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I often think of a pack of cards as made up of 13 suits of four cards, in the obvious way. (The 2s 3s etc). Though come to think of it I can think quite easily of COMBIN(13,4) 4 card “suits” each from a different suit, though {A.A,A,A}, {K,K,K,K,} does seem the most obvious

COMBIN is the EXCEL function, COMBIN(n,k)=nCk for the mathematicians (or exes), . POWER is another EXCEL function POWER(n,m) raises n to the nth power. Another obvious set of “suits” that you might be able to notice if you were to think about it is {2S,3S,4S,5S}{6S,7S,8,S,9S},(ST,SJ,SQ,SK}{H2,H3,H4,H5}{H6,..}{SA,HA,DA,CA}, 4333 hands with one A one and only K,Q or J in each suit (plus a couple of other cards)

Now fairly clearly the commonest distribution in the 2s,3s 4s 5s etc“suits” is 23456789TJQKA. in fact almost never all from the same suit.

But there are POWER(13,4)=67108864 cases of this holding, about .01% of all (52)C(13) bridge hands you might be dealt.

So round about once on 10000 hands you should see a thirteen card straight. Suit wise its most probably a 4432 hand. So if Kit has played 50000 tournament boards he might expect to have seen 5 13 card straights.

If you take another of the (52)C(15) ways of choosing fthirteen sets of four cards from a pack and in a small blank space write A on each card in the first set of four, K in the space on card in the second set of four, etc you will have a straight 13 just as often with these four card “suits” as before. But you won't notice it very often, but it will happen pretty often.

The thing is in bridge as in life, some things are just a lot more noticeable than others, and you miss many things that are logically basically the same.

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I've seen 7600 and 8500. Tom Townsend who is the Bridge Correspondent of the UK Daily Telegraph reported on a deal from a UK Teams competition including this 0805 hand , which was held in first seat at green. void-AKQTxxxx-void-KQTxx. At one table 1C was opened overcalled 1S and the very solid 7C was reached, partner having xxx-x-AJxx-AJ9xx. At the other table 2C was opened, there was no overcall from the next hand with AQ9xxx-x-K9xx-xx. and this time the very solid 7H was bid.

The story is that the hand does not meet the EBU rules to be described as strong (it has less than 5 controls and less than 16 points) so was ruled illegal and the score adjusted.

Tom doesn't say what actually happened, but I assume the opponents called the director other than the cll the call ws ruled illegal.

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Philip Martin’s first problem can be “solved’ much the same way as `i use below. If you are not mathematically minded the point I am making is that Vacant Space problems are often not really bridge problems at all, they are combinatorial problems. Simple sounding problems can be difficult to calculate in full and it is easy to fall into some trap or other when trying to solve them.

Many inferences can be drawn at the table because we know the opponents are following the rules of bridge quite separately from any counting questions. . For example suppose we are missing Q6543 in a suit and we know that LHO always (or even usually) plays his cards upwards (lowest first). We lead low towards AKJ of that suit in dummy, he follows with the 5. We know (probably) that RHO has 34 of the suit and when we play the Ace RHO follows with the 4. We cross to hand and lead towards dummy again and West follows with the 6. We know the suit is breaking 3-2 or 2-3 so we might finesse for safety even if for other reasons the odds are that LHO holds the queen.

Mathematically we write nCr forethe number of ways of choosing r objects from n. When n=2m+1 ,ie n is odd. nCm=nC(m+1). This is obvious when you think about it. A bit more generally (assuming k less than n) nCk=nC(n-k) because when you choose k elements from n you can think of that as selecting the other (n-k).

Less obviously. nC(m+1)+nCm=(n+1)C(m+1). For any n and m. So for example 14C7=3432 And 13C7=13C6=1716 And 1716+1716=3432 as expected.

This turns out to be exactly what you need for Lauritz’s last problem.

In that problem, after West has cashed HAK and South has ruffed HQ, he leads a spade to dummy’s Q and a spade back on which East plays the J. South is at the position when he needs to decide whether to cover the SJ . Either West holds SA9, 2 hearts and 5 minor cards (in which case he should cover) and East holds eight minor cards or W holds SA (bare) and 6 minor cards and East holds S9 and 7 minor cards in which case he should not cover.

I represent the cards LHO-RHO as Case1 1: A9(2H)(5m)-(8m) Case 2: A(2H)(7m)-9(6m) A,9 are spades, and 2H are the remaining two hearts in West’s hand, (6m) means 6 minor suit cards etc

Case 1. There are 13C5 =1287combinations of 5m from the 13 minor cards Case 2 There are. 13C6=1716 (=13C6) combinations of 7m. from the 13 minor cards. The spade 9 is in LHO (case 1) and RHO (Case 2). There are 13C5+13C6=14C6(=3003) combinations of 13 minor cards and the spade 9 in the two cases

13C5/14C6=6/14 13C6/14C6=8/14

(if you are at all familiar with combinations you can do these in your head). So as calculated the probability that West has the SA single is 8/14 as per Vacant Places.

If you say that West could not have a six card minor that only rules out 8 combinations (one when diamonds are 6-0 and seven when clubs are 6-1). This hardly alters the ratios at all.

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I considered I played “Acol” until around 5 years ago. However it would never have occurred to me to quote any 1970s Acol book, (or even Bird and Reese's “Acol in the 90's”) as an oracle, not back then and not many years later.

Opening bids changed somewhat , I never used to open 1NT (weak) with a five card major, though I was an early adopter of Reese's multi 2D, I knew I had diverged from ancient Acol before thinking about any particular bidding sequence.

But Acol was never really prescriptive in the way that referring back to a particular book implies, and I don't remember anyone ever referring back to those books. There were principles (4 card majors, weak NT, “low and high ”reverse requirements" and then partnership agreements).

Also I think books then thought the real game was imps or even rubber, so any pairs consideration get barely a mention.

You might refer to a book to see what it had to say about a particular sequence, because with luck the author might have put some thought into explaining why non-forcing was a good idea for example, but usually they do not.

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I agree with David Burn's initial comment. On the other hand it is reasonable to work out what the various possibilities will give in imps assuming identical choices at the other table, especially with a deal as a hypothetical bidding problem where we can be sure the bidding is identical to the point where the critical decision is made ie to double or not!

Surely we want our action to gain imps ie have a positive expectation, and Kit's analysis is that double does, though pretty marginally.

But we compare scores with what actually happened at the other table. Unless. the opposition are playing a very similar system we can be pretty sure the situation is different in the other room on this hand. West is likely to have passed and he if opened 1NT it might be 12-14 and the opposition have better information than we do about the expectation of defeating 2S, ie a rather better chance.

More likely than West having 12 points is that he has 10 or 11 and passes initially, from East's viewpoint North opens say 1NT and based on what happened in our room it may well be that South transfers to 2S (partner doubling 2H perhaps) and they play there. A penalty double is more or less out of the question in the other room, and passing in our is likely to give a flat board - maybe that is right for the match situation.

This is different to a lot of competitive situations where you might expect similar bidding at the other table where the real life analysis is much closer to Kit's original analysis.

Even when the opposition play a mini-NT North may act differently and either pass or make an overcall for example. If North passes what we now need to do from an analysis viewpoint is vary the North South hands, North South have 5 cover cards and North will have four about 13% of the time, so mostly with the East West hands as is North passes.

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Hi Richard, scoring was in fact pairs if that matters. I thought 3S was pretty near forcing but I expected a retreat to 4C rather than the 4S bid which I thought showed only 4 cards.

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The double of 5D was pointless, but though I use DOPI if 4N RKCB is overcalled, I wondered if people have some standard way utilising a double of the reply. I generally find my partners never think the way I do unless we have an agreement in advance. Agree that 4N was not ideal but if it's my lucky day he has two aces anyway. But if I go 5D instead of 4N what does he do with A98x K9xxx xxx x

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EBU land does have it problems. Tom Townsend reported this hand which was opened 2C announced as “strong” from the 2019 Middlesex Swiss Teams: void AKQT8643 void KQT54 Partner had both minor aces and 7H was bid and made. Afterwards the 2C bid was ruled illegal (strong=16+ or 12+with 5+rc) and the score adjusted. But neither the 18 point example nor this distributional monster come up very often, much harderto cope with is the randomness of preempts these days.

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I like 2D=an A or K. 2H=no A or K 2S+ as you like Mostly responder will bid 2D but when she bids 2H. Asking by opener if still slam interested after 2H. Begins with Qs,, eg Roman Blackwood for Q.

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At the time as South having shown some interest in hearts with my 2C bid i thought I would get a heart lead. Partner took the majority view and lead CJ. Sadly dummy had CQT9Xx I had Kx and declarer Axxx. So they made 2Cx - we won 3spades and partners two aces.

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Opp longest % 8 22.7 9 47.3 10 23.4 11 5.9 12 0.8 13 0.0 weighted average 9.1 Given our longest suit is 10 cards the above is the distribution for the opponents longest suit. If our longest is 9 cards then weighted average for theirs is 8.8, and if our longest is 8 cards the weighted average for theirs is 8.2. So for 8 and 9 card fits 2x our trump length is not too bad a guess for Total Trumps.

The sum of the lengths of our two longest suits is always the same as the sum of the lengths of their two longest suits so Steve Bloom's SF (Second Fit) and without interrogating GIB (I don't know how I'm afraid) it does seem that SF=17 often provides 19 or 20 Total Tricks and we may know this more easily that what opponents have in their longest suit.

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Where did you get your stats from Nigel? Sounds like they would be worth publishing. However it seems clear that LoTT is at best a Rule of Thumb(RoT). It could never really be close to 100% right and probably nothing else can either. The question is, is it close enough to be useful?

I call the 26 cards in my hand and partner's the TH. Those in the opponents are OTH. Say I have a cards in one suit b in another etc I write abcd for the TH. The commonest TH pattern is 8765 - the 24 permutations of the suits accounting for about 1% each of all TH , about 24% in total. OTH is also 8765 (up to permutation) . To Total Tricks is always 16 for any 8765 TH. I looked at the outcomes for a randomly selected set of 33 boards played at my club a while ago. There were 12 8765 TH and Total Tricks (reported by the Hand Analyser) were: Total Tricks No of Deals 15 5 16 5 17 1 18 1 Not too bad for a RoT if this were typical.

Not all TH are like 8765 , i.e. the same up to permutations, e.g. 9665. OTH is 4778 (or 8774 after reordering). Notice all the suit lengths are different between TH and OTH. All TH fall into two classes like this.

Of course the problem with LoTT is that the distribution of OTH is independent of our distribution.

The oddest thing in the 8765 hands I looked at was the one with 18 Total tricks with points split 19-21, where there were no singletons (and the side that could make 4C could also make 3NT. The hand was played 9 times 5 times in Hearts each making 8 tricks but the contracts were 2H(3 times) , 3Hx and 4H. The other 4 contracts were 2S (a 5-2 fit) making 7,8 or 9 tricks (analyser says 8 in spades) I imagine LoTT may be adjusted for TH shape where you have a good idea about it, but often you will not.

I meant to add that if we have a 10 card fit about 30% of the time they have 10 card or bigger fit, and about 70% of the time they have a 9 card fit or less.

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The hand was played 9 times and opened 3 times. Points are 20-20 and South was 3442 shape, but NS can make 1NT or 2S and EW can make nothing. -1.5 imps to me.

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Call the 26 cards in the partnership the TH, with suit lengths abcd a spades b hearts etc. The the conditions Philip specified in the first place imply a and b are at most 7, and c and d are at most 8 and if c or d is 8 the suit splits 4-4, e.g. 5678 with clubs 4-4. About 15% of TH have all a,b,c,d at most 7. About 45% of TH have abcd at most 8, but quite a lot of them have a or b 8. I calculate that around 18% of TH abcd have the a,b<8, at least one of c or d=8. These 8 cards split 4-4 about 33% of the time. (Philip has removed 5-3 fits from consideration). So less than 6% of all TH satisfy the shape constraint. To first approximation that with be at most 6% of 22 point TH, at most 6% of 23 point TH etc.

With a 5678 TH split 1534-4144 bidding might go 1H 1S 2C 3C so these splits (if affording slam with 2 singletons) may well be handled straightforwardly.

So I think we are looking at things like 5678-4324-1354 and 5678-4234-1444 with a NT opening. These seem like 27-28 occasionally having the right cards and 29-32 more often. But it's not very often and a 4-1 trump split may sink it.

Finally of course there will be times when for slam we need to play in the 4-4 fit rather than the 5-3 fit we discover first!

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I assume that Pavlicek did a full calculation - the page that Greg Lawler mentioned gave the sample size (6.35E+11) as the full set of all possible (13 card) hands. In fact with a bit of effort you can do this in a spreadsheet and that gives 4.13 too.

When I saw the original post I thought that whenever I feel like I've averaged 8 points the hand sheets my club gives out (with 33 boards) say I've averaged 10, But I found 4 sheets lying around and the North averages were 8.7, 10.1 10.0 8.6 (in date order between Feb and April), I usually sit East and those averages were 9.7,10.4,10.0,9.7.

I could look online and get more data, and even plot out a whole year to see a nice normal distribution.

If my annual average of averages were 8 then I guess I would be the unluckiest person on the planet.

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Maybe Richard will put a summary of Mosso on Bridgewinners, but the 1C opening bid is (Forcing for one round) and either a)13+ unbalanced with 4+ clubs or b)17+ with 5+ spades or c) balanced with 12-14,18-20 or 23+ points. A balanced 12-14 hand might have 5 hearts but not usually 5 spades.

So not really that complex. 1D/H are also forcing so there is no need for a 2/1 type 2C bid. One of the areas of controversy in the UK over the years has been what “strong” means anyway.

1D in response to 1C is (usually) negative and is bid on nearly all hands with 0-6 points. It can be stronger with no 4 card major.

After opening 1C with a weak NT hand and getting a 1D response opener is looking for the safest low bid and so may bid 1H or 1S with 3 cards when nothing better is available. 1C 1D 1N shows the 18-20 balanced type.

I'm slightly surprised by your comment that 1S would be announced - in the UK it certainly would not - so do Precision 1H/S opening bids get announced in the US? 1S shows 5+ spades 9-16 points, denies 15-17 balanced (which would be opened 1NT).

I don't play Mosso but have adopted a couple of things - Mosso replies to 2C/2H openings.

Keith George

COMBIN is the EXCEL function, COMBIN(n,k)=nCk for the mathematicians (or exes), . POWER is another EXCEL function POWER(n,m) raises n to the nth power.

Another obvious set of “suits” that you might be able to notice if you were to think about it is

{2S,3S,4S,5S}{6S,7S,8,S,9S},(ST,SJ,SQ,SK}{H2,H3,H4,H5}{H6,..}{SA,HA,DA,CA}, 4333 hands with one A one and only K,Q or J in each suit (plus a couple of other cards)

Now fairly clearly the commonest distribution in the 2s,3s 4s 5s etc“suits” is 23456789TJQKA. in fact almost never all from the same suit.

But there are POWER(13,4)=67108864 cases of this holding, about .01% of all (52)C(13) bridge hands you might be dealt.

So round about once on 10000 hands you should see a thirteen card straight. Suit wise its most probably a 4432 hand. So if Kit has played 50000 tournament boards he might expect to have seen 5 13 card straights.

If you take another of the (52)C(15) ways of choosing fthirteen sets of four cards from a pack and in a small blank space write A on each card in the first set of four, K in the space on card in the second set of four, etc you will have a straight 13 just as often with these four card “suits” as before. But you won't notice it very often, but it will happen pretty often.

The thing is in bridge as in life, some things are just a lot more noticeable than others, and you miss many things that are logically basically the same.

Keith George

At the other table 2C was opened, there was no overcall from the next hand with AQ9xxx-x-K9xx-xx. and this time the very solid 7H was bid.

The story is that the hand does not meet the EBU rules to be described as strong (it has less than 5 controls and less than 16 points) so was ruled illegal and the score adjusted.

Tom doesn't say what actually happened, but I assume the opponents called the director other than the cll the call ws ruled illegal.

Keith George

Many inferences can be drawn at the table because we know the opponents are following the rules of bridge quite separately from any counting questions. . For example suppose we are missing Q6543 in a suit and we know that LHO always (or even usually) plays his cards upwards (lowest first). We lead low towards AKJ of that suit in dummy, he follows with the 5. We know (probably) that RHO has 34 of the suit and when we play the Ace RHO follows with the 4. We cross to hand and lead towards dummy again and West follows with the 6. We know the suit is breaking 3-2 or 2-3 so we might finesse for safety even if for other reasons the odds are that LHO holds the queen.

Mathematically we write nCr forethe number of ways of choosing r objects from n. When

n=2m+1 ,ie n is odd. nCm=nC(m+1). This is obvious when you think about it.

A bit more generally (assuming k less than n)

nCk=nC(n-k) because when you choose k elements from n you can think of that as selecting the other (n-k).

Less obviously.

nC(m+1)+nCm=(n+1)C(m+1). For any n and m.

So for example

14C7=3432

And 13C7=13C6=1716

And 1716+1716=3432 as expected.

This turns out to be exactly what you need for Lauritz’s last problem.

In that problem, after West has cashed HAK and South has ruffed HQ, he leads a spade to dummy’s Q and a spade back on which East plays the J. South is at the position when he needs to decide whether to cover the SJ .

Either West holds SA9, 2 hearts and 5 minor cards (in which case he should cover) and East holds eight minor cards or

W holds SA (bare) and 6 minor cards and East holds S9 and 7 minor cards in which case he should not cover.

I represent the cards LHO-RHO as

Case1 1: A9(2H)(5m)-(8m)

Case 2: A(2H)(7m)-9(6m)

A,9 are spades, and 2H are the remaining two hearts in West’s hand, (6m) means 6 minor suit cards etc

Case 1. There are 13C5 =1287combinations of 5m from the 13 minor cards

Case 2 There are. 13C6=1716 (=13C6) combinations of 7m. from the 13 minor cards.

The spade 9 is in LHO (case 1) and RHO (Case 2).

There are

13C5+13C6=14C6(=3003) combinations of 13 minor cards and the spade 9 in the two cases

13C5/14C6=6/14

13C6/14C6=8/14

(if you are at all familiar with combinations you can do these in your head).

So as calculated the probability that West has the SA single is 8/14 as per Vacant Places.

If you say that West could not have a six card minor that only rules out 8 combinations (one when diamonds are 6-0 and seven when clubs are 6-1). This hardly alters the ratios at all.

Keith George

Keith George

Opening bids changed somewhat , I never used to open 1NT (weak) with a five card major, though I was an early adopter of Reese's multi 2D, I knew I had diverged from ancient Acol before thinking about any particular bidding sequence.

But Acol was never really prescriptive in the way that referring back to a particular book implies, and I don't remember anyone ever referring back to those books.

There were principles (4 card majors, weak NT, “low and high ”reverse requirements" and then partnership agreements).

Also I think books then thought the real game was imps or even rubber, so any pairs consideration get barely a mention.

You might refer to a book to see what it had to say about a particular sequence, because with luck the author might have put some thought into explaining why non-forcing was a good idea for example, but usually they do not.

Keith George

Surely we want our action to gain imps ie have a positive expectation, and Kit's analysis is that double does, though pretty marginally.

But we compare scores with what actually happened at the other table. Unless. the opposition are playing a very similar system we can be pretty sure the situation is different in the other room on this hand. West is likely to have passed and he if opened 1NT it might be 12-14 and the opposition have better information than we do about the expectation of defeating 2S, ie a rather better chance.

More likely than West having 12 points is that he has 10 or 11 and passes initially, from East's viewpoint North opens say 1NT and based on what happened in our room it may well be that South transfers to 2S (partner doubling 2H perhaps) and they play there. A penalty double is more or less out of the question in the other room, and passing in our is likely to give a flat board - maybe that is right for the match situation.

This is different to a lot of competitive situations where you might expect similar bidding at the other table where the real life analysis is much closer to Kit's original analysis.

Even when the opposition play a mini-NT North may act differently and either pass or make an overcall for example. If North passes what we now need to do from an analysis viewpoint is vary the North South hands, North South have 5 cover cards and North will have four about 13% of the time, so mostly with the East West hands as is North passes.

Keith George

Keith George

Keith George

Keith George

Keith George

Agree that 4N was not ideal but if it's my lucky day he has two aces anyway.

But if I go 5D instead of 4N what does he do with

A98x

K9xxx

xxx

x

Keith George

void

AKQT8643

void

KQT54

Partner had both minor aces and 7H was bid and made. Afterwards the 2C bid was ruled illegal (strong=16+ or 12+with 5+rc) and the score adjusted.

But neither the 18 point example nor this distributional monster come up very often, much harderto cope with is the randomness of preempts these days.

Keith George

Mostly responder will bid 2D but when she bids 2H. Asking by opener if still slam interested after 2H. Begins with Qs,, eg Roman Blackwood for Q.

Keith George

I had Kx and declarer Axxx. So they made 2Cx - we won 3spades and partners two aces.

Keith George

8 22.7

9 47.3

10 23.4

11 5.9

12 0.8

13 0.0

weighted average 9.1

Given our longest suit is 10 cards the above is the distribution for the opponents longest suit.

If our longest is 9 cards then weighted average for theirs is 8.8, and if our longest is 8 cards the weighted average for theirs is 8.2. So for 8 and 9 card fits 2x our trump length is not too bad a guess for Total Trumps.

The sum of the lengths of our two longest suits is always the same as the sum of the lengths of their two longest suits so Steve Bloom's SF (Second Fit) and without interrogating GIB (I don't know how I'm afraid) it does seem that SF=17 often provides 19 or 20 Total Tricks and we may know this more easily that what opponents have in their longest suit.

Keith George

However it seems clear that LoTT is at best a Rule of Thumb(RoT). It could never really be close to 100% right and probably nothing else can either. The question is, is it close enough to be useful?

I call the 26 cards in my hand and partner's the TH. Those in the opponents are OTH. Say I have a cards in one suit b in another etc I write abcd for the TH. The commonest TH pattern is 8765 - the 24 permutations of the suits accounting for about 1% each of all TH , about 24% in total. OTH is also 8765 (up to permutation) . To Total Tricks is always 16 for any 8765 TH. I looked at the outcomes for a randomly selected set of 33 boards played at my club a while ago.

There were 12 8765 TH and Total Tricks (reported by the Hand Analyser) were:

Total Tricks No of Deals

15 5

16 5

17 1

18 1

Not too bad for a RoT if this were typical.

Not all TH are like 8765 , i.e. the same up to permutations, e.g. 9665. OTH is 4778 (or 8774 after reordering). Notice all the suit lengths are different between TH and OTH. All TH fall into two classes like this.

Of course the problem with LoTT is that the distribution of OTH is independent of our distribution.

The oddest thing in the 8765 hands I looked at was the one with 18 Total tricks with points split 19-21, where there were no singletons (and the side that could make 4C could also make 3NT.

The hand was played 9 times 5 times in Hearts each making 8 tricks but the contracts were 2H(3 times) , 3Hx and 4H. The other 4 contracts were 2S (a 5-2 fit) making 7,8 or 9 tricks (analyser says 8 in spades)

I imagine LoTT may be adjusted for TH shape where you have a good idea about it, but often you will not.

I meant to add that if we have a 10 card fit about 30% of the time they have 10 card or bigger fit, and about 70% of the time they have a 9 card fit or less.

Keith George

Keith George

The the conditions Philip specified in the first place imply a and b are at most 7, and c and d are at most 8 and if c or d is 8 the suit splits 4-4, e.g. 5678 with clubs 4-4. About 15% of TH have all a,b,c,d at most 7. About 45% of TH have abcd at most 8, but quite a lot of them have a or b 8. I calculate that around 18% of TH abcd have the a,b<8, at least one of c or d=8.

These 8 cards split 4-4 about 33% of the time. (Philip has removed 5-3 fits from consideration). So less than 6% of all TH satisfy the shape constraint. To first approximation that with be at most 6% of 22 point TH, at most 6% of 23 point TH etc.

With a 5678 TH split 1534-4144 bidding might go 1H 1S 2C 3C so these splits (if affording slam with 2 singletons) may well be handled straightforwardly.

So I think we are looking at things like 5678-4324-1354 and 5678-4234-1444 with a NT opening. These seem like 27-28 occasionally having the right cards and 29-32 more often. But it's not very often and a 4-1 trump split may sink it.

Finally of course there will be times when for slam we need to play in the 4-4 fit rather than the 5-3 fit we discover first!

Keith George

When I saw the original post I thought that whenever I feel like I've averaged 8 points the hand sheets my club gives out (with 33 boards) say I've averaged 10, But I found 4 sheets lying around and the North averages were 8.7, 10.1 10.0 8.6 (in date order between Feb and April), I usually sit East and those averages were 9.7,10.4,10.0,9.7.

I could look online and get more data, and even plot out a whole year to see a nice normal distribution.

If my annual average of averages were 8 then I guess I would be the unluckiest person on the planet.

Keith George

(Forcing for one round) and either

a)13+ unbalanced with 4+ clubs or

b)17+ with 5+ spades or

c) balanced with 12-14,18-20 or 23+ points. A balanced 12-14 hand might have 5 hearts but not usually 5 spades.

So not really that complex. 1D/H are also forcing so there is no need for a 2/1 type 2C bid. One of the areas of controversy in the UK over the years has been what “strong” means anyway.

1D in response to 1C is (usually) negative and is bid on nearly all hands with 0-6 points. It can be stronger with no 4 card major.

After opening 1C with a weak NT hand and getting a 1D response opener is looking for the safest low bid and so may bid 1H or 1S with 3 cards when nothing better is available.

1C 1D 1N shows the 18-20 balanced type.

I'm slightly surprised by your comment that 1S would be announced - in the UK it certainly would not - so do Precision 1H/S opening bids get announced in the US? 1S shows 5+ spades 9-16 points, denies 15-17 balanced (which would be opened 1NT).

I don't play Mosso but have adopted a couple of things - Mosso replies to 2C/2H openings.