Ask Jerry (ACBL Bulletin)

[TimG]

In the July Bulletin Jerry helms writes (p 40) about opening 1♦ with 4=4=3=2 shape.

 

He says that 4=4=3=2 shape makes up 1.796% of all hands. (98.204% are not 4=4=3=2.)

 

He goes on to say that "if you further calculate that a one-level opening bid shows between 12 and 21 HCP (and factor out the 15-17 HCP with this shape that are opened 1NT), the probability of a four-card or longer diamond suit for a 1♦ opener increases to 99.56%."

 

Won't the percentage of 12-14 and 18-21 point hands that are 4=4=3=2 stay pretty close to the 1.796% of all hands that are 4=4=3=2? (My guess is that slightly more than 1.796% of 18 HCP hands are 4=4=3=2, in part because it is impossible to have 18 HCP in some extreme shapes).

[helene_t]

As you cite it, it is indeed wrong.

 

If the probability of being 4=4=3=2 is 1.8%, then probability of being 4=4=3=2 given that we opened 1♦ must be substantially more.

 

4=4=3=2 is one of the 11 balanced shapes that open 1♦. If they are all appr. equally frequent and a little less than half of our 1♦ openings are balanced, the probability of being 4=4=3=2 given that we open 1♦ is appr. 4%

[Free]

There's no direct link between "all possible distributions" and "opening 1♦". You don't open all distributions 1♦, so a hell of a lot hands are irrelevant. The number of hands that are opened 1NT, compared with the irrelevant hands, is very small. This proves helene's quote that "probability of being 4=4=3=2 given that we opened 1♦ must be substantially more". I always thought it was around 5%, but haven't calculated it so I could be wrong.

[Lobowolf]

In the July Bulletin Jerry helms writes (p 40) about opening 1♦ with 4=4=3=2 shape.

 

He says that 4=4=3=2 shape makes up 1.796% of all hands.  (98.204% are not 4=4=3=2.)

 

He goes on to say that "if you further calculate that a one-level opening bid shows between 12 and 21 HCP (and factor out the 15-17 HCP with this shape that are opened 1NT), the probability of a four-card or longer diamond suit for a 1♦ opener increases to 99.56%."

 

Won't the percentage of 12-14 and 18-21 point hands that are 4=4=3=2 stay pretty close to the 1.796% of all hands that are 4=4=3=2?  (My guess is that slightly more than 1.796% of 18 HCP hands are 4=4=3=2, in part because it is impossible to have 18 HCP in some extreme shapes).

If I'm understanding your post, you're questioning that apparently a disproportionately large percentage of hands are removed, when the 1.8% drops down to 0.4%?

 

His numbers seem plausible to me, without doing the math. If you get rid of all 0-11 point hands, all 15-17 point hands, and all 22+ point hands, I think you could lose about 75-80% of the hands (1.4% of the 1.8%).

 

 

Edit: Was only considering what he was presenting, which, as Helene points out, is a different question that he purports to answer. This all would suggest that 0.4% of hands would be 4-4-3-2 and open 1♦ , but not that a given 1♦ opener has a 0.4% change of being 4-4-3-2.

[barmar]

Tim's concern is correct. It doesn't matter what percentage of all possible hands are 4432 and opened 1♦. What matters is the percentage of hands that are opened 1♦ that are 4432.

 

An analogy: there are millions of animal species on the planet, and only a tiny fraction of them are dogs that bark. But if you encounter a dog, there's probably at least a 95% chance that it barks. This is a much more extreme difference than the 1♦ opening, but it demostrates how different the two ways of looking at the situation are.

[aguahombre]

isn't lobo's figure of "0.4%" off by a tad? not being a rocket scientist, i would think 4% (one in 25) is the number.

[ArtK78]

The question should be: "What percentage of 1♦ opening bids contain 3-card diamond suits?" or "What percentage of 1♦ opening bids contain 4-card or longer diamond suits?" Obviously, an answer to either of these questions answers the other (assuming that one cannot open 1♦ on a 2-card or shorter holding).

 

I am assuming a Standard American bidding structure.

 

Having defined the question properly, one then goes on to determine the criteria for opening 1♦ on 3-card diamond suits, 4-card diamond suits, and so on up the line. After defining the criteria, one can attempt to determine how many hands (or what percentage of all hands) match the criteria, and then one can answer the question.

 

Without going through all of the hoops, I would guess that about 10-20% of all 1♦ openings contain a 3-card diamond suit. This is purely a guess based on my experience.

 

Certainly, the number of hands which qualify for a 1♦ opening (specificially a 4-4-3-2 distribution and not a 1NT opening (15-17 HCP) and not too strong to open at the one level (20+ HCP)) is very small as a percentage of all hands. But, compared to all hands that would otherwise open 1♦, the percentage is not that small.

 

Also, one is assuming that one would never open 1♦ on a hand with 4-3-3-3 or 3-4-3-3 distribution because of the relative honor strength of the minor suits. Many players would open 1♦, rather than 1♣ holding KQx Qxxx AKx xxx, but many other players always open 1♣ with this distribution.

[jdonn]

Without going through all of the hoops, I would guess that about 10-20% of all 1♦ openings contain a 3-card diamond suit. This is purely a guess based on my experience.

I would have estimated something more like 3%.

[ArtK78]

Without going through all of the hoops, I would guess that about 10-20% of all 1♦ openings contain a 3-card diamond suit.  This is purely a guess based on my experience.

I would have estimated something more like 3%.

Maybe 10-20% is an overbid, but it is certainly a lot more than 0.4%.

[Lobowolf]

isn't lobo's figure of "0.4%" off by a tad? not being a rocket scientist, i would think 4% (one in 25) is the number.

I'm just taking the flip side of the 99.56% figure in the original post (rounding down from 0.44%).

[TimG]

I think what the column wanted a figure for was: what percentage of all opening bids are a 1D opening with 4=4=3=2 shape? Not: what percentage of 1D openings are made with 4=4=3=2 shape? But, I don't think either is 0.44%. I think that number is what percentage of all hands are opened 1D with 4=4=3=2 shape.

[Lobowolf]

Without going through all of the hoops, I would guess that about 10-20% of all 1♦ openings contain a 3-card diamond suit.  This is purely a guess based on my experience.

I would have estimated something more like 3%.

Maybe 10-20% is an overbid, but it is certainly a lot more than 0.4%.

The point of the previous posts is that the 0.4% figure (if it's accurate) isn't the percentage of all 1♦ openers (as Helms stated), but the percentage of all hands. (that are specifically 4=4=3=2 and open 1♦).

 

That is, 1.8% of the time you're dealt ANY hand, you'll be 4=4=3=2. That 1.8% breaks down as follows:

 

1.4%: You're either too weak to open, or you have a 1NT opener, or you have a 2♣ opener.

 

0.4% You have a 1♦ opener.

 

 

 

But if you just look at how many 1♦ openers you have, then a much higher percentage of them are 4=4=3=2 (for the reasons given by others, above).

[helene_t]

Tim is probably right.

[peachy]

In the July Bulletin Jerry helms writes (p 40) about opening 1♦ with 4=4=3=2 shape.

 

He says that 4=4=3=2 shape makes up 1.796% of all hands. (98.204% are not 4=4=3=2.)

 

He goes on to say that "if you further calculate that a one-level opening bid shows between 12 and 21 HCP (and factor out the 15-17 HCP with this shape that are opened 1NT), the probability of a four-card or longer diamond suit for a 1♦ opener increases to 99.56%."

 

Won't the percentage of 12-14 and 18-21 point hands that are 4=4=3=2 stay pretty close to the 1.796% of all hands that are 4=4=3=2? (My guess is that slightly more than 1.796% of 18 HCP hands are 4=4=3=2, in part because it is impossible to have 18 HCP in some extreme shapes).

Playing strong NT, 1D opener with 4-4-3-2 is more frequent than a 2C opener, in my experience. If I play weak NT [sometimes I do] the 1D opening with three is rare since much of the hands are opened 1NT and only 15-19HCP 4-4-3-2 are opened 1D.

 

Not going into percentages, that is not my strong suit :)

[Mbodell]

The math in this article was terrible with respect to the conditional probability. I likewise laughed when reading it.

 

I just ran a couple of quick sim that suggests playing a standardish strong nt 5 card major system, but with a very strict balanced nt (no 5422, no 6322) that you'll open 1♦ about 11.23% of the time in first seat. About 0.52% of hands you'll open 1♦ with a 3 card suit. But the more important thing is when you open 1♦ you'll have only 3 diamonds 4.61% of the time. This shifts your expected diamonds for a 1♦ from 4.86 (require 4+) to 4.76 (require 3+).

 

If instead we make 5m422 and 6m322 hands open 1nt when in range then it shifts up only slightly with us opening 1♦ 11.00% of the time with about 0.53% of hands having you'll open 1♦ with a 3 card suit but now the conditional probability of 1♦ having only 3 is up to 4.81% of the time. And the expected diamonds doesn't really change.

 

So really, if someone plays 1♦ promises 4+ or 4432 then almost 1 in 20 times when they open 1♦ it is 4432.

 

I'd, personally, rather have 4432 in the 1♣ because then I know the 1♦ is 4+ and the 2 card club suit since the clubs are always suspect.

[mike777]

I thought the point of the article was that playing a short club in a 2/1 framework was not worth it. If you open 1d you have 4+d almost 100% of the time and responder should assume such. If you open 1c you promise 3+ clubs 100% of the time.

[TimG]

I sent a note to Jerry and got a reply which started with: "You are 100% correct that the math was flawed." He also said he is planning a correction.

 

I ran a simulation with some very basic conditions (44 in minors open 1D and 33 in minors open 1C; 1N = 15=17, 2N = 20-21; opening bid = 12-21; 5M332 always opened 1M, 5m332 always opened NT when in range; 54 and any 6-card suit never opened NT) and came up with 4.28% of 1D openings were 4=4=3=2.

[awm]

Another issue is that the diamond length matters most in auctions where responder is interested in raising diamonds. These are hands where responder has four diamonds (and considers raising) or where responder has five diamonds (and considers raising to the three-level).

 

The conditional probability Pr[1D opener has only 3 diamonds GIVEN responder has 5 diamonds] will unfortunately be substantially higher than the a priori probability. So the "could be only three" situation hurts even more than these statistics would indicate.

[PrecisionL]

I sent a note to Jerry and got a reply which started with: "You are 100% correct that the math was flawed."  He also said he is planning a correction.

 

I ran a simulation with some very basic conditions (44 in minors open 1D and 33 in minors open 1C; 1N = 15=17, 2N = 20-21; opening bid = 12-21; 5M332 always opened 1M, 5m332 always opened NT when in range; 54 and any 6-card suit never opened NT) and came up with 4.28% of 1D openings were 4=4=3=2.

I did such an analysis on 1♦ openings in a Precision context in 2004. Using a strong NT and 5-card major system my 'adjusted' calculations show that when you open 1♦ it will only be a 3-card suit 5.6 % of the time.

 

Larry

[aguahombre]

Anyone run an analysis based on opening 1C with balanced 44 in minors and 33 in minors(with a 15-17 NT)? This eliminates 2344 and 3244 from the relevent cases where 1D is opened. 1345, 3145, and 0445 with less than reverse strength added to the relevent cases? Whether you approve of the methods or not, lots of people do, and I think this changes the percentage of 1D openings which are exactly 4432, are opening bids, and are too weak to open in NOTRUMP.

[PrecisionL]

Anyone run an analysis based on opening 1C with balanced 44 in minors and 33 in minors(with a 15-17 NT)? This eliminates 2344 and 3244 from the relevent cases where 1D is opened.  1345, 3145, and 0445 with less than reverse strength added to the relevent cases?  Whether you approve of the methods or not, lots of people do, and I think this changes the percentage of 1D openings which are exactly 4432, are opening bids, and are too weak to open in NOTRUMP.

Eliminating (32)-4-4 removes 3.6 % of hands from opening 1♦ and raises the precentage of 1♦ being only 3-cards from 5.6 % (my previous post, recalculated to 5.9 %) to 6.8 %.

 

f(3♦  only) = 1.667 / (28.23 - 3.6) = 6.77 %

 

Larry (Who opens 1♦ with minimum hands and 4-4 or 4-5 or 5-4 in the minors.) [This is a statement, not a question.]

 

Edited 7/4/09 7:06 pm EDT (Sorry about the confusing ending to my unedited post)

[fred]

For those of you quoting odds, how are you computing odds of hand types in which both distribution and HCPs are defined?

 

To the best of my knowledge there is no easy way to do this using pen and paper (with or without the help of a calculator). Of course it is not that hard to write a computer program to do the math.

 

Larry - there exists a significant minority of experts in North America who favor opening 1D on some minimum hands that include 4-5 in the minors. Eric Kokish is a strong believer in this approach. I tend to do this myself with 0445 distribution and (very) occasionally with other shapes if my diamonds are especially strong and if my clubs are especially weak.

 

No doubt there are other parts of the world where people bid this way as well, but as far as I can tell this practice is considered bizarre in much of Europe (because it is considered "normal" to rebid 5-card minors - something that experts on this side of the Atlantic tend to try to avoid).

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[PrecisionL]

Fred,

 

I have many spread sheets and have calculated the frequency of various opening bids allowing for distribution and hcp for many years. Before bridge odds complete by Frost et al, I calculated with a mechanical calculator the hcp up to about 24 hcp. Math was my minor in my Engineering studies.

[fred]

Fred,

 

I have many spread sheets and have calculated the frequency of various opening bids allowing for distribution and hcp for many years.  Before bridge odds complete by Frost et al, I calculated with a mechanical calculator the hcp up to about 24 hcp.  Math was my minor in my Engineering studies.

Thanks for explaining, Larry. Cool that you were able to create spreadsheets that can solve these problems :unsure:

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[Mbodell]

For those of you quoting odds, how are you computing odds of hand types in which both distribution and HCPs are defined?

 

To the best of my knowledge there is no easy way to do this using pen and paper (with or without the help of a calculator). Of course it is not that hard to write a computer program to do the math.

Computer for me. I was basing mine above on two different 1,000,000 deal hands (looking only at South for convenience on each hand) using the deal program. I was opening constructively any hand that was extended rule of 21 (HCP + 2 longest suits + QT >= 21) and then removing the other bids like 1NT and 2♣ and what not. I was assuming that xy45 was opened 1♣ and that ♦ was opened for all xy44 and clubs for all (43)33 that weren't a 1nt, 2nt or 2♣ strength. My second run was putting the 2245 into 1nt when in range, but (13)45 and out of nt 2245 would be bidding clubs. If you open 1♦ with those then you lower the expected number of diamonds in a 1♦ opening, but also lower the number of times it is exactly 3.