[pclayton]

Scoring: MP

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You play in 4♠ after a transfer response. The lead is the ♣2, x, A, x; ♣K, 5, x.

Now what? Remember its matchpoints.

[ArtK78]

EDITED SEVERAL TIMES.

Sorry. Misread the problem.

Without the club lead I would have eliminated the red suits and played the ♠K and then a spade to the J. With the club lead, this is no longer required.

Is that enough of a reason to eschew the percentage play in the spade suit of cashing one top honor and leading to the other top honor?

It appears that clubs are 3-3 (assuming that the 2 lead was from 3). If that is the case, the whole hand is who has the ♠Q. In any event, I would still play the top spade honors. If the ♠Q does not come down, eliminate the red suits and throw in the opponent with the ♠Q in the hope that he is out of clubs.

I don't see any reason not to take this line.

[Echognome]

The opponents clubs seem to be breaking 3-3, although it'd be nice to know their actual lead agreements. If they are leading 2nd/4th, then it's possible clubs are 4-2, in which case I would eliminate the red suits and play ♠A and finesse West for the ♠Q, so that I get the extra chance that East will have to give me a ruff and sluff if he wins.

This is basically the same idea that Art had, but had misread the play to the first couple of tricks.

[FrancesHinden]

Why are we assuming clubs break 3-3? Doesn't anyone play standard leads any more?

There's actually something very weird going on in the club suit, which might become slightly less so if you were to tell us which club RHO played back on the second round, I assume it's an honour (in fact, it's very likely to be precisely the jack to avoid too much in the way of weirdness).

[cherdano]

FrancesHinden, on Nov 11 2008, 03:14 PM, said:

Why are we assuming clubs break 3-3?  Doesn't anyone play standard leads any more?

There's actually something very weird going on in the club suit, which might become slightly less so if you were to tell us which club RHO played back on the second round, I assume it's an honour (in fact, it's very likely to be precisely the jack to avoid too much in the way of weirdness).

I guess we have to deduce lead agreements, as well as RHO's play to trick two, from the fact that it is a worthwhile play problem...

[kenrexford]

Win the club.

Then, play a spade to dummy's high spade. A 4-0 with the length to the right seems more likely.

Then, assuming no 4-0, pull, a second spade. If they do not split, eliminate the diamonds first, in case hearts are 5-2.

[awm]

I am impressed that we have seven clubs between the two hands, we played two rounds of clubs with everyone following, and none of the ♣Q, ♣J, or ♣T has appeared. It seems that there are fourteen clubs in the deck, so perhaps the correct answer is to call director....

Okay, joking aside, knowing exactly which club spots have been played by each of the two opponents could easily help us figure out how the suit is dividing.

[Lobowolf]

awm, on Nov 11 2008, 04:32 PM, said:

I am impressed that we have seven clubs between the two hands, we played two rounds of clubs with everyone following, and none of the ♣Q, ♣J, or ♣T has appeared. It seems that there are fourteen clubs in the deck, so perhaps the correct answer is to call director....

Okay, joking aside, knowing exactly which club spots have been played by each of the two opponents could easily help us figure out how the suit is dividing.

P'rhaps RHO played an honor the second time...I see 7 played clubs.

[mikeh]

As others have observed, we lack info.

If clubs are ostensibly 3-3 (and we'd all know if they led 3rd best or 4th best), then I can't see why we'd think of playing for a hook in spades.. altho I would cash the A first, because of a faint, almost invisible inference that LHO may have 4 trumps...not enough to warrant playing him for Qxx.

If clubs are ostensibly 4=2, with length on my left, I suppose I have a tiny missing spaces argument in favour of playing rho for 3 trump... I'd like to know if Frances' deduction was correct.. that rho returned the club J.

But it being mps makes me wonder... the entire field (in most events) will be playing for the drop, so taking a finesse is staking a top or bottom decision on the tiniest of percentages. I tend to go with the field on this one... but in a really tough field, in which I need every good board I can get, and where playing for averages on this kind of hand won't get it done, and where a lot of declarers aren't wedded to '8 ever 9 never', I'd hook rho... and I would definitely hook rho at imps if clubs were 4=2.

[pclayton]

The opponentrs lead 4th best.

If I have an extra spot in there, I wasn't aware of it.

[jdonn]

I play king of spades since LHO is more likely to be short given clubs 4-2. If no one showed out I play spade to the ace and go for the elimination if someone had Qxx. I really don't see much more to this, am I missing something?

The club spots do matter of course, if clubs are 3-3 you want to play ace of spades first I think.

[cherdano]

pclayton, on Nov 11 2008, 08:37 PM, said:

The opponentrs lead 4th best.

If I have an extra spot in there, I wasn't aware of it.

You forgot to tell us which card RHO returned.

[Lobowolf]

cherdano, on Nov 12 2008, 12:05 AM, said:

pclayton, on Nov 11 2008, 08:37 PM, said:

The opponentrs lead 4th best.

If I have an extra spot in there, I wasn't aware of it.

You forgot to tell us which card RHO returned.

I was starting to think it was just me.

[pclayton]

Sorry. Thanks for bearing with me.

RHO returns the Jack.

[FrancesHinden]

Well, it was obviously a two-part play problem. The first part was to deduce that RHO returned the CJ....

Just because they are playing 4th highest doesn't actually prove that clubs are 4-2, either LHO has led from Q52 and RHO has AJ10, or LHO has Q1052 and RHO AJ doubleton.

However, I think I'm prepared to go with clubs being 4=2, because Qxx club lead looks wildly unattractive given that RHO probably has some diamond length, plausibly to an honour.

So we are being to decide between three possible lines:
i) play two top rounds of trumps
ii) cash one top spade from hand, eliminate the red suits, and finesse trumps into RHO
iii) finesse trumps through RHO on the basis that he only has 2 clubs

(i) makes 11 when the spade comes down, or when RHO has Qxx and a doubleton club
(ii) makes 11 any time clubs are 2-4 and trumps not 0-4, or when LHO has the SQ, assuming no costing red suit ruff
(iii) makes 11 if RHO has the Q of trumps

I can't be bothered to work out the relative odds, but I think these must be quite close. Line (ii) is most flash, so I'd probably go with that. (Spade length on our left is actually reasonably likely given that LHO has led a 4-card club suit)

[ArtK78]

FrancesHinden, on Nov 13 2008, 06:22 AM, said:

(Spade length on our left is actually reasonably likely given that LHO has led a 4-card club suit)

Huh?

[FrancesHinden]

LHO decided to make a very aggressive club lead, at matchpoints, from either Qxx or Q10xx into a strong NT.

If he has a singleton spade, what does that make the rest of his hand? I don't think one spade and 3 clubs is really likely.

Obviously this depends a bit on LHO's leading style, I think a lot of people would prefer to lead from pretty much any of the possible 5-card diamond suits than a 4-card club suit. Maybe we should give this as a lead problem:

x
Jxx
Q9xxx
Q10xx

would there be a unanimous club lead?

while if he has longer spades, he is much less likely to have an alternative potential side suit lead.

Also a trump lead from a low doubleton is quite likely, but not from 1 or 3 trumps.