How do you play?

[Little Kid]

[hv=d=n&v=n&n=s3hkq52dat875ca52&s=saq8h643dkj9ckq93]133|200|Scoring: MP

1♦-(Pass)-3NT

 

Lead: ♠2 (3rd, 5th)[/hv]

East rises with the ♠K and now it is up to you.

[JLOGIC]

Win the spade, cross to the club ace, run the DT.

 

On top of RHO being more likely to have the DQ since LHO has 5 spades, this might give LHO a problem of what to lead back if he doesn't have the JT of spades.

[gnasher]

On top of RHO being more likely to have the DQ since LHO has 5 spades

I'm not very good at this stuff, but I don't think that's a valid inference (or at least it's a much weaker inference than you suggest).

 

LHO would probably always lead his longest suit. Therefore all that we know from the lead is that LHO's longest suit is of exactly five cards. The average length of the longest suit in any hand is about 4.9, so the lead doesn't tell us anything significant about the diamond distribution.

[Bbradley62]

Barry Crane Rule... ♠A, ♦K, ♦J

[3for3]

I agree that the spade length inference is of little value on this hand. However, you definitely don't want another spade played, and your RHO is close to 100% to continue spades, your LHO is not

[JLOGIC]

On top of RHO being more likely to have the DQ since LHO has 5 spades

I'm not very good at this stuff, but I don't think that's a valid inference (or at least it's a much weaker inference than you suggest).

 

LHO would probably always lead his longest suit. Therefore all that we know from the lead is that LHO's longest suit is of exactly five cards. The average length of the longest suit in any hand is about 4.9, so the lead doesn't tell us anything significant about the diamond distribution.

I have had this exact conversation with Eugene, pretty sure I'm right.

 

Of course it's not a strong inference, but it is definitely valid, it is rule of empty spaces. Knowing that someone has a 5 card suit (and the other guy has only 4) makes it more likely they have short diamonds.

 

All that leading a 5 card suit means is that he probably doesn't have another 6 card suit. That doesn't mean much at all, but it does mean that rule of empty spaces is still our strongest clue. If he had led a 5 card suit and his partner had 5 of the same suit, it would be marginally better to play him for the length because RHO can be 6-5 and LHO cannot (and if LHO was 5-5 he might have led the other 5 card suit), but with 5-4 it's definitely better to play LHO for the shortness.

 

I know gnasher knows this but for others who might not understand this, if LHO led from a 4 card suit and RHO had 3 for example, it is still right to play LHO for the side queen despite rule of empty spaces. This is because LHO would probably lead from a 5 card suit over a 4 card suit, and that is the stronger clue. However there is not much information to be gained about the rest of LHOs hand when he leads from a 5 card suit.

 

There was once an article written about this that said that knowing LHO has a 5 card suit does not mean anything (basically) since they usually have a 5 card suit. That article was wrong.

 

All this said 5 to 4 is not a strong clue at all, I didn't mean to imply it was, but we're talking about a 2 way guess for the queen and I think that it's strong enough to outweigh the normal advantage of hooking the other way (we can pick up stiff Q).

[W Kovacs]

Am I the only one who doesn't see a problem here? Even if the ♦ finesse fails, you have 9 tricks for the taking (2♠, 3♣, 4♦). A successful finesse gives you 11 (2♠, 3♣, 5♦, 1♥).

 

I rise with the A♠, low club to the A♣, and 10♦. If the finesse works, the low ♦, K♦ and drive out the A♥.

 

If the finesse fails, you have to give up on a heart trick, unless the defense helps you out.

[Pict]

I'd assume LHO has spades by default - I can't find out. Why not cash 2/3 clubs seeing what happens and finesse RHO for diamonds.

 

Edited L and R.

[JLOGIC]

It would be nice if we could cash the CK and CA and then run the diamond ten if LHO didn't show out.

 

That might give us some entry problems later on but I think it might be worth it. If LHO has a stiff club we can hook diamonds the right way. Another downside is it might make it more obvious to LHO to continue spades if he was going to have a problem.

[JLOGIC]

Am I the only one who doesn't see a problem here? Even if the ♦ finesse fails, you have 9 tricks for the taking (2♠, 3♣, 4♦). A successful finesse gives you 11 (2♠, 3♣, 5♦, 1♥).

 

I rise with the A♠, low club to the A♣, and 10♦. If the finesse works, the low ♦, K♦ and drive out the A♥.

 

If the finesse fails, you have to give up on a heart trick, unless the defense helps you out.

In bridge, taking more than the minimum amount of tricks earns you a higher score for the board. In certain forms of scoring, this is a significant difference!

[JLOGIC]

I'd assume LHO has spades by default - I can't find out. Why not cash 2/3 clubs seeing what happens and finesse RHO for diamonds.

 

Edited L and R.

Cashing 3 clubs ending in dummy is not really possible, we have too many entry problems that way and might never score our spade queen (eg club club club DT, diamond to jack, DK, heart to king holds and they take the last 3).

[Pict]

Poorly expressed, two clubs ending in North and review your options with diamond finesse if nothing particular in clubs- why not?

[W Kovacs]

Am I the only one who doesn't see a problem here?  Even if the ♦ finesse fails, you have 9 tricks for the taking (2♠, 3♣, 4♦).  A successful finesse gives you 11 (2♠, 3♣, 5♦, 1♥). 

 

I rise with the A♠, low club to the A♣, and 10♦.  If the finesse works,  the low ♦, K♦ and drive out the A♥. 

 

If the finesse fails, you have to give up on a heart trick, unless the defense helps you out.

In bridge, taking more than the minimum amount of tricks earns you a higher score for the board. In certain forms of scoring, this is a significant difference!

So it's not an issue of making the contract, but of trying for overtricks. I'll buy that. But you don't have enough of any suit to try and divine the opps distribution. RHO is more likely to hold the missing ♦ length, so he has a greater chance of holding the Q♦. Therefore I finesse him for it.

 

If ♦ split 4-1, I need two finesses to take 5 diamond tricks. If they break 5-0, then I have a diamond loser no matter how I play it.

 

For the record, I knew that overtricks are golden. But these problems are usually set up to be difficult to make the contract; hence my confusion.

[Trumpace]

Am I the only one who doesn't see a problem here? 

 

<snip>

 

For the record, I knew that overtricks are golden.  But these problems are usually set up to be difficult to make the contract; hence my confusion.

Hi,

 

I see that you just joined recently. Welcome :-)

 

When people present a hand in this forum, they have the option of specifying the scoring.

 

In this case the scoring was specified as MP (to the left of the hand), which is MatchPoints, so overtricks are important.

[gnasher]

Instead of running ♦10, I think you should play a diamond to the jack. That's what I'd do if I had ♦KJ doubleton, whereas with K9 doubleton and not many entries to dummy I'd probably play diamonds from the top. Hence if LHO has ♦Qx he may not realise that the diamonds are running.

 

This won't stop me picking up Qxxx on the right - I can cross to a club, take another finesse, then set up a heart entry. If it turns out that LHO ducked with Qxx, I had no chance of winning this event anyway.

[gnasher]

There was once an article written about this that said that knowing LHO has a 5 card suit does not mean anything (basically) since they usually have a 5 card suit. That article was wrong.

OK, I'm convinced, though I had to use a simulation (below) to prove it to myself.

 

Typical results for LHO's [edited - previously said "East"] diamond length were:

 

0 1779

1 16517

2 37711

3 30519

4 8968

5 4506

 

Now I just need to understand what was wrong with my original argument.

 

[Edit: throughout, assume "east" means "west".]

 

source format/none

# Counters
set d0 0
set d1 0
set d2 0
set d3 0
set d4 0
set d5 0

north is 3 KQ52 AT875 A52
south is AQ8 643 KJ9 KQ93

main {
 if {[longest east] != 5} {
     reject
   }

 set d [diamonds east]

 if {$d == 0} {
   incr d0
   accept
   }

 if {$d == 1} {
   incr d1
   accept
   }

 if {$d == 2} {
   incr d2
   accept
   }

 if {$d == 3} {
   incr d3
   accept
   }

 if {$d == 4} {
   incr d4
   accept
   }

 if {$d == 5} {
   incr d5
   accept
   }
}

deal_finished {
 puts " 0                $d0"
 puts " 1                $d1"
 puts " 2                $d2"
 puts " 3                $d3"
 puts " 4                $d4"
 puts " 5                $d5"
}

proc longest {hand} {
 set longest [clubs $hand]
 set d [diamonds $hand]
 set h [hearts $hand]
 set s [spades $hand]
 
 if {$longest < $d} {
   set longest $d
   }

 if {$longest < $h} {
   set longest $h
   }

 if {$longest < $s} {
   set longest $s
   }

 return $longest
 }

Edited August 19, 2010 by gnasher

[655321]

OK, I'm convinced, though I had to use a simulation (below) to prove it to myself.

 

Typical results for East's diamond length were:

 

0                1779

1                16517

2                37711

3                30519

4                8968

5                4506

 

I don't understand these results.

 

Firstly, in fact these results do give East the long diamonds only 44% of the time (West 56%).

 

Secondly, diamonds 5-0 happened 1779 times compared to 0-5 4506 times (i.e. in 5-0 breaks, East had the long diamonds about 2.5 times as often as West), yet

diamonds 4-1 happened 16,517 times compared to 1-4 8968 times (in 4-1 breaks, East had the long diamonds about 0.5 times as often as West).

 

if {[longest east] != 5} {

     reject

   }

Is there a confusion between West (5 spades, no 6 card suit) and East (4 spades, can have longer suit)?

 

Also (I am not a mathematician, so I could be wrong, but) shouldn't the simulation only count hands where West does have exactly 5 spades - it looks to me that all hands where East (West?) has no 6 card or longer suit are counted.

Edited August 19, 2010 by 655321

[cherdanno]

Andy, I don't think your simulation is useful. We didn't only learn that West has a 5-card suit, we also learned that one of his 5-card suits is spades. Since we don't have the same number of combined total cards in spades as in other suits, there is no symmetry between learning West having 5 hearts and West having 5 spades.

 

I would say the way to think about this is restricted choice. We learnt that spades split 5-4. But we also learned that West does not have a 6-card side suit, and that he is less likely to have a 5-card heart or club side suit (as he might lead the other). But since 55 and 56 hands are so rare, the "also" part isn't actually that important.

[MarkDean]

The lead of the 2 playing 3rd/5th inference definitely depends on many factors. For example, if you and pd have a combined 13 clubs, then the fact that LHO has 5 spades, and presumably at most 5 hearts means he must have at least 3 diamonds. The more balanced the combined holdings in both hands, the more the 5 card suit implies less length in other suits. Another thing that must be taken into account is that if LHO is 55 in the majors, he will choose hearts some of the time: it is a different kind of restricted choice situation, so any simulations or calculations have to make approximations for that as well.

 

About once a year I get a hand like this, spend about 15 minutes at the table trying to do the math, invariably guess wrong, and then spend an hour later doing the math in Excel to see if I was right in my odds. Yes, I am that nerdy.

[gnasher]

Is there a confusion between West (5 spades, no 6 card suit) and East (4 spades, can have longer suit)?

Yes, wherever I said East I meant West. I'll edit the original post, to avoid further confusion.

 

Also (I am not a mathematician, so I could be wrong, but) shouldn't the simulation only count hands where West does have exactly 5 spades - it looks to me that all hands where East (West?) has no 6 card or longer suit are counted.

If we did that, we would obviously find that RHO's average diamond length is greater than LHO's. My argument was that doing that is equivalent to thinking that Monty Hall is offering you a 50-50 guess.

[shyams]

I had read this article a few weeks ago. Found it interesting -- but I think the experts and math buffs will derive much more value out of this than I did.

 

http://sites.google.com/site/psmartinsite/...monty-hall-trap

[gnasher]

Actually, comparing this hand with the Monty Hall trap makes it clear why my argument was wrong. Monty Hall had a free choice of doors to open, but on this hand the suit that LHO tells us about will nearly always be spades.

 

If Monty Hall always had to open the leftmost door, we'd have a 50-50 guess. If he nearly always had to open the leftmost door, we'd get only a slight advantage from switching.