Play 3NT

[vuroth]

I'm always a bit nervous posting problems, because it's always possible I've overlooked something that makes it a poor hand to share. (After all, I get so many of these wrong at the table...) Of course, I feel that way whenever I post my answers to others' problems, and it never stops me so...

 

[hv=d=s&v=e&n=sa842ht2dkt2caq43&s=sk97hakq9dq3c9876]133|200|Scoring: IMP

1♣ 1♠

1NT 3NT

 

5♦ led to the T, J, Q.

 

Plan the play.[/hv]

[hanp]

I would play a spade to the king and attempt to duck a spade into RHO's hand. Suppose RHO wins and they duck a diamond. Then I would test spades (if they split, 9 tricks), play hearts from the top (if the jack drops, 9 tricks), or else take the club finesse.

 

The combined chances seem best this way.

 

If LHO plays an honor on the second round of spades I will win and play a third round. If RHO, upon winning a spade doesn't play diamonds but (for example, hearts) then I will win high in hand, unbocking the 10, and play clubs, probably running a low club if it isn't covered.

[kgr]

If LHO plays an honor on the second round of spades I will win and play a third round. If RHO, upon winning a spade ....

It is also possible that LHO wins the 3th ♠. You are confident enough that he also has ♦A?

[Siegmund]

I'm going with the plain simple "lead clubs out of my hand and cover whatever LHO puts up" line. It's same idea as ducking spades to East, but repeatable one more time before West threatens to gain an entry.

[hanp]

If LHO plays an honor on the second round of spades I will win and play a third round. If RHO, upon winning a spade ....

It is also possible that LHO wins the 3th ♠. You are confident enough that he also has ♦A?

Confident enough that I thought this is the best line. Obviously I can go down but for example the alternative line that Siegmund suggests is so much worse when ace-fifth of diamonds is onside that I will take that risk.

[vuroth]

Interesting. Han's line is a revelation to me.

 

I tried the math, and the lines were pretty much a wash. Maybe someone with better math can puzzle it out for me.

 

My reasoning is that, if the A♦ is on your left (more or less 50%, I guess a touch more), Siegmund's line is better. Unless LHO is KJT, KJTx or KJTxx in clubs, you're always getting 2 club tricks, and the KD is always coming home (regardless of what the hearts do) So 87.5%. Han's line is a little over 80% in this case, but not quite as good.

 

If the A♦ is on your right, and we assume that RHO is never leading ♦s if he gets in (maybe this is unreasonable?), then both lines are longshots, but han's is better. Siegmund safely gets 2 club tricks around 40% of the time, and will have to try hearts for his 9th trick - which works maybe 25% of the time? With only 1 club safe club trick, Siegmund has no play I think. han will find spades 3-3 36% of the time, but 30% of THAT time LHO will have QHx and he'll be down straight away. If spades are 3-3 he still needs a 9th trick form hearts, and if spades aren't 3-3 he needs hearts PLUS the club finesse. I have it at something like 32%?

 

EDIT: Ok I have it a dead heat.

 

I took Siegmund's line at the table. I think a B/I who took either line would be above the B/I average. :(

 

V

[vuroth]

Obviously I can go down but for example the alternative line that Siegmund suggests is so much worse when ace-fifth of diamonds is onside that I will take that risk.

Not sure what you mean by onside, but LHO always has 4 or 5 diamonds, right? If LHO has the A♦, Siegmund's line seems quite secure.

 

If RHO has A-fifth in ♦s, the opening lead sure fooled me.

[hanp]

If LHO has ace-fifth of diamonds and Jx or 10x in clubs, you go down using Siegmund's line? They'll get 5 tricks before you get 9.

[vuroth]

Good point. If LHO has A-5th, you can't lose 2 club tricks even if you promote 2. So you lose to

v KJTxx

x KJTx

xx KJT

Hx KHx

Hxx KH

 

Or 10 of 32 cases, succeeding 68.8%.

 

If LHO has A-4th, Siegmund's like works unless clubs are

 

xx KJT

x KJTx

v KJTxx

 

or 4 of 32, succeeding 87.5%

 

That's something like a 78% chance combined for the Ace being with LHO, so han's line is slightly better either way.

 

Thanks, han.

[Fluffy]

I like playing ♠7 without cashing the spade King.

 

It is technically incorrect, but LHO will often missplay with ♠Q10x or ♠J10x

[Siegmund]

I admit I didn't worry as much as I should have about LHO being good enough to duck trick 2 and maintain communication. (The live defenders at my club, as a rule, miss that play.) If there's no chance of LHO having six diamonds, han's line may well be the better one against good defenders.

[hanp]

If LHO has A-4th, Siegmund's like works unless clubs are

 

xx KJT

Not this one, you'll lose only 2 club tricks.

 

Also, from your 78% it seems like you think that ace-fourth is about equally likely as ace-fifth, this seems a big assumption. To compute these chances is extremely hard, because you will have to take into account LHO's motivation to lead a diamond. With ace-fifth of diamonds he would almost certainly lead a diamond, but with ace-empty-fourth he might not. That makes the ace-5th holding more likely. Moreover, ace-fifth is positively correlated with LHO holding short clubs, UGH!!

 

There is also the matter of the diamond 5. For LHO to hold 5 diamonds he needs to hold the 54. For LHO to hold 4 diamonds he needs to hold the 5 and RHO needs to hold the 4. Here at least it is easy to see that the number of 5-card holdings is exactly equal to the number of 4-card holdings. Had LHO led the ♦6 though, then there would be exactly twice as many 5-card holdings as 4-card holdings, a HUGE difference. (This calculation is surprisingly simple and it seems to me that it is useful on many hands, I will start a new thread)

 

So what can we conclude from this endless calculation? Perhaps that computing the best line for this seemingly simple hand is next to impossible at the table. Besides, if your line on this particular hand is 2% better or worse than the alternative, that is not going to determine your succes as a bridge player. A good declarer (and I'm guessing here!) would probably think of most of these factors and then estimate only roughly which line is best.

[vuroth]

With ace-fifth of diamonds he would almost certainly lead a diamond, but with ace-empty-fourth he might not. That makes the ace-5th holding more likely. Moreover, ace-fifth is positively correlated with LHO holding short clubs, UGH!!

Good point about the bias on lead.

 

There is also the matter of the diamond 5. For LHO to hold 5 diamonds he needs to hold the 54. For LHO to hold 4 diamonds he needs to hold the 5 and RHO needs to hold the 4. Here at least it is easy to see that the number of 5-card holdings is exactly equal to the number of 4-card holdings. Had LHO led the ♦6 though, then there would be exactly twice as many 5-card holdings as 4-card holdings, a HUGE difference. (This calculation is surprisingly simple and it seems to me that it is useful on many hands, I will start a new thread)

 

Good point. I had that the 4 was equally likely to be in each hand, but hadn't extended it to other leads. That's a useful at-the-table calculation. Thanks.

 

In general, I'm probably over-calculating lately. Still, it's a useful exercise not only from the point of view of helping to remember the numbers (i.e. 3/3 is around 36%), but to challenge some of my incorrect notions and tendencies. Even if you only do a few calculations in your life on combining chances, it quickly becomes obvious that this is a vastly beneficial strategy.

 

I also mentally hand-wave at too many problems that I should give serious thought to. I wondered about trying spades on this hand, but dismissed it as unlikely - the 3-3 fit isn't overly likely. Still, by dismissing it, I overlooked the possibility of combining chances.

 

Anyways, thanks for the feedback han. As always, I appreciate you taking the time to correct my oversights. :)