The "Bridge Methods" Blog

My 08th December 2009 entry engendered some correspondence on this topic and at the time, I suggested I would do a separate response on this at some time.  Here it is.

Let’s first recap on the semi-POS responses to 1♣ themselves.  1♠ shows hearts, 1NT shows BAL, 2♣ shows spades and at least one minor (possibly both), 2♦ shows diamonds possibly also clubs, 2♥ shows a 6+ club single-suiter, 2♠ shows a 6+ spade single suiter excluding those short in hearts and 2NT shows a 6+ spade single suiter short in hearts.

The first question to answer, IMO, is whether there is a sufficiently strong case for playing 2NT as natural in any relevant auction where it is available?  It is my judgement, on balance, that not sufficiently strong case can be made, so within DIP it is my intention to devote 2NT where available for artificial purposes.

I can see a case for playing a quantitative 2NT as natural in 1NT 2NT auctions, or those that might be analagous (e.g. DIP could use 1♣ 1NT 2NT this way if so desired) but with alternative ways of doing this in DIP it won’t be required.  For the others, I think the 6+ suits more often than not lend themselves to play in the suit when 2NT might be plausible (and when they don’t, how would you be able to tell anyway)?  2♣ and 2♦ are the more interesting cases.  I suspect many hands that might suggest 2NT opposite 2♣ might play OK in 3m once the minor suit(s) have been diagnosed.  Indeed, there may be a case for 2NT natural after such exploratory diagnosis, if it turns out to be a misfit of sorts.  This can be considered more as 1♣ 2♣ auctions are expanded.  Similarly, I think it would typically be too committal to bid 2NT after 1♣ 2♦, as once again, 3m is likely to play well after hand type diagnosis.  A secondary 2NT after such diagnosis may also have merit.

So, we are back to (direct, at least) 2NT as artificial, probably irrespective of the presence or otherwise of interference.  So, how to use this bid?  When I was younger, I had a mantra in this area that was along the lines of … “Lebensohl if you are forced to call, Rubensohl if you choose to call”.  However, in all the situations we are considering, except after 1♣ 1NT 2NT, at least one suit will have been shown.  Further, it will usually be the 1♣ opener rather than the semi-POS bidder that will be taking the next action.  For both these reasons, I think Rubensohl is likely to be less than optimal.

I have perhaps also been influenced by the work of Abraham and Johnson at the following link.

www.users.on.net/~mabraham/systems/misc/TOS_2NTinComp.doc

To summarise one of the key understandings from the above, you need to understand whether the partnership is in a strong (good) situation, in which case 3 level bids are also strong (not giving LHO a chance to further preempt) and going via 2NT is weak (bad).  Conversely, if the partnership is in a potentially weak situation, 3 level bids are also weak (not giving LHO a two-way shot to double you) and going via 2NT is strong.  In this 1♣ then semi-POS  scenario, the partnership can be regarded as being in a strong situation.

Though there are other times in DIP where “bad 2NT” will apply, for Part II of this blog, we’ll carry the “good 2NT” forward, see where it applies, and see where the exceptions lie.

Regards, BM

For anyone who uses a lot of relay sequences in their methods, Fibonacci numbers are important, even if you don’t think about them explicitly.  If you want to know why, look here

http://www.jimloy.com/algebra/fibo.htm

or in the previously linked to document of Bo-Yin Yang’s.  Fibonacci numbers will become relevant below.

Column readers may recall on my 10/11/2009 post that I had originally intended to get rid of Stayman and (Jacoby) transfers over DIP’s 1NT opening.  In the end I had to go back to a form of Stayman and transfers to make things work as I would like.  That’s OK – I don’t mind being proven wrong by going down a  technical dead-end.  What is more interesting is that I didn’t know why, in a sort of theoretical proof sense, why I couldn’t make my preferred approach work.  Until now that is.

In the February 2010 The Bridge World there is an article from Teng-Yuan Liang entitled “Puppet Plus”.  In Bridge World Standard 2001 (BWS2001), the latest version, 2NT is played as a transfer to diamonds, with three level bids showing both minors.  Note, it is acceptable to open 1NT in BWS2001 with a five card major.  The premise of Puppet Plus is that it is better to switch the meanings of the 2NT response (to 1NT) with that of the 3♦->3♠ responses (you lose the 3♣ weak with both minors functionality).  Why?  Because the 2NT strong with both minors response can be sensibly played initially as a five card suit ask, reverting to focusing on the minors if a 5-3 major suit fit was desired, but not found.  This way, you don’t need to play the 2♣ response as Puppet Stayman, which many find undesirable.

You can read the article itself if this interests you for your own BIG-like method.  It doesn’t do much for me, as you may recall:  I’m not a fan of five card majors systemically in a 1NT opening.  But it opened my eyes firstly as to why I went back to Stayman in response to DIP’s 1NT opening and secondly to consider the discovered principle as a general case.

The reason why Stayman works better than the line I was thinking along (more extensive use of tranfers) is that the non-step responses (i.e. 2♥+ rather than 2♦) do more good (eliminate the need for responder to show certain hand shapes/types as they are known to no longer be relevant) than bad (consume space preventing the showing of shapes/types which may still be relevant).  I can’t prove this to you explicitly, and in some case it relies on the ability to fully relay out shape/controls on the slam interested hands, but I am fairly certain this is the underlying reason it works better.  Let’s look at the general case.

In general, but to take as a specific example, a 1NT opening, a response “S” can show the same number of shapes/types as “S+1″ and “S+2″ combined.  So 1NT 2NT can show as many shapes/types as 1NT 3♣ and 1NT 3♦ combined.  You see this principle approximately at work in Puppet Plus.  Previously, I had implicitly considered these sequences as largely equal.  However, it is now apparent to me that that is not the case.  If 1NT 2NT 3♦ (that is, not bidding the step response) shows something than eliminates the need to show a bunch of other shapes/types, then1NT 2NT is more useful than 1NT 3♣ and 1NT 3♦ combined, notwithstanding Fibonacci principles.  In Puppet Plus, it is the case that a 3♦+ response by opener obviates the need to search for a minor suit fit on a number of 5431, 5530 and 6430 hands, either because a 5-3 major suit fit has been found, or because 3NT is more attractive once a five card major is known opposite responder’s shortage.

There is a second, lesser factor, in considering how to slice and dice S vs S+1 and S+2 responses.  In those times where the considerations are generally equal above, i.e. there is no particular advantage that could accrue from not bidding the step response (e.g. as per the earlier example, but where 1NT 2NT 3♦ costs more than it gains) then it makes sense to slice and dice based on what would happen in competition.  Carrying this principle to its conclusion, you want the S+1 and S+2 steps to shows as different shapes/types as possible given the options you have at your disposal.

In short, distributing shapes and types according to Fibonacci principles in relay (and other) methods is a good place to start, but there are pragmatic reasons why one might slice and dice, possibly even in an ostensibly inefficient way, how the system construction is finished.

Regards, BM

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.