Suit combination with some holes

[cherdano]

In a f2f pairs tournament, top bracket, we got an 85% score when declarer took a clearly bad line in the trump suit:

[hv=n=shj9dc&s=shk86542dc]133|200|Scoring: MP

You are in 4♥, no losers outside of hearts.[/hv]

Can you do better than my opponent? How many tricks are you playing for? Which line do you take? How do you like your chances?

 

Since you are playing against me and my partner, you should assume world class defense (just kidding).

 

Arend

[Free]

I would play small to the 9, safety play and hope you get some extra on the way :blink:

[scoob]

i'll play the J from dummy, but then i'm notoriously bad at playout :blink:

[luke warm]

playing the jack first seems to almost (i said almost) guarantee 3 tricks... but this *is* mps :blink:

 

hell, i'd play that anyway

[cherdano]

playing the jack first seems to almost (i said almost) guarantee 3 tricks... but this *is* mps :)

 

hell, i'd play that anyway

I think the goal has to be 4 tricks.

Btw, what does everyone do on the second round?

 

Arend

[luke warm]

i'd lead the jack anyway... play to 2nd trick depends on what happened 1st

[Fluffy]

small to the J for me, but might change my mind over opponent's level and tempoes.

[wvlaker]

What was the obviously bad line that the opp took?

[HeartA]

small to J (or 9). If you run J from dummy and your lho shows off, you have trouble.

[luke warm]

small to J (or 9). If you run J from dummy and your lho shows off, you have trouble.

you mean your rho? if your rho is void it means lho has A,Q,T,7,3 and you're in trouble anyway, no matter what you do

[PriorKnowledge]

The problem with leading the J is that it does not help you. There are no arrangement of the missing honors that leading the J wins.

 

Leading small towards the 9, unnecessarily loses 3 tricks to 10x in RHO

 

Leading the 9 to the K gains only when RHO has the A. The Q in either hand can wait for the now stiff J.

 

The correct play is:

 

Lead towards the J. If LHO has the Q, the J will force out the ace (1st or 2nd round). If the J loses to RHO's Q, then get to dummy and lead towards the K. This gives you 2 chances to find either honor onside so that you only lose 2 tricks if trump break 3-2.

 

If trump break 4-1 with both honors in one hand, leading towards the J will only lose 3 tricks either way. Leading the J will lose 4 tricks if LHO has 4.

 

If trump break 5-0 either way, you are screwed. You are losing 4 tricks even if you stand on your head.

[luke warm]

The problem with leading the J is that it does not help you. There are no arrangement of the missing honors that leading the J wins.

since it's equally as likely for any 3/2 or 4/1 split to be on the left as the right, it seems to me that leading the jack certainly can't be wrong... maybe leading *to* the jack (or 9?) isn't wrong either, i don't know

 

If LHO has the Q, the J will force out the ace (1st or 2nd round). If the J loses to RHO's Q, then get to dummy and lead towards the K. This gives you 2 chances to find either honor onside so that you only lose 2 tricks if trump break 3-2.

 

if lho has A,Q he'll play the Q on your lead to the J.. now you say lead the 9 toward the king (do you play the king?).. but that loses 2 more tricks, to the A and T.. taking the same split, lead the jack then lead the 9 and let it ride (unless covered with T)

If trump break 4-1 with both honors in one hand, leading towards the J will only lose 3 tricks either way.

 

i don't think that's true.. give rho AQT7 and lead toward the jack.. he takes the queen and exits.. you enter dummy and lead the 9 and he covers, you win the king... now he has A,7 over your 8,6 eh?

 

maybe i'm looking at it wrong

[PriorKnowledge]

This is not that difficult to work out and realize that leading the J never gains. It is a 100% losing play.

 

A wonderful site to help you with this type of problem is:

http://www.automaton.gr/tt/en/OddsTbl.htm

[PriorKnowledge]

If LHO has the Q, the J will force out the ace (1st or 2nd round). If the J loses to RHO's Q, then get to dummy and lead towards the K. This gives you 2 chances to find either honor onside so that you only lose 2 tricks if trump break 3-2.

if lho has A,Q he'll play the Q on your lead to the J.. now you say lead the 9 toward the king (which i agree with).. but that loses 2 more tricks, to the A and T.. taking the same split, lead the jack then lead the 9 and let it ride (unless covered with T)

 

uh... If LHO has AQx and plays the Q, you don't play your J... Now the J forces the A and the K is good. This line of play only loses 3 tricks to LHO=A10x and RHO=Qx.

 

If trump break 4-1 with both honors in one hand, leading towards the J will only lose 3 tricks either way.

i don't think that's true.. give rho AQT7 and lead toward the jack.. he takes the queen and exits.. you enter dummy and lead the 9 and he covers, you win the king... now he has A,7 over your 8,6 eh?

 

maybe i'm looking at it wrong

You won the K. You can't lose 4 tricks to a 4 card suit if you win one of the tricks.

[luke warm]

sorry, was thinking 3, not 4... my error

[cherdano]

My opponent lead the 9 to the king, which is clearly not optimal.

 

I am pretty sure that leading the J is best (covering if RHO plays the Q). If it looses to the Q, you have a complete guess: either lead the 9 to the K, or run the 9. Surprisingly, the only 3-2 splits against which this line (say by running the 9) loses is exactly AQT or QT on your left.

On the other hand, leading low the J then the 9 to the k loses to ATx or Ax on your left, i.e. 4 instead of 2 of the 20 3-2 splits.

 

I think it is better to lead the J than to lead the 9, because a singleton queen never helps you, but leading the J suceeeds if LHO has the singleton 10.

 

Arend

[cherdano]

If trump break 4-1 with both honors in one hand, leading towards the J will only lose 3 tricks either way. Leading the J will lose 4 tricks if LHO has 4.

That is, of course, wrong. Basically, you can never lose 4 tricks against a 4-1 split if you never play the K on the same trick as the J or 9 (unless it covers an honor). Play the J, 9, K, 8 in any order and the 7 can't make a trick.

 

Arend

[Stephen Tu]

This is not that difficult to work out and realize that leading the J never gains. It is a 100% losing play

 

This statement is 100% wrong.

 

The 2 lines of running the J, then leading to the K or running the 9 if it lost to the Q on the left, have identical success rates, one losing to QT doubleton on the left, the other to

AQ doubleton on the left.

 

Leading to the J, if you lose to the Q on the right, you also have choice of running the 9 or low to the K second round, which are also identical success rate, losing to either A7/A3 on the left, or T7/T3 on the left.

 

So let's just compare the two lines where second round you run the 9 if you lost to the Q. Leading to the J gains in 2 combinations, AQT on the left, and QT on the left. But it loses with AT7/AT3/T7/T3/7 on the left. Clearly inferior.

[PriorKnowledge]

my last post trying to help you - pls try to think

 

The real simple case:

If you lead the J and AQ108 is on your left, you will lose 4 tricks.

If you lead toward the J and the AQxx is in either hand you cannot lose 4 tricks.

 

Let me repeat - There is NO arrangement of outstanding cards where leading the jack gains over leading toward the J - NONE

[cherdano]

my last post trying to help you - pls try to think

 

The real simple case:

If you lead the J and AQ108 is on your left, you will lose 4 tricks.

If you lead toward the J and the AQxx is in either hand you cannot lose 4 tricks.

 

Let me repeat - There is NO arrangement of outstanding cards where leading the jack gains over leading toward the J - NONE

If AQT8 is on my left, I lose no tricks at all, because I call the director. I have the 8 myself, too. Maybe that is what you overlooked?

 

Arend

[MickyB]

Let me repeat - There is NO arrangement of outstanding cards where leading the jack gains over leading toward the J - NONE

Stephen has already given several combinations where leading the jack does work, eg ATx with LHO.

 

Even if declarer didn't have the eight, Suitplay says that leading the jack would be the best line for 3 tricks or max tricks (but not for 4 tricks).

[Echognome]

Hmm. When I put the combo into Suitplay it gave 2 lines which are identical for 3 tricks, 4 tricks, or max tricks. Both lines involved starting with the Jack.

[Stephen Tu]

my last post trying to help you - pls try to think

Maybe you should follow your own advice?

 

The real simple case:

If you lead the J and AQ108 is on your left, you will lose 4 tricks.

If you lead toward the J and the AQxx is in either hand you cannot lose 4 tricks.

 

In the problem, declarer has the 9 in dummy & 8 in hand, which are quite significant cards. As long as you don't do something completely irrational, you can never lose 4 tricks. The question is whether you lose 3 tricks or 2 tricks.

 

If LHO has AQTx, you will lose 3 tricks either way. If RHO has AQT7, same thing. But if RHO has AQT3, leading the J gains over leading toward the J. Leading toward the J, you lose the first trick to the Q. Then when you lead toward hand, RHO ducks or plays the ace, sooner or later you lose a trick to the T and the ace. In contrast, if you lead the J, pinning the singleton 7, you will hold your losses to 2 tricks. If RHO plays ace first round, you later go to dummy & run the 9 if not covered. If covered, you win K, and your 8 drives out defender's 2nd and last trick. If RHO ducks, you duck & lead toward the K later. If RHO covers with the Q, you win the K, and your 9 & 8 will drive out defender's remaining A/T.

 

Let me repeat - There is NO arrangement of outstanding cards where leading the jack gains over leading toward the J - NONE

 

It doesn't matter how many times you repeat a false assertion, it won't become true.

 

Besides the singleton 7, you also gain when LHO has ATx, and either Tx or Ax depending if you are leading toward the K or running the 9 after losing to RHO's Q. Try it! Running the J, if LHO has ATx, you lose to just LHO's A & T whether RHO covers first round or not. Leading toward the J, you lose first trick to RHO's Q, then the next two to the A & T. If LHO has Tx, running the J loses only 2 tricks, as RHO has to play ace or Q 1st round. Leading toward the J, you lose to the Q, then when you lead back to the K you have to guess. If you duck, you lose to the doubleton T on the left and the third trick to the ace. If you go up, you pick up that case, but instead you lose a third trick when RHO started with QTx. Leading the J picks up all these cases for 2 losers only.

 

Leading toward the J does pick up some cases leading the J does not, but only AQT on the left and one of QT/AQ on the left (depending if K is played 2nd round or not by people who run the J 1st round), which is fewer cases.

[PriorKnowledge]

sorry, you r right, i am wrong. i did not notice the 8 and even so, did not do enuf thinking.

 

After further analysis (with the right hands) - lead the J is best. If the J loses to A, then run the 9. If the J loses to Q, then 9 to K.

 

Leading the J is better when:

x AQ10x

Ax Q10x

A10x Qx

A10xx Q

Which occurs 22%

 

Leading to the J is better when:

Q A10xx

AQ 10xx

AQ10 xx

Which occurs 9.6%

[Stephen Tu]

Leading the J is better when:

x AQ10x

 

Actually, it's better only with 7 AQT3

With 3 AQT7, it doesn't help you. as the T7 will later from a tenace vs. the 8.

 

Ax Q10x

A10x Qx

A10xx Q

 

Leading the J in the last case doesn't help you either, again because of the 7.

 

So you gain in 16.39% vs. leading toward the J.

 

Leading to the J is better when:

Q A10xx

AQ 10xx

AQ10 xx

 

The first of these is not right. You'll still lose 3 tricks. Q wins. J forces the A,

RHO still has Tx and you have no cards in dummy to lead to finesse through him. Or he could just duck the J. You'd only be able to pick this up if the rest of the

hand allowed for a coup.

 

So loss is 6.78%, net 9.61% gain for leading the J.

 

My first post was already comprehensive on the topic; you should check your work more carefully.