Read the distribution

[Fluffy]

[hv=d=e&v=n&n=skq108ha863dj74c109&w=sj9732h1072dq9caj3]266|200|Scoring: MP[/hv]

 

If you aren´t used of this forums you may have problems to see the position: You are SOUTH, dummy is WEST your cards are at the bottom of diagram, and you are the one to lead.

 

The bidding:

 

W E

1NT

2♣-2♦

3NT-Pass

 

The play starts:

 

♠3 (2nd/4th)-K-6-5

♦J - 2(even number off cards)

 

By now you should be able deduce the distribution.

 

You have a defence to make, but that is another problem.

[Gerardo]

Partner has 1=4=4=4, declarer 3=2=4=4

[Flame]

I don't think you can know the exact distribution.

[inquiry]

I think declearer is 3-3-4-3 or 3-2-4-4

[jfy18]

Well, let's think about it.

 

At the start of the hand, there are 4♠ 6♥ 6♦ and 8♣ outstanding in the 2 hands we cannot see, adding to 26 cards.

 

After the first two tricks, since East's 1NT bid shows at least 2 of each suit, those numbers will have been reduced to 2♠ 6♥ 4♦ and 8♣.

 

North's opening discard shows an odd number of ♠ and his second discard shows an even number of ♦.

 

{By the way, is it just I or do others think the usual convention is the other way around ie. that playing low signifies an odd number of cards in the suit?? Your comments are most welcome here, especially, Moderator}.

 

What else do we know?

i) By the Stayman bidding, East has less than 4 cards in each of ♥ and ♠

 

 

Clues in, let's first think about ♠. East must start with an odd number (since North has an odd number and they add to 4), which must be 3 ♠ since by their 1NT bid they cannot have less than 2. Therefore, North started with 1 ♠.

 

Continue by thinking about East. Other than 3 ♠, they have 10 cards, including an even number of ♦. So East's possible ( ♥,♦,♣ ) combinations are:

(2,6,2)

(2,4,4)

(3,4,3)

(2,2,6)

(3,2,5)

 

Once we know which, we can easily work out North's hand; so one of the above 5 options must be the full answer. I cannot work out which can be eliminated, however; I suppose we could (somewhat arbitrarily) eliminate the hands with 6 in a suit on the grounds that East would not have bid 1NT with such a hand. Even if we do this, however, there are still 3 possibilities, all of which seem plausible.

 

Any ideas?

[1eyedjack]

Personally I am with you ify18, but many players will deny two doubletons when opening 1N

[Flame]

"North's opening discard shows an odd number of ♠"

> The first card played by partner unless agreed otherwise is attitude and not count signal.

 

"By the way, is it just I or do others think the usual convention is the other way around ie. that playing low signifies an odd number of cards in the suit?? Your comments are most welcome here, especially, Moderator}."

> the count signaling is high for even and low for odd, and when attitude high for incarage, low for discarage. The opposite way is called UDCA= upside down count and attitude and in need to be alerted.

 

"1NT bid shows at least 2 of each suit"

> a normal 1NT is one of : 5332, 4432,4333, and not any hand with atleast doubleton in each suit like 7222

[Fluffy]

Hi again guys!

 

Maybe I am too used of my 'advanced' opponents to try free fineses, or just as used of my partner ducking smothly with ♠Ax that I had no doubt of wich was the ♠ position. (partner held ♠A6)

 

I was right to deduce the ♠ position, but not to deduce it wasn´t as obvious as I though.

 

Was suposed to be an easy count exescise for beginners, oh well, I didn´t pick the right one :-( .

 

Anyway after it there comes a defence problem: what to play after you win ♦Q? (you will win it either first or second round of ♦)

 

 

a) play a ♠ to the ace, the problem I saw about it is: if we could manage not to play ♠ at all after the end, declarer will posibly play it himself, and then he is favourite to missguess putting the Q wich could be vey helpful at matchpoints. But if you play ♠ inmediatelly declarer is forced to play the 10 or 8 and succerd.

 

;) play a ♥ can hardly be right since 4th ♥ in dummy will become good.

 

c) play 9♦ back, it may give a free entry to dummy, as well as a free guess of ♦9.

 

d) play a ♣, very risky, giving declarer 1 trick, but maybe getting 2 in reward when partner has H8xx

[jfy18]

"North's opening  discard shows an odd number of ♠"

> The first card played by partner unless agreed otherwise is attitude and not count signal.

 

"By the way, is it just I or do others think the usual convention is the other way around ie. that playing low signifies an odd number of cards in the suit?? Your comments are most welcome here, especially, Moderator}."

> the count signaling is high for even and low for odd, and when attitude high for incarage, low for discarage. The opposite way is called UDCA= upside down count and attitude and in need to be alerted.

 

"1NT bid shows at least 2 of each suit"

> a normal 1NT is one of : 5332, 4432,4333, and not any hand with atleast doubleton in each suit like 7222

Re. Flame's helpful comments:

 

> The first card played by partner unless agreed otherwise is attitude and not count signal.

 

> the count signaling is high for even and low for odd, and when attitude high for incarage, low for discarage. The opposite way is called UDCA= upside down count and attitude and in need to be alerted.

 

> a normal 1NT is one of : 5332, 4432,4333, and not any hand with atleast doubleton in each suit like 7222

 

 

In that case, the situation in the example given becomes clearer in one sense (ie. that the hands East can have are limited to 5332, 4432 or 4333) but less clear in another sense (ie. because the opening trick has no count information from North).

 

So what do we know:

i) N and E between them have 4♠ 6♥ 8♦ 8 ♣.

ii) On the second trick, North's low card shows he has an even number of ♦ according to Fluffy who set the problem.

iii) Therefore, East must also have an even number of ♦.

iv) East has no more than 3 cards in either major suit.

 

So, what are the combinations East can have that are either 5332, 4432 or 4333?

 

In terms of (♠, ♥, ♦, ♣) with even ♦ I make East's possible holdings:

 

(2,3,4,4)

(3,2,4,4)

(3,3,2,5)

 

But how can we now work out which one of these it is? Perhaps you could let us know your line of thinking, Fluffy; especially if I have missed something important.

 

Cheers,

Jfy

[jfy18]

Regarding the best defensive lead after winning the Q ♦ on, say, Trick 2:

 

Let's think about the points declarer and dummy together have: between 25-27 one should think. They are very likely to make their 3NT. So the question probably is: how can the defence find their fourth trick?

 

Since South has 8 points, this also therefore gives North 1-3 points in addition to his A♠.

 

I agree that playing a ♠ now jeopardises a plausible source of a trick later on, if North has A6 ♠.

 

I am not so sure why playing a ♥ necessarily sets up a trick in dummy. Consider:

i) East may have doubleton ♥;

ii) Even if East has 3 ♥, if two of them are K and Q then they will make 4 [H] tricks whatever we lead now, so in that case we would lose nothing by playing one.

 

My conclusion, however, is that the ♦ suite is now probably a lost cause. North does not have enough room, points-wise, for the A and is more likely to have an honour in another suit (since the MAXIMUM other honour North can have is a K).

 

Therefore, leading a ♦ back is unlikely to give anything away, since the only time it does give something away is when North has the K (which is unlikely). Does this seem reasonable?