Jump to content

Suit combination with constraints


cherdano

Recommended Posts

It looks to me like the player on vuegraph came closer to getting this right than the commentators:

[hv=n=stxxx&w=s&e=s&s=saq87]399|300|[/hv]

You are in dummy, and have two more entries to get there. What's the right line for 3 tricks?

Now, assume you know RHO is more likely to have length, because you know one suit split 2-2, and one suit split 5-3 or 6-2 (you don't know yet). What's best now?

 

Arend

Link to comment
Share on other sites

It looks to me like the player on vuegraph came closer to getting this right than the commentators:

[hv=n=stxxx&w=s&e=s&s=saq87]399|300|[/hv]

You are in dummy, and have two more entries to get there. What's the right line for 3 tricks?

Now, assume you know RHO is more likely to have length, because you know one suit split 2-2, and one suit split 5-3 or 6-2 (you don't know yet). What's best now?

 

Arend

I am sorry, but it actually matters if this was a suit contract or a NT contract, and how you know one suit was 5-3 or 6-2.

 

If its a NT contract, and LHO led from his 5 card suit and thats how you find out that suit is 5-3, and another suit is 2-2, then this suit you care about is much less likely to be 1-4 than you expect, since that would give the opening leader a second 5 card suit, which he might have led instead, so by restricted choice he is more likely to have only 4 cards in the other suit than 5.

 

Phil Martin wrote a great article years ago, called "The Monty hall Trap", about the problem of treating biased information (the fact that you got this information is directly correlated with the information itself) as random information. This article can now be found in the Granovetter's collection, For Experts Only.

 

So in this case, LHO's expected length in the given suit depends somewhat on some of these other issues. But yes, LHO is expected to be slightly shorter in the given suit than RHO, but the liklihood of a singleton varies.

Link to comment
Share on other sites

Well, thanks for the lessons in probabilities. :) The suit splits 5-3 or 6-2 because it was overcalled 1, and his partner followed to two rounds of hearts. You can read nothing into the signals, as there was singleton A in dummy. Since you opened 1, his possible other 5-card suit, there is no inference from the lack of a two-suited ovvercall. His hcp strength was unsuited for 2 in their style. He wouldn't lead a singleton, as he had Qx. Is that pure enough for you? How about thinking about the problem instead? ;)

 

(Yes I know about common misconceptions of probabilities. Playing the opening leader vs NT for shortness because he has led a 4-card suit, where his partner has 2 or 3 cards, is certainly the most popular one.)

 

Arend

Link to comment
Share on other sites

OK, given those constraints:

If this suit was 1-4 then:

Starting with the ace is right if the stiff is the K J or 9. So if you play the ace and if:

a. The K or J comes down, you lead toward dummy's T (picks up all the 4-1's plus KJ tight)

b. the 9 comes down then go to dummy and run the T, this loses to J9 doubleton and picks up singleton 9 or K9 doubleton

c. if a low card comes down , you need the suit to be 3-2

case c1: RHO also plays low. Go to dummy and lead a low card and play the Q if RHO plays low. This picks up 9x or Jx offside but loses to Kx offside.

case c2: RHO follows with the J or 9. This case you are playing RHO for KJ9...

 

 

This line also pick up Kx or KJ or Jx onside.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...