dellache

There's a line that always work (even if West is 1462) provided that we can read West's length in Diamonds (let's suppose he has 6). Note : if East has the stiff ♦Ace, West cannot be 1462 (he would have led a Diamond). The principle is [234] cash Spades, [56]Ruff one club, [7]Heart to dummy, [8]Ruff club. We now have West's complete count, and West has kept 5 red cards : 1. if he's kept 2 diamonds only (he still has 3 hearts, and was 1462), we cannot strip him out of Hearts, but we can just play [9]♦King from Hand in order to set up one diamond. 2. if he's kept 3 diamonds, we can strip him out of Hearts [9-10] ruff a Heart, and [11] play Diamond toward K10x, covering East's card. Playing this way, I think we almost always score 11 tricks.

The 3♠ bid is quite strange, and I suspect the Spades are 6-2. Anyway I don't see what I can do with that. Main plan is to play another Spade and play a Diamond up to the King. If South has the ♦Ace, I can succeed only if (1) he keeps the Heart Guard (QJ9? or pard discards a "useless" Heart in 9xx?) and (2) I read the position correctly to triple squeeze him. Anyhow, I'll play a Spade back, and decide what to do after cashing the clubs and one top Heart.

Now he doesn't get a chance to do that, but you just go down. If I'm correct, you have just reached (lead in dummy) : [hv=pc=n&s=sqhakqdcq&w=shdc&n=shjdc8765&e=sh875dckj]399|300[/hv] You now lack an entry to your own hand, don't you ? (or maybe we should claim at this point :))

Hmmm Phil, it doesn't work, does it ? : 1. If East has the ♥Ten, he can overruf one of your spades ; 2. If West has the ♥Tx, curiously enough, East can promote his partner's 10, if he defends correctly (your ♥64 is to small to prevent killing defense when you ruff the 3rd Spade : East can discard a Diamond and... :rolleyes: ). Cheers.

There's no real hope of making if this is a (probable anyway) singleton lead, unless West has specifically KJxxxx xxxx Kx x. So I'll just duck this to my queen, hoping for a doubleton lead, or maybe KJxxxx x Kxx Kxx, or maybe something else with 3-2 clubs.

Untrue (see my post above). Say West is 2=3=4=4 : West discards a Heart on the 3rd spade (trick 4). When you play Heart (trick 5), East takes the Ace and plays the 4th spade (trick 6), allowing West to throw his last Heart : This leaves : [hv=pc=n&s=s7hkqdqj83c&w=shdt976ct92&n=sh87dakck86&e=shdc]399|300[/hv] ... w/ the lead in dummy and you are toast : no way to score ONE(single) Heart AND the trumps separately (an alert East will see that playing the master Spade kills you). Try it. You don't have THIS problem (but other problems :)) if you start with an early Heart.

To me this is a MOST difficult hand to analyse : I've no definite conclusion, but finally I'd like to share my thoughts. 1. I don't think the Spade finesse is right : it is definitely LESS than 50% (when it scores, there's still lot of work), when playing on Spades is somewhere between 36% (3-3) and 36+24=60% (24%=2-4 spades, but then we need some guesswork in those cases, see 4. below). 2. I think that playing Heart early is usually right, but not for the reason that Fluffy pointed out (good oppos will duck this most of the time and moreover, playing twice toward KQ is dangerous wherever the Ace may be -- anyway that's part of the difficulty to analyse). We may need to SCORE a Heart trick early before someone can discard Hearts (West). 3. What makes computing odds difficult is also the choice of lead : West rates to have JT9(x+) or even QJT(x+) [deceptive] hence long clubs to justify leading in dummy's suit. That consideration clearly influences the distributions in other suits, and by consequence the choice of the right line. 4. I think that the general line is to ruff a spade LOW at SOME point (obviously making w/ 3-3 spades) and try to survive when West has only two of them (we don't care about East being 2-2 in the pointed suits for the time being : see point3, West has long Clubs). One of the point to analyse is the following : what happens if West is 2x4x ? Then in order to make we will need some cross-ruff with the good *timing*. Example : West has the simple xx xxx T9xx JT9x In that case, we need to cash a Heart early, before playing any spade ([2]K♣ discarding Heart, [3] Heart toward King, East ducks, [4] proceed to ruff spades and guess in which rounded suit(s) West has kept cards). If we don't cash a Heart quickly, and start with Spades, here's what happens (provided East has the Heart Ace) : West discards a Heart on the 3rd Spade, and when we play a Heart from Dummy, East plays the Ace and fires back the 4th Spade : now West discards his last Heart and the cross-ruff is doomed. But if West had xx xxxx T9xx JT9 (not probable Club distribution), we must start the same way [123] and exit with a Heart at trick 4 : we must plan to ruff two cards from dummy before West has got any chance to discard on the 3rd Spade. It seems that there's some guesswork involved (can we read the rounded suit distribution at trick 2 and 3 ?) as well as the interpretation of the lead in order to choose the right timing. If the diamonds are 3-2, we can play on the same line, and we have more leeway for guessing. Analysis depends on multiple factors difficult to assess : - the lead ; - ability to read the parity of the oppos ♥♣ suit early ; - what will they do when I play a Heart ; - what will they do after they cash the Heart Ace ? I think computing a %age is nearly impossible. Anyway, I'd start like Jack (kgr post) and think about continuation at trick 4 (♠ now probably, Heart maybe, never finesse) (apologies if this post is to confuse).

3♦, then I'll raise any bid from pard, unless he bids 3NT (unlikely). Over 3NT I'll bid 4♠ (no strong feelings though).

Yes you are of course right, I didnot reread last minute modifications of my post. The maths and results remain the same. I'll edit my post.

It's hard to believe that those 2 distributions can have the same probabilities (3 cases for ♠QJx/xx, and only one case for ♠xxx/QJ, other suits being the same). It looks like there's something wrong doesn't it ? Or do I miss something obvious ?

I'm not sure that counting card combinations like gnasher did is enough, because East has got to have 12-13 HCP, and that has an influence on the whole distribution probabilities. BTW, you didnot say who has the ♦J, important card. CASE 1: West had ♦J As East is balanced, he has got all the missing honor cards (12 HCP), and the combinations have to be made on the 'x' slots. WHOLE EAST DIST --- SLOTS E ------ SLOTS W QJxx xxx AKx Qxx -- xx xxx . xx -- x xxxx xxx xxxxx QJxx xxxx AKx Qx -- xx xxxx . x -- x xxx xxx xxxxxx QJx xxxx AKx Qxx -- x xxxx . xx -- xx xxx xxx xxxxx If you count the combinations on the major suits only regarding the 'x' cards, you get the same for the 3 distributions. Only the club suit leads to a difference, and we have 4333=3433 > 4432. The ratio 4432/3433 = B(7,1)/B(7,2) = 1/3. Hence 4333=3433=3/7, and 4432=1/7. CASE 2 : East had ♦J. Now all the honors are known except the place of the ♠J who acts "as a small card". The table has become : WHOLE EAST DIST --- SLOTS E ------- SLOTS W Qxxx xxx AKJ Qxx -- xxx xxx x xx -- x xxxx xx xxxxx Qxxx xxxx AKJ Qx -- xxx xxxx x x -- x xxx xx xxxxxx Qxx xxxx AKJ Qxx -- xx xxxx x xx -- xx xxx xx xxxxx We now clearly have 3433 > 4333 > 4432. 4333/3433 = B(4,1)/B(4,2)= 2/3 4432/4333 = 1/3. Looks like (3433,4333,4432) = (9/17,6/17,2/17) now [edited] All these figures don't tell us what to do in reality if the ♣Queen falls on the King. [edited in red on feb the 15th]

[hv=d=w&v=0&b=8&a=1hp1sp4cp4sppp]133|100[/hv] 4♣ is an underbid but I don't like alternatives (fake 3♦). 4♦/4♣ would be exagerated I think : a lot of work to do (including the real hand) ! Hence 4♠. Over 4♠ we have far more chances to find a weakish hand that means danger in 5♠ than the good hand for slam. Hence West makes a statistical Pass. This is imho one of the tribute to pay to very wide range 1♥ openings if you don't have the artificial arsenal to cope with "end of the spectrum hands".

The obvious line "declarer has the ♣7 and need glasses" makes sense. At the table though, good cooperation may suffice : - I would discard AH, then AC, then JH, then JC, and after trick 11, keep just ♥K, ♣K. Before discarding at trick 11, pard knows my cards. - Now (1) if he was missing one of the relevant Tens (♥♣), he just discards the whole suit and (2) otherwise he keeps both Tens. All I have to do is to keep the King in which there are the less remaining cards left : it obviously works in case (1), and in case (2), partner will also know what to do.

[hv=pc=n&n=s7654hd732ck65432&d=e&v=b&b=10&a=1hd4hpp5d5h]133|200[/hv] What do you do and why ?

[hv=pc=n&n=sat42ha3d7432ckj2&d=s&v=b&b=7&a=p1hd]133|200[/hv] Do you agree with North's double ?

[hv=pc=n&n=s52hk632dat93cj64&d=s&v=n&b=15&a=1np2cp2dp]133|200[/hv] IMP. Do you bid 3NT or not ?

[hv=pc=n&s=s5ha54daqjt532ckq&n=sa72hk762dk7ca652&d=s&v=b&b=7&a=1dp1hp3dp3nppp]266|200[/hv] IMPs. The result is ridiculous, and at least one of the players seem to have underbid (please vote). BTW, what do you suggest as a bidding sequence with those hands ?

1. If they bid no higher than 3♠, we can show our distribution by bidding 2♣/4♥ 2. If they bid to 4♠, we will probably play at least at the 5-level. In that case we will want to play in the best fit, and ♣ is a priori the best target. Hence, playing "natural", I would never consider anything else than 2♣. Actually with xtruel we play a 5542 system where the 3♠ jump-cue shows that kind of hand (rounded suits 5+5+, 6- losers hand). If they bid 4♠ now, I'll show my hand by bidding 5♣ on the next round.

Declarer has certainly : 1. ♥ AKQxx+ after the 1st trick 2. ♠ xxx because of second trick (playing like he did with Kxx is nonsense, pard showed count, and if he has xxx, ♠ is safe) 3. he has certainly at most 2 diamonds (otherwise he would draw trump for fear of 6 diamonds by opener). Declarer surely has problem to create his 10th trick and : 4. is trying to set up 3 spades -- this is compatible w/ the play so far. 5. has few chances to set up extra tricks in the minors unless we help him. So I certainly will not play a club (AJx by declarer is a disaster), and will surely play a Diamond. what about a "safe" Heart ? It's dangerous if declarer has xxx AKQxx xx AJx. Declarer will draw trump, test the Spades (bad news), ruff the last spade, and might think about cashing the last Heart. The ♦7 begins to look enormous compared to your 642, doesn't it ? It looks like pard will have no good option at trick 8 : [hv=pc=n&s=shdaqjcq92&w=shdk73ct75&n=shd642ck84&e=shkd85caj3]399|300[/hv] Declarer plays his last trump and : - if he keeps 2 clubs, declarer discards a Diamond and just plays small to the Ten ; - if pard keeps 2 Diamonds, declarer discards a Club, and plays a Diamond : -o- if pard ducks, declarer exits with a Diamond, and defense is endplayed ; -o- if pard takes his Ace, the 7 of Diamonds is the 10th trick. Maybe difficult to see at the table, but anyway, the ♦6 returns looks like the most "hygienic" return. NB: We may have to hope that declarer doesn't have the ♦Jx, otherwise it is still not over : pard will have to be in great shape in order to definitely kill the impending squeeze by playing the ♦Q next... Bof.

I'd bid 4♥, then DBL. I agree that 2♠ first is clever, but we don't have this agreement in our pair and I don't want do have to deal with 2♠ (X) Pass (4♠).

Agree with Rainer that ♣finesse is totally obvious (and leads to 12 tricks 99% of the time via the upcoming squeeze). Long version: In practice if West doesn't have both honors he will have x xxxxx Q KJT987 for his discards (discarding 78♣ is easy). In theory we lose if West has x Qxxxxx Q JT987. Totally unlikely ?... Hmmmmmm.... if West comes from Jupiter he could make the following reasoning in 10 steps : 1- South has KQx+♠ (no 2nd round finesse) ; 2- South doesn't have KQJx (he would have left the double) ; 3- East knew that "small spade" costs nothing : he has 7 Spades ; 4- South has ♠KQx exactly 5- West "sees" 5+13+5 = 23 HCP in W-N-E hands respectively. 17HCP are missing. 6- South has ♥Ace by counting HCPs. 7- If South has ♣K, South has 11 tricks from the top (2+2+4+3), and 12 thanks to an obvious Squeeze ; 8- So West hopes that pard has the ♣K, and South upgraded a 14HCP count because he had 5♦ ; 9- So West hopes that south has the actual hand KQx Ax AJT9x xxx 10- West thinks like Rainer, and discards 2 clubs from JT9xx to induce the OBVIOUS ♣finesse. Aspirin please.

Over 2♥, I'd bid DBL, followed by 4♠. Over 3♥, I'd bid 4♠ directly to avoid 2 problems : - having to rebid over 5♣ ; - avoid uncertainty in pard's mind when I rebid 4♠/4m. Still with all those controls, I'm not so happy about that.

2♦, no second choice. Then we'll think again.

Indeed. ♥Ace is a dubious lead at any rate. We need to evaluate how dubious it was depending on the trump holding. QTx is rarely going to be a trick, but now it is if declarer plays for the drop ;)

This is puzzling. I can't believe East would underlead the ♥10 (dangerous !) with two perfectly safe leads in spades and diamonds xxx+. So the following might not apply to East... Anyway, as East I would certainly lead a spade/diamond with xxxx of them, hence East is 3x3z, hence East has long clubs. I finesse against him.