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Endplay or ?


nielsfoged

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[hv=pc=n&s=sa98hdaqj74cat874&n=st52hakjt3d92ckq2&d=s&v=e&b=3&a=1dp1h1s2cp2sp3nppp]266|200[/hv]

Regional Matchplay Championships

Lead Q-2-3-8; 7-5-J-A; 4-3-Q-4!

What now?

If you continue 2-3-Q-K; then LHO returns with 6.

If you then play 6-K-2-7; and continue , it goes 9-T-A-5.

What now?

Your opponents are national open team level!

Does that influence your plan?

 

Kind regards from Niels, who - obviously! - chose the wrong continuation.

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I would cash a diamond pitching small heart. If they break I have lots of tricks.

 

If not I will play RHO for HQ. I know he started 5440 so I cash two clubs ending in dummy. If I see two heart pitches then hearts should be good from the top. Else exit a spade to endplay RHO.

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[hv=pc=n&s=sa98hdaqj74cat874&n=st52hakjt3d92ckq2&d=s&v=e&b=3&a=1dp1h1s2cp2sp3nppp]266|200|

Regional Matchplay Championships

Lead Q-2-3-8; 7-5-J-A; 4-3-Q-4!

What now?

If you continue 2-3-Q-K; then LHO returns with 6.

If you then play 6-K-2-7; and continue , it goes 9-T-A-5.

What now?

Your opponents are national open team level!

Does that influence your plan?

Kind regards from Niels, who - obviously! - chose the wrong continuation.

[/hv]

After winning A, lead a If RhO shows out then cash 3 clubs and 2 diamonds before exiting in to endplay RHO in . Otherwise finesse 7 Even if it loses, you make at least 1 X , 2 X , 3 X and 3 X .

 

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Thank you Adam

 

I think your plan is good. If break you will most likely make 1+2+4+3=10 tricks (maybe even more), and if are 2-4 you will have 9 trick on an endplay ensuring you 3 tricks independently of the position of Q, but the opponents will have their 2+0+2+0=4 tricks.

 

Niels

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Hi Niels

 

Cashing the third diamond can't be wrong, can it?

 

If diamonds are not 3-3, awm's plan works also with W having the Q, because we have full count.

 

Tempting to finesse J when west returned a heart, because east's spade discard suggests east doesn't have Qxxx or Qxxxx. But that is probably too big a play.

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Hi Michael

 

You and Adam (and probably also Nigel) hit the head of the nail: caching is at least almost unlikely to be wrong.

However, in the heat of (MP-)play, my analysis stopped at the excitement of endplaying a RHO holding a 5-4-4-0 distribution for the 9th trick.

So after the first 6 tricks (Q-2-3-8; 7-5-J-A; 4-3-Q-4, 2-3-Q-K; 6-K-2-7; and 9-T-A-5, I erroneously played 9 and hoped for something like the following distribution (where the position of Q is irrelevant):

[hv=pc=n&s=s9hdj74cat8&w=sh974dcj965&n=sthkjt3dcq3&e=sk6hq85d86c]399|300[/hv]

Actually LHO pitched 6 when I played the third , and after RHO had catched his 2 tricks, I had to settle for 9 tricks, where at least 10 would have resulted from just continuing , which were distributed 3-3.

What would have been even more emberassing was that if LHO had held 8 (iso 6) and pitched 4 and 5 on his partners two tricks (thereby keeping -/9/8/J96), I would have finessed , when RHO continued with 6! - 1 down!! :unsure:

 

PS: Michael, I am impressed with your logic conclusion that RHO early discard shows that he does not hold Qxxx (or Qxxxx) in . It is a wonderful simple argument, but I missed it - of course it would be easy for him to discard 1 (respectively 2) with one of those holdings, rather than dropping a potential -trick. Actually RHO had KJ643/98742/T83/-.

So only if RHO made an exotic/unlikely -discard from a holding of KJ643/9872/T863/- or KJ643/Q982/T863/-, we will end up having to guess whether to endplay LHO or RHO, respectively. Apparently, you would go for the first, and Nigel for the second endplay. Still, both of these situations are unlikely with the early -discard.

 

Best regards Niels

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Note that if East is 5440, we will always get 9 tricks.

 

After the third diamond, seeing the split, we play A club to dummy. If east keeps 2 hearts, we will endplay him in spades. If east has kept only 1 heart, we play K, J. Whatever west has kept, he will have to give us trick 13 in dummy (T) or in hand (T). They get one trick in each suit only.

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Note that if East is 5440, we will always get 9 tricks.

 

After the third diamond, seeing the split, we play A club to dummy. If east keeps 2 hearts, we will endplay him in spades. If east has kept only 1 heart, we play K, J. Whatever west has kept, he will have to give us trick 13 in dummy (T) or in hand (T). They get one trick in each suit only.

 

Not necessarily "always" 9 tricks if East if 5440 - at least not with me as Declarer :P

Remember this is MP, so what are you discarding from Hand in trick 5 (on the return from LHO)?

Admittedly, I pitched 7, thereby destroying the chance of playing two more rounds of before setting up T on an endplay. My reason for not discarding a from hand was the obvious chance of making 4 tricks, but maybe I could have discarded 9 instead, though I in the actual situation thought I may need to execute the endplay from hand, while keeping the communications in open. Probably, that was a mistake, too!

 

PS: Based on your previous valuable comment on the most likely distribution of , I now fully realize that unless RHO is a really tricky guy (=super-expert ...or beginner?), his discard rather than an easy on the first round of in trick 3 places him with either exactly Qxx or 9xxxx(x) in . The former (=5-3-5-0) can be excluded after having seen at least 2 in LHO, so it appears that the actual distribution (RHO: 5-5-3-0) was an open book already in trick 3! I look forward to the day, when I see that at the table and not 5 days later :D

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