Today's suit combo

[whereagles]

[hv=n=saj9xx&w=s&e=s&s=sxxxx]399|300|Play for 4 tricks.[/hv]

My instinct is low to the 9.

- If that loses to the 10, cash ace next.

- If it loses to an honor, finesse jack next.

 

Now that we're in it, here's another:

[hv=n=saj9xx&w=s&e=s&s=sxxxx]399|300|Play for 4 tricks.[/hv]

Should you finesse an honor 1st round? Or cash the ace, see what happens and then lead towards the queen? Remember you only lead towards the hand once.

[Gerben42]

First one: When you lead a low one and the missing small card is played, putting in the 9 is best.

 

Second one: With only one entry to dummy I would cash the Ace and then lead the 10.

[MickyB]

Second one: Cash the ace then run the ten. Cashing the ace first allows you to pick up Kx onside, running the ten (as opposed to small to the queen) gives you the extra chance of KJxx onside.

 

First one: Hmm. I don't like it when I disagree with Suitplay :P

[MickyB]

Ok, I don't feel so bad now - the first one is a very nice combination. More hidden below.

 

 

If you knew LHO would play x from Tx, it would be right to play the ace. That gains against KQ doubleton offside (6.8%) and loses to KQTx onside (4.8%).

 

So, what should you do when LHO plays the ten? The jack gains against KQT (or KQTx, if he has that), but what if he has played the ten from Tx, KTx or QTx? Having finessed, you will be left with a guess on the next round, so in your attempt to pick up KQ doubleton you could be encouraged to lose to not just KQTx, but also KTx and QTx. If LHO *always* played the ten from Tx, KTx and QTx but not KQTx, then it would be right to put the ace on the ten, because KQ doubleton is slightly more frequent than KQT. But as long as LHO mixes it up a bit, declarer is best off giving up on KQ doubleton offside and catering to all the other combinations that can be picked up.

 

[Free]

1st one seems right to play small to the 9, but A followed by small to the J might also be a good one.

 

2nd is easy: you play for MAXIMUM tricks, so cashing the Ace is no solution. Maximum tricks = 5, so run the 8

[foo]

Since we can't bring Suitplay or any other aid to the table (and you should not be using such things when playing online either), let's talk about how to work these out ATT

 

AJ9xx

xxxx Play for 4 tricks.

 

There are 4 cards out, which means there are 2^4= 16 possible layouts to deal with:

KQTx:- 1 x -> AJ9, cover, repeat theme

-:KQTx 1 (this one we can't do anything about, so we ignore it)

HTx:H 2 play A and then play x -> J9 OR x -> AJ9, cover, repeat theme

H:HTx 2 (another one we can't do anything about)

HT:Hx 2 x -> AJ9, play J, play A

Hx:HT 2 x -> AJ9, play J or 9, play A

KQ:Tx 1 x -> AJ9, cover, play anything to crash H+T

Tx:KQ 1 x -> AJ9, cover, play A

 

What becomes clear is that the line of play that covers the most layouts is

x -> AJ9, cover, repeat theme

 

 

T8x

AQ9xx Play for max tricks, only 1 entry in dummy.

 

32 possible layouts:

KJxxx:- 1

-:KJxxx 1

KJxx:x 3

x:KJxx 3

Kxxx:J 1

J:Kxxx 1

Jxxx:K 1

K:Jxxx 1

KJx:xx 3

xx:KJx 3

Kxx:Jx 3

Jx:Kxx 3

Jxx:Kx 3

Kx:Jxx 3

KJ:xxx 1

xxx:KJ 1

 

Using the method shown for the 1st problem, find the line of play that leads to Max tricks for as many of these 32 layouts as possible assuming you can only get to T8x once to take any finesses.

 

Answer posted later if necessary. :P