Probability: how many clubs has partner

[kgr]

You open short clubs. Style 5=5=4=2 (and you open clubs with 4-4 in the minors).

1C-(p)- What is the probability that partner has 2,3,4+ clubs

1C-(2X)- What is the probability that partner has 2,3,4+ clubs

[EarlPurple]

NT range?

Weakest hand on which you'd open?

Do you ever open 1♣ with 5-5 in blacks? Is that your normal opening?

 

Apart from all this, you also have the knowledge of your own hand, which must influence the probabilities.

[Gerben42]

what about 4-4 and 4-5 minors? :rolleyes:

 

But the answer is: still usually 4+♣

[helene_t]

Assuming the answers to Earl's and Gerben's questions are "I flip a coin", assuming you never open 1NT (sorry) and assuming that the opp's pass does not give you any information, the answers are:

 

Two: 4.7%

Three: 18.4%

Four: 40.1%

Five: 24.8%

Six: 9.9%

Seven: 1.2%

(Edited, at first I didn't notice that you said you always open 1♣ with 4-4 minors

[kgr]

Thanks.

The question was triggered because somebody at the club did bid after

1C-(1S) .. 2C with a 5-card C and 11 HCP.

He said that they play 5=5=4=2, so that partner could only have two clubs and that he didn't want to bid 3C (limit) because of that.

I told that I estimated the probability of 4+ to more then70%. Apparantly it is even 70% without the 1S bid.

[helene_t]

The question was triggered because somebody at the club did bid after 1C-(1S) .. 2C with a 5-card C and 11 HCP.

If he didn't want to bid 3♣ he could have made a cue-bid. Unless he judged that they shouldn't be in 3NT with 14 HCPs and a ballanced hand by partner. In that case 2♣ is all right. First 2♣ and then double describes his hand reasonably.

 

But of course, if he really thought 2♣ was an underbid he shouldn't bid it just because he was worried about a doubleton clubs.

[kenberg]

Despite Helene's providing an answer, I would be reluctant to agree without some further stipulations (the NT range is relevant, since it determines how many balanced hands with two clubs will not be opened 1C even though the pattern allows it). Even if that could be sorted out there is still a substantial issue (arising after H's post):

 

The question posed was: What is the probability distribution of the number of clubs?

 

The application in mind (I was wondering what it could be) is: How concerned should a player be that his partner only holds two when contemplating support?

 

For this, the relevant question is: What is the conditional probability that partner holds only two clubs, given our agreements AND given that I hold five. The answer is likely to be significantly different from (and larger than) what it would be without the second stipulation.

 

 

This seems truly academic to me however. (Of course as a retired academic I am not dissing academic questions.) If I have five clubs and partner opens clubs, I will assume three even if he need not have them. After which I will try to get him to stop playing that 1C can be on two.

 

Ken

[Sigi_BC84]

For this, the relevant question is: What is the conditional probability that partner holds only two clubs, given our agreements AND given that I hold five. The answer is likely to be significantly different from (and larger than) what it would be without the second stipulation.

I get the following:

 

Rule of 20 opener, 1♣ has been opened according to original post, responder holds exactly 5♣, probability for opener holding only 2♣ (i.e. we have no fit):

 

with a 15-17 NT: 8.7%

with a 14-16 NT: 8.9%

with a 12-14 NT: 6.5%

 

Conclusions: weak NT stabilizes (no surprise), fit very likely (Ken, why don't you like opening 1♣ on two?)

 

--Sigi

[Gerben42]

After which I will try to get him to stop playing that 1C can be on two.

 

I've said it before and I'll say it again. ANY system in which 1♣ promises at least 3 cards in the suit is gametheoretically suboptimal.

 

It is not for nothing that almost all top pairs play some kind of system where 1♣ "could be short" or has more than one possible meaning.

 

The most prominent counterexample is the pair Gitelman - Moss, who have a very natural bidding philosophy and anything other than a natural 1♣ would not fit into this.

 

However we all know that system is not decisive for success. Playing well is much more important :)

 

This claim is based on personal investigation on this subject as well as observation. I am sure that great theorists like Garozzo have given this much more thought than me.

[kenberg]

For this, the relevant question is: What is the conditional probability that partner holds only two clubs, given our agreements AND given that I hold five. The answer is likely to be significantly different from (and larger than) what it would be without the second stipulation.

I get the following:

 

Rule of 20 opener, 1♣ has been opened according to original post, responder holds exactly 5♣, probability for opener holding only 2♣ (i.e. we have no fit):

 

with a 15-17 NT: 8.7%

with a 14-16 NT: 8.9%

with a 12-14 NT: 6.5%

 

Conclusions: weak NT stabilizes (no surprise), fit very likely (Ken, why don't you like opening 1♣ on two?)

 

--Sigi

I would have bet a large sum of money that the probabilities would be monotone as you moved downward through the nt range. Maybe I'll do my own calculation, although it seems a lot like work.

 

 

As for opening one club on two, it probably doesn't much matter. I certainly agree to play that way if someone prefers. I have rearely if ever found a problem with the D opening not promising four, however. Say the auction goes 1D-1M-1N. Partner has at least four diamonds. Same with 1D-1H-1S. And so on. In fast competitive auctions it could be a problem, but I can't recall when it ever was. So I guess my main reason is that I see little gain from the agreement to open 1C on 4-4-3-2. Or much harm either.

 

Ken

[Sigi_BC84]

I would have bet a large some of money that the probabilities would be monotone as you moved downward through the nt range. Maybe I'll do my own calculation, although it seems a lot like work.

The local maximum for 14-16 surprises me as well. I am not claiming that my results are correct, but I've checked my conditions quite carefully.

 

If you want to do your own experiments, you can work with the scripts I've posted here.

 

The changes that were needed for this problem:

 

open_one_diamond = ( open_one_suit and shape{west, 4-4-4+x:d>c + 5+M6+d(xx):d>s,d>h} )
open_one_club = ( open_one_suit and not open_one_heart and not open_one_spade and not open_one_diamond )
club_response = ( open_one_club and clubs(east) == 5 )
condition club_response

I'd certainly be interested to hear if you find glitches in my setup (since I'm using this as a foundation to run all my experiments).

 

--Sig

[Sigi_BC84]

I am not happy with my one_diamond condition (not to say that I think it's plain wrong :-), before you go to lengths to point that out. I guess I have to think it over.

 

--Sigi

[kenberg]

I might give it a try. My original thought was pretty much inuitive. Here is a more formalized version. Warning to children: Math symbols ahead.

 

We take three quantities:

 

X= the number of hands with exactly two clubs, and which by the shape rules but forgetting hcps would be opened 1C.

 

Y=the number of balanced hands with three or more clubs that, by shape and ignoring hcps, would be opened 1C

 

Z=the number of unbalanced hands that would, by shape and ignoring hcps, be opened one club (and so automatically having long clubs).

 

 

A first approximation to the probability of opener having 2 clubs is X/(X+Y+Z).

 

 

Now bring in hcps, say requiring 12 to 20. Roughly (not exactly, I know), this can be modeled by multiplying each variable by some constant j, where j is somewhere between 0 and 1.

The new estimate of the probability is jX/(jX+jY+jZ) . The j cancels so it's a wash.

 

But now require also that balanced hands be opened 1NT. In fact, for these purposes take the definition of unbalanced to be "would open in a suit regardless of the stated NT range". This is modeled by multiplying X and Y but not Z by some k. Our new fraction becomes kX/(kX+kY+kZ) which is equal to X/(X+Y+(Z/k)). As k goes down from 1, Z/k goes up, the denominator goes up, the fraction goes down.

 

Now what is k? For weak nt, maybe 1/2, maybe less. It's probably known (given some assumptions), and certainly can be estimated. Surely (?) it decreases as the NT range decreases.

 

The argument is not exact, probably not exact enough for decent numerical accuracy, but it seems to me to be exact enough to be confident the probabilities will decrease as the NT range goes down, whether we are doing the original problem or the problem conditioned by seeing partner's hand with the five card club suit. Anyway, that's the long version of my original thought that the monotonicity is obvious. "Obvious" and "right" have an annoying habit of coming out different.

 

Ken

[Sigi_BC84]

Ken, I think your reasoning is perfectly valid, and my experiment must have been flawed.

 

I have now made my conditions pretty water-tight:

 

diamond_condition = ( diw >= 4 and diw >= clw and not ( diw == 4 and clw == 4 ) )
open_one_club = ( open_one_suit and not open_one_heart and
       not open_one_spade and not diamond_condition )

('diw' and 'clw' are the number of diamonds and clubs in openers hand, respectively)

 

Also I have found that my results for the 15-17 NT range are not stable, probably because there are not enough valid hands being found among the 1 million I'm usually generating, so I have increased the sample size to 10 million and repeated the experiment a few times. The main culprit seemed to have been the small sample size and not the flawed condition for the club opener.

 

I now get 6.8/8.8/9% (for 12-14/14-16/15-17).

 

I have also changed my opening condition from rule-of-20 to 12-21 HCP (probably more realistic, since minimum 4333s are not opened using rule-of-20 and strong NT) and then I get 7.4/8.4/8.5%.

 

--Sigi

[kenberg]

Theory and practice joined together. As Yogi Berra did or didn't say: In theory, practice should be the same as theory. In practice, they aren't.

[helene_t]

Yes, theory and practice joint together. Great. I nominate Sigi and Kenberg for the "Thread of the year" price.