finessing 3nt.

[matmat]

you bid to 3nt via

 

[P] - 1NT - [P] - 2♣

[X] - 2♦ - [P] - 3NT

all pass

 

you get a small club lead and manage to guess what combination of ♣ to play to not go off off the top:

 

T1: ♣5-8-T-J

 

[hv=n=skt65h6432daj4ck5&s=saj7hak8dkt865cj7]133|200|[/hv]

 

 

(sorry if i got the forum wrong).

[Jlall]

DK, DA, then go for four spade tricks is the intuitive line.

 

This picks up Qx or stiff Q of diamonds anywhere, or Q Qx/Qxx of spades on right.

 

Another option is A then king of spades followed by the DJ. This picks up stiff Q or Qx of spades somewhere or Qx, Qxx, Qxxx, and Q9xx of diamonds on right.

 

Line A picks up diamonds ~33 % of the time and spades ~28 % of the time.

 

Line B picks up diamonds ~45 % of the time and spades ~18.5 % of the time.

 

So line A is ~52 % and line B ~55 % if I did the math right here.

[655321]

Short answer, play the Ace and King of diamonds, then if the Queen doesn't drop, play for 4 tricks in spades.

 

Once we have played off the top 2 diamonds, we don't have the option of playing RHO for 9x or 8x of spades (♠A, then ♠J-Q-K-9, run ♠7) because there will be no entry to the last spade in dummy. So even though we might prefer to play LHO for the ♠Q, instead play RHO for the Queen. So ♦K, ♦A, Spade to J. Makes when the ♦Q comes down, or RHO has Qx or Qxx in spades.

 

It might be worth cashing a couple of hearts to see if we can get more information, but most likely we won't learn anything useful.

[gnasher]

I calculate that:

 

A: Top diamonds, then a spade finesse against East, works against:

♦Q in either hand: 28 x 1/5 = 6%

♦Q(x) in either hand: 68 x 2/5 = 27%

♠Q(x)(x) on the right: (48 x 1/6 + 36 x 1/2 + 15 x 1/12) x 67% = 18%

 

Justin had the second figure as 28%, but I think that's a typo - I agree with his overall total of 51%.

 

B: Top spades, then ♦J works against:

♠Q(x) in either hand: 48 x 1/3 + 15 x 1/6 = 18.5%

♦Qx(x)(x) on the right: (68 x 1/2 + 28 x 4/10) x 81.5% = 37%

 

Justin's figure of 45% for the diamonds appears to be because he counted double for the 4-1 breaks. He seems to have corrected this in arriving at the total of ~55%, with which I also agree.

 

Thus it seems to be better to cash the spades first, which seems counterintuitive.

[rhm]

I calculate that:

 

A: Top diamonds, then a spade finesse against East, works against:

♦Q in either hand: 28 x 1/5 = 6%

♦Q(x) in either hand: 68 x 2/5 = 27%

♠Q(x)(x) on the right: (48 x 1/6 + 36 x 1/2 + 15 x 1/12) x 67% = 18%

 

Justin had the second figure as 28%, but I think that's a typo - I agree with his overall total of 51%.

 

B: Top spades, then ♦J works against:

♠Q(x) in either hand: 48 x 1/3 + 15 x 1/6 = 18.5%

♦Qx(x)(x) on the right: (68 x 1/2 + 28 x 4/10) x 81.5% = 37%

 

Justin's figure of 45% for the diamonds appears to be because he counted double for the 4-1 breaks.  He seems to have corrected this in arriving at the total of ~55%, with which I also agree.

 

Thus it seems to be better to cash the spades first, which seems counterintuitive.

It does not look counterintuitive to me.

 

Your major chance is the finesse, not the drop.

But you require 4 tricks from the finesse.

 

In ♠ that is far less likely than in ♦, where only a singleton ♦ queen with East will hurt you.

 

So play for the drop in ♠ and finesse in ♦.

 

If you were in 2NT and needed only 3 tricks from the finesse it would be correct to play for the drop in ♦

Rainer Herrmann

[pooltuna]

you bid to 3nt via

 

[P] - 1NT - [P] - 2♣

[X] -  2♦  - [P] - 3NT

all pass

 

you get a small club lead and manage to guess what combination of ♣ to play to not go off off the top:

 

T1: ♣5-8-T-J

 

Dealer: ????? Vul: ???? Scoring: Unknown

♠ KT65 ♥ 6432 ♦ AJ4 ♣ K ♠ AJ7 ♥ AK8 ♦ KT865 ♣ 7

 

 

(sorry if i got the forum wrong).

Well I suspect MP play will be different than IMP play so I will assume IMPs. So your decision is, do I try for 4!S by taking the finesse or playing for Qx (or Q alone) leaving the diamonds in reserve if your play for the drop fails or if the finesse works but no 33 split. Or do I play for the drop of the Q in the ♦ suit and hold the spade finesse and 33 split in reserve. Somehow my gut tells me to go with the (to me) counter intuitive Qx♠ but I don't have any numbers to confirm this.

[655321]

My line was wrong and Justin's was right (no surprises there. :P)

 

♠Q(x)(x) on the right: (48 x 1/6 + 36 x 1/2 + 15 x 1/12) x 67% = 18%

 

Justin had the second figure as 28%, but I think that's a typo - I agree with his overall total of 51%.

♦Qx(x)(x) on the right: (68 x 1/2 + 28 x 4/10) x 81.5% = 37%

 

Justin's figure of 45% for the diamonds appears to be because he counted double for the 4-1 breaks.  He seems to have corrected this in arriving at the total of ~55%, with which I also agree.

 

I think Justin's numbers (28% and 45%) are the percentages before multiplying by the chance of the queen of the first suit not dropping.

 

Also these a priori numbers might change a little if we assume that clubs are Qxx opposite Axxxxx (possibly Qxxx opposite Axxxx).

Edited October 12, 2009 by 655321

[Jlall]

I calculate that:

 

A: Top diamonds, then a spade finesse against East, works against:

♦Q in either hand: 28 x 1/5 = 6%

♦Q(x) in either hand: 68 x 2/5 = 27%

♠Q(x)(x) on the right: (48 x 1/6 + 36 x 1/2 + 15 x 1/12) x 67% = 18%

 

Justin had the second figure as 28%, but I think that's a typo - I agree with his overall total of 51%.

 

B: Top spades, then ♦J works against:

♠Q(x) in either hand: 48 x 1/3 + 15 x 1/6 = 18.5%

♦Qx(x)(x) on the right: (68 x 1/2 + 28 x 4/10) x 81.5% = 37%

 

Justin's figure of 45% for the diamonds appears to be because he counted double for the 4-1 breaks.  He seems to have corrected this in arriving at the total of ~55%, with which I also agree.

 

Thus it seems to be better to cash the spades first, which seems counterintuitive.

hmm, I meant if diamonds are 45 % to come in normally by playing the jack off dummy, and if spades are 18.5 % to come in normally then I did:

 

45+(18.5*.55)

 

And if diamonds are 33 % to come in by playing it one way and spades are 28 % then I did:

 

33+(28*.67)

 

Sorry if I represented this badly or if this is a weird way to do it, it makes sense to me and since gnasher's numbers agree with mine minus rounding errors then I think it's an ok way to do it.

[gnasher]

Sorry if I represented this badly

No, I'm just too stupid to understand.

[gwnn]

I think this is a very interesting hand and I definitely learned something. I wonder though does Meckwell or Helgeness do these calculations and go for the .55 line in lieu of the .52 line?

[Jlall]

I think this is a very interesting hand and I definitely learned something. I wonder though does Meckwell or Helgeness do these calculations and go for the .55 line in lieu of the .52 line?

I can tell you 100 % no that they do not go through these calculations. It is very very rare in bridge when you need to know exact percentages, usually you can just view combinations against themselves.

 

So someone like Meckstroth is going to be very good at figuring out the relevant combinations, and then it's usually easy to see which is more likely without knowing any numbers.

 

This hand for instance at the table I would analyze like I did here that one line picks up Q(xx) of spades on right, or Q(x) of diamonds. One line picks up Q(x) of spades or Qx(xx) or Q9(xx) of diamonds on right. I would have no idea what the numbers are, but usually they are not that close so it's easy to tell, or they have more in common so it's easier to compare.

 

At the table I can tell you I would have gone wrong (hence when I said line A was intuitive), because my math intuition was wrong.

 

It is entirely possible that Helgemo or Meckstroth's bridge math intuition is better than mine and they would have gotten to that point and said obv line B is better, but there is no chance they would know the exact numbers and then crunch them.

 

You can go your entire bridge career without knowing any numbers other than the basic finesse = 50 %, 3-2=68 %, 3-3=36 % etc and be the best declarer in the world. Maybe you will lose 1 % on close lines here or there, but that's not what it's all about.

 

Another funny story about Meck, my dad asked him a similar hand to this once and he looked at it for maybe 2 seconds and said "I don't know ask Rosenberg." Fair enough.

[bid_em_up]

Doesn't line B also win when diamonds are 5-0 on the right?

 

(ok, so its not much, but an extra 2% is an extra 2%).

[woefuwabit]

If you consider that west is leading 4th best and therefore has minimum 4 clubs and rule of 11 giving minimum 2 clubs to east, the odds will favour line B more. This is my simulation:

 

north is "KT65 6432 AJ4 K8"
south is "AJ7 AK8 KT865 J7"

sdev lineA
sdev lineB

source format/none

main {
reject unless {[clubs west] > 3}
reject unless {[clubs east] > 1}

if { ([west has QD] && [diamonds west] < 3) || ([east has QD] && [diamonds east] < 3) || ([east has QS] && [spades east] < 4)} {lineA add 1}
if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] < 5)} {lineB add 1}
accept
}

deal_finished {
puts "Line A = [lineA count]"
puts "Line B = [lineB count]"
}

 

Results (1 million trials):

Line A = 505884

Line B = 576482

 

 

Without taking the 4th best lead into consideration, figures look about right where jlall and gnasher have calculated.

Line A = 508196

Line B = 554000

[Jlall]

Doesn't line B also win when diamonds are 5-0 on the right?

 

(ok, so its not much, but an extra 2% is an extra 2%).

Yes you are right, I miscounted the tricks and thought we only had 8 with 5-0 diamonds on right. Oddly gnasher made the same mistake!

 

Since I want to blame someone else obv, minor gripe but I prefer if the whole hand is posted and the tricks are explained unless it's an end position. I don't think of this as a 12 trick end position though I guess you could, it's more natural to me if all 13 cards are presented and we're just told how trick 1 went.

[woefuwabit]

Doesn't line B also win when diamonds are 5-0 on the right?

 

(ok, so its not much, but an extra 2% is an extra 2%).

when you play the 3rd round ♦4 from dummy, you will have T86 over east's 974, you can't pick up all 3 remaining tricks.

[bid_em_up]

Doesn't line B also win when diamonds are 5-0 on the right?

 

(ok, so its not much, but an extra 2% is an extra 2%).

when you play the 3rd round ♦4 from dummy, you will have T86 over east's 974, you can't pick up all 3 remaining tricks.

I dont need all of them. A total of 4 diamond tricks will suffice (or two of the remaining three).

 

4 diamonds, 2 spades, 2 hearts and the club from trick 1 is 9.

 

The point was that a 5-0 break wasn't included in any of the calculations above, and I thought it needed to be.

[gwnn]

If you consider that west is leading 4th best

west can have Qxx or Qxxx, east doubled for the lead.

[woefuwabit]

Good points. I reran the simulation with the 5-0 break, and without club restrictions, results as expected:

Line A = 508765

Line B = 570156

 

However, giving west 3-4 clubs and east 5-6 clubs...

Line A = 530659

Line B = 519285

 

Well, that's a significant change!

Here's my code if you want to check it:

north is "KT65 6432 AJ4 K8"
south is "AJ7 AK8 KT865 J7"

sdev lineA
sdev lineB

source format/none

main {
reject unless {[clubs west] > 2}
reject unless {[clubs east] > 4}

if { ([west has QD] && [diamonds west] < 3) || ([east has QD] && [diamonds east] < 3) || ([east has QS] && [spades east] < 4)} {lineA add 1}
if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] > 1)} {lineB add 1}
accept
}

deal_finished {
puts "Line A = [lineA count]"
puts "Line B = [lineB count]"
}

[matmat]

Without taking the 4th best lead into consideration, figures look about right where jlall and gnasher have calculated.

Line A = 508196

Line B = 554000

when RHO doubles for a club lead a 4th best lead from LHO is somewhat unlikely.

[matmat]

Since I want to blame someone else obv, minor gripe but I prefer if the whole hand is posted and the tricks are explained unless it's an end position. I don't think of this as a 12 trick end position though I guess you could, it's more natural to me if all 13 cards are presented and we're just told how trick 1 went.

I did that specifically to annoy you ;)

 

(fixed)

[bid_em_up]

if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] > 1)} {lineB add 1}

 

should be:

 

if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] > 0)} {lineB add 1}

 

 

 

The way it is coded, East must hold 2 or more diamonds, if I read it correctly, and that is not the case.

 

(Unless the language is smart enough to know that since East has to hold the diamond Q, he must hold at least 1 diamond).

 

I don't use this language so am not really sure.

[woefuwabit]

I don't think you can play diamonds for 4 tricks with a singleton Q with east if you start with the ♦J