following schedule

• 1♥ = ♠, 1/S or 2/S or 3/S
• 1♠ = ♥, 1/S or 2/S
• 1NT = BAL
• 2♣ = ♦, 1/S or 3/S
• 2♦ = ♣, 1/S
• 2♥+ = both minors

Except, we don’t like to wrong-side the notrumps like this.  The simple way to avoid this (and how I suggested bidding in the previous blog entry) is to juggle 1♠ and 1NT respones around so that

• 1♠ = BAL or ♥/♦
  • then [3/4] 1♣ 1♠ 1NT 2♣ shows ♥/♦ and 2♦+ shows BAL as if it had went [3/4] 1♣ 1NT 2♣ 2♦+ in the unjuggled method)
• 1NT = ♥ or ♥/♣
  • then [3/4] 1♣ 1NT 2♣ 2♦+ is the same as if it had went [3/4] 1♣ 1♠ 1NT 2♦+ in the unjuggled method)

This is certainly a reasonably effective and simple switch, but there is another option (possibly slightly superior) you may prefer, which lets 1♠ remain showing all the heart hands.  This version juggles the 1NT+ responses instead.

• 1NT:  ♣ or ♣/♦ or ♣/♦/M
• 2♣:  BAL, type 1
• 2♦:  BAL, type 2
• 2♥+:  ♦, 1/S

Actually, you can slice’n'dice your three minor suit options (♣ or ♦ or ♣&♦) around whichever way you like in the above structure, as long as the ♣/♦/M ones remain in 1NT.  The key thing is, because you are immediately breaking out your balanced hands into two types, it is desirable that these two bids are internally consistent.  What does this mean you need to do in practice?

Simplistically, you could just equate the following

1. [3/4] 1♣ 2♦ 2♥ 2♠+ with [1/2] 1♣ 1♦ 2♣ 2♦ 2♥ 2♠ 2NT+, i.e. the same as normal but down one step, and similarly
2. [3/4] 1♣ 2♣ 2♦ 2♥+ with [1/2] 1♣ 1♦ 2♣ 2♦ 2♠+

Doing this gives two problems.  Sequence (2) above wrong-sides the majors if left untouched.  So, we have to flip the majors around and show spades first.  So, after 1♣ 2♣ 2♦, we should play

• 2♥:  4+♠, <4♥
• 2♠:  4♥, <4♠
• 2NT:  5♥332
• 3♣:  4♠4♥32
• 3♦:  3=3=3=4
• 3♥+:  3=3=4=3

OK, so that sorts out the important first problem.  The less important second one is that [3/4] 1♣ 2♣ would promise a major, except when 4m333.  This appears a little ungainly.  If you add 5m4m22’s into the overall balanced hand mix, this problem goes away.  This would leave

[3/4] 1♣ 2♦ 2♥

• 2♠:  both minors
• 2NT:  5♣332
• 3♣:  4m333
• 3♦+: 5♦332 resolving

[3/4] 1♣ 2♣ 2♦

• 2♥:  4+♠, <4♥
• 2♠:  4♥, <4♠
• 2NT:  5♥332
• 3♣:  4=4=2=3
• 3♦+:  4=4=3=2

Note how the above does two ostensibly good things

1. [3/4] 1♣ 2♦ 2♥ 2♠+ is the same as [1/2] 1NT 2♣ 2♦ 2♥ 2♠+
2. Makes [3/4] 1♣ 2♣ always promise at least one major and [3/4] 1♣ 2♦ always deny one.

Having done the above, it is probably then worth juggling the [3/4] 1♣ 2♣ sequence around to the following more optimal structure (everything out by 3♠)

• 2♥:  4♠432
• 2♠:  4♥432, <4♠
• 2NT:  5♥332
• 3♣:  4M333
• 3♦+:  5♠332

All in all, I don’t think I’ve achieved the Grand Unification Theory hoped, but I think I’ve almost convinced myself to include 5m4m22’s in BAL hands irrespective of how the auctions starts.  I’ll have a think about it some more.

Regards, BM