[bluecalm]

---

NS are Brink-Drijver.
W is Helgemo.

1NT was probably 9-12 but maybe 10-13 or maybe 12-14 (cc of Brink-Drijver from Bermuda Bowl says 9-12).
2♣ = clubs.

First trick:
5♣, 6♣, K♣, A♣.

We play J♠ and see a 5 on the left. Well ?

How strong 1NT have to be to make it a finesse ? Or maybe it's finesse (or drop) regardless of the range ?
I have some thoughts but I am not sure if I am doing it correctly.
The difference between getting it right and wrong is 4 tricks so 450 points.

[gszes]

all we know so far is lho started with the club Q and rho the club K. Now we
have to guess. We can eliminate lho starting with AK of KQJ of dia since a club
lead while safe is probably unproductive at best and a dia would look a ton better.
So rho probably has at least the dia K. That is 8 out of the 20 the opps own and
even if lho started with only 9 (and their p an unlikley 11 with 6 clubs and still bidding
only 2c) the odds still favor lho having the spade Q by 7:5 and the stronger lho is the
more likely they are to have the Q.

Take the finesse.


Have a quiet chat with your p after the session and tell them to please trust that you
would not have bid 2n if short in spades so at the very least bid 3s vs passing since
3s rates to be a ton safer.

[rhm]

I wish all my math problems were so simple.

We and the opponents have 20 HCP. 5 HCP are probably accounted for in clubs, North having the queen.
Assuming a 9-12 notrump range, North has between 7 and 10 HCP of the 15 outstanding HCP in the other suits.
So as a first approximation North is somewhere between 7/15 to 10/15 favorite to hold the spade queen.
This is an approximation.
Better would be to look at all combinations of the outstanding honors there are comprising 7 to 10 HCP and which portion of these contain the spade queen.
You can assume that South holds at least one diamond honor because North would not lead a club with ♦AKQ.
But the above approximation will be good enough.

As usual there are a lot of other indications, all pointing in the direction of the finesse:
South ran from 1NT doubled when their combined assets were 20 HCP and likely having a long club suit.
If North were at the minimum of their notrump range South would know that their side had at least half the deck
and RDBL would be attractive, unless being short in spades.
This argues that North is more likely to be at the top of their notrump range.
South has more incentive to run the weaker he is or with shortage in spades.
He is more likely to account for more cards in clubs than his partner, leaving less free slots for spade cards.
Last but not least, North did not fancy 3♣ with club support at favorable vulnerability.
With ♠Qxx, 3♣ would be far less attractive than with ♠xx.

I am pretty sure Helgemo finessed. What I am not sure is, whether the patient survived.

Rainer Herrmann

[helene_t]

Finesse. Empty spaces. Opener has fewer clubs, quite likely clubs are 3-6 since with 4 he might have raised at these colours.

[dake50]

Are you asking the percents?
Or enough to decide best play?

[bluecalm]

Some ballpark estimation would be nice. I am trying to arrive at it but I am unsure if I am doing it right.

[han]

Brink-drijver cover with an honor, I'd play for the drop.

[rhm]

Besides the spade queen declarer is missing one ace, 2 kings, one other queen and one jack.
We can put this into a tuple (1,1,2,1,1) where the first position is for the spade queen, the next positions for the missing ace, kings, queen and the last is for the jack.
The following vectors are subsets in the 7-10 range
(1,1,1,0,1)(1,1,1,0,0)(1,1,0,1,1)(1,1,0,1,0)(1,1,0,0,1)(1,0,2,1,1)(1,0,2,1,0)(1,0,2,0,1)(1,0,2,0,0)(1,0,1,1,1)(1,0,1,1,0)
(0,1,2,0,0)(0,1,1,1,1)(0,1,1,1,0)(0,1,1,0,1)(0,1,1,0,0)(0,1,0,1,1)(0,0,2,1,1)(0,0,2,0,1)
So if I have not overlooked a honor combination 19 such honor combinations.
But the combinations with one king must be counted twice.
So we get 27 different honor combinations, of which 15 contain the spade queen.
If we discount the two where North has ♦AKQ (from the opening lead) we are left with 25 different combinations, of which 15 have the spade queen.
If we assume all these combinations equally likely, which is approximately true, the likelihood that North has the spade queen if North opened a 9-12 NT is 15/25=60%.

From further Bridge clues (see my first post) the actual probability must be higher.

Rainer Herrmann

[han]

Sometimes when north has the spade queen, it will be doubleton and it drops anyway.

This math problem seems too difficult to do away from the table.

[helene_t]

Yes, it is xxx or xx (4 combinations) versus Qxx or Qxxx(4 combinations). But with a void spades, responder might have competed over 2♠. Maybe even with a singleton?

So: given that opener has two or three small spades, what is the chance that he also has the queen?

[jonottawa]

The matching singleton rule never fails. Finesse!

[bluecalm]

Quote

ut with a void spades, responder might have competed over 2♠. Maybe even with a singleton?

I think he might be reluctant to compete seeing probable misunderstanding of the opponents (and 3S likely being better contract than 2N).
Let's ignore this possibility for the purpose of analysis.

Quote

(1,1,1,0,1)(1,1,1,0,0)(1,1,0,1,1)(1,1,0,1,0)(1,1,0,0,1)(1,0,2,1,1)(1,0,2,1,0)(1,0,2,0,1)(1,0,2,0,0)(1,0,1,1,1)(1,0,1,1,0)
(0,1,2,0,0)(0,1,1,1,1)(0,1,1,1,0)(0,1,1,0,1)(0,1,1,0,0)(0,1,0,1,1)(0,0,2,1,1)(0,0,2,0,1)
So if I have not overlooked a honor combination 19 such honor combinations.

My quick script shows that there are 27 such combinations and 14 of them contain Qs (so that confirms your calculations almost). Those are:
(A, K, K, Qs, Q, J)

(0, 0, 1, 1, 1, 0) hcp: 7 Qs HERE
(0, 0, 1, 1, 1, 1) hcp: 8 Qs HERE
(0, 1, 0, 1, 1, 0) hcp: 7 Qs HERE
(0, 1, 0, 1, 1, 1) hcp: 8 Qs HERE
(0, 1, 1, 0, 0, 1) hcp: 7
(0, 1, 1, 0, 1, 0) hcp: 8
(0, 1, 1, 0, 1, 1) hcp: 9
(0, 1, 1, 1, 0, 0) hcp: 8 Qs HERE
(0, 1, 1, 1, 0, 1) hcp: 9 Qs HERE
(0, 1, 1, 1, 1, 0) hcp: 10 Qs HERE
(1, 0, 0, 0, 1, 1) hcp: 7
(1, 0, 0, 1, 0, 1) hcp: 7 Qs HERE
(1, 0, 0, 1, 1, 0) hcp: 8 Qs HERE
(1, 0, 0, 1, 1, 1) hcp: 9 Qs HERE
(1, 0, 1, 0, 0, 0) hcp: 7
(1, 0, 1, 0, 0, 1) hcp: 8
(1, 0, 1, 0, 1, 0) hcp: 9
(1, 0, 1, 0, 1, 1) hcp: 10
(1, 0, 1, 1, 0, 0) hcp: 9 Qs HERE
(1, 0, 1, 1, 0, 1) hcp: 10 Qs HERE
(1, 1, 0, 0, 0, 0) hcp: 7
(1, 1, 0, 0, 0, 1) hcp: 8
(1, 1, 0, 0, 1, 0) hcp: 9
(1, 1, 0, 0, 1, 1) hcp: 10
(1, 1, 0, 1, 0, 0) hcp: 9 Qs HERE
(1, 1, 0, 1, 0, 1) hcp: 10 Qs HERE
(1, 1, 1, 0, 0, 0) hcp: 10

So it's almost 14/27 = 51.85%
If we eliminate AKQ of diamonds (2 possibilities with or without side jack) it gives: 14/25 = 56%
(I made a mistake before in this post, now corrected)

I don't think that's the end of the story though as we need to consider how often opener has 3 or 4 spades opposite him having 2 and those doesn't seem that easy to estimate in independent way.
I may end up just dealing those hands and counting occurrences but it would be nice to estimate it without help of the dealer program to get some intuition about what are major factors influencing the odds here.

[bluecalm]

Quote

Brink-drijver cover with an honor, I'd play for the drop.

Maybe Helgemo thoughts so as he played for a drop and it was down 3.
In all seriousness I think Helgemo is one of the very few people who are good at this game and every time he does something and I doubt his decision I am assuming my intuition is wrong until I see very convincing evidence to the contrary. I haven't so far.

[rhm]

bluecalm, on 2012-August-24, 10:07, said:

Maybe Helgemo thoughts so as he played for a drop and it was down 3.
In all seriousness I think Helgemo is one of the very few people who are good at this game and every time he does something and I doubt his decision I am assuming my intuition is wrong until I see very convincing evidence to the contrary. I haven't so far.

I am not convinced. Even the best players make a lot of mistakes and I believe this is one.
reports about sensational deals give a wrong impression

Rainer Herrmann

[bluecalm]

Quote

I am not convinced. Even the best players make a lot of mistakes and I believe this is one.
reports about sensational deals give a wrong impression

Just in case you missed it, my calculations were wrong and yours were correct. Sorry for confusion.
As to sensational deals. I have different way of assessing this. For example I have a script which pulls all cardplay misplays from vugraph (in double dummy sense) for given player and put them into .lin files. I then go through them all.
Helgemo is one of those guys who impressed me to such extent I am always careful to call "mistake". I am sure he makes those I just want to make sure as too many times my intuition was wrong.

[gszes]

the overall hand starts with the 8 ever 9 never theory of the 22 split
being superior to trying to guess if its right to play for 31. The problem
is that if there is 31 split you have already found it (ie LHO did not show
out) and now a new set of calculations must take place. Helene_t points
out that at the point of decision open spaces dictates that lho actually
has a greater chance of having a 3rd spade (and thus the Q) than rho has
of starting with a doubleton. Playing for the 22 split was originally favored
by almost 14% but at this point the finesse is favored 57-43. Even this
% is based primarily on the probability the Rho holds at least 6 clubs and that
might be wrong.



It is quite possible that the evidence of the auction was not considered
critical enough to sway declarer away from this origina 14% distribution
advantage. IMO the (theoretical) evidence supplied by the bidding
and opening lead --where we are 7:5 favorites at least ---carries
more weight in this decision then the hard probabilty of 14%
advantage due to distributional probabilites. That's why we do not
have duplicate results everywhere.

[bluecalm]

I used the following dealer script:

produce 10000
predeal west SJ97, H964, DJ953, CAJ2
predeal east SAKT432, HAQ2, D642, C6

hcp(north) >= 9 &&
hcp(north) <= 12 &&
shape(north, any 4333 + any 5332 + any 4432 +any 5422) &&
clubs(south) >= 5 && clubs(south) <=6 &&
hascard(north, QC) &&
hascard(south, KC)

action printcompact

And then count spade holdings for N.
The results:

Qxx - 30.72%
xx - 27.74%
Qx - 26.29%
xxx - 7.96%
Qxxx - 7.29%

Or 64.3% for Qs.

I feel that those conditions favor playing for drop too as they don't contain that max range is more likely for N (as rhm pointed out), allow for AKD♦ in N hand and for 5-4-2-2's including hands with 5 hearts.
Still 64.3% is more than point count alone suggests. Assuming I didn't make any mistake in the conditions, does anybody has an idea how to arrive at such estimation at the table ?

Anyway it's much higher than I expected. I now think Helgemo blundered on this hand.

[kayin801]

Since the spade J could be stiff I bet they'd cover from Qx (partner could have 9xxx) but not with any Qxx. With Q8xx they will probably cover too hoping partner has the 9 singleton or doubleton.

(Post edited to make more sense)

[FrancesHinden]

helene_t, on 2012-August-24, 04:10, said:

Finesse. Empty spaces. Opener has fewer clubs, quite likely clubs are 3-6 since with 4 he might have raised at these colours.

Opener will basically never raise immediately as responder could be 4441 with short clubs planning to redouble.

[jogs]

Why didn't he just pass 2♠? The difference between right and wrong would be only one trick and it would be East's problem.