4603 facing a 2nt opener

[benlessard]

[hv=d=n&v=b&s=sjxxxhat9xxxdcajx]133|100|Scoring: IMP[/hv]

 

2Nt-------3D

3H--------3S

3NT-------4C

4D--------???

 

 

3S was slammish at least 4 card.

4D was 1st round.

[hotShot]

4H

I think I've shown 6♥, 4♠ and the A of ♣ all that with a shaped hand.

Partners ♦ control is not very encouraging.

Next move is up to partner.

[Cascade]

Tend to be optimistic.

 

Partner is much more likely to have useful fitting cards than lots of stuff wasted in diamonds.

 

In a 1000000 hand simulation for a 20-22 2NT requiring a balanced hand here are the probabilities of the strong 2NT having various honours opposite this hand:

 

AS 87%

AD 83%

 

KH 79%

KC 75%

KS 72%

KD 66%

 

QH 61%

QC 56%

QS 52%

QD 46%

 

Edit:

 

These numbers will be lower for hearts (and spades) given that partner has not shown a fit. Nevertheless we should be more optimistic than you might imagine.

[Ant590]

I would be interested in the simulations %s for the heart and spade pips also.

[jdonn]

4H

I think I've shown 6♥, 4♠ and the A of ♣ all that with a shaped hand.

Partners ♦ control is not very encouraging.

Next move is up to partner.

I agree with everything you said.

 

The sim is interesting, but I don't think all that helpful. Aside that partner didn't show a fit in either major and cuebid diamonds, it's the odds of all the various combinations of fitting cards that matters, taken in conjunction with partner's possible shapes, not the individual odds of any particular card. Also he is certainly entitled to move over 4♥ by us.

[maggieb]

4♥, my trump suit is too big of a liability to move again if partner can't do it.

[Cascade]

I would be interested in the simulations %s for the heart and spade pips also.

I am not sure exactly what you mean since we have the heart 10 and 9.

 

♠ 10 - 20%

♠ 9 - 20% - all spots should be the same in each suit since they don't count towards the hcp total

 

♥ J - 43%

♥ 8 - 25%

[benlessard]

Qxx

Kx

AKQJT

KQx

 

Even 4 is in danger. AK of S lead. club switch. K of H and now you have to pitch 2S even if they are winners before playing H to the T.

 

Over 4C would you bid 4D or since you have like the worst hand possible bid 4H ?

 

We werent bidding this hand but i find it painful, if you bid 4D its forward going but if you dont partner will just be more excited because hes thinking you dont have the D ace. I think its not that easy to stop in 4H.

[gwnn]

I would be interested in the simulations %s for the heart and spade pips also.

I am not sure exactly what you mean since we have the heart 10 and 9.

 

♠ 10 - 20%

♠ 9 - 20% - all spots should be the same in each suit since they don't count towards the hcp total

 

♥ J - 43%

♥ 8 - 25%

how is it 20%? isnt it supposed to be 33% or very close to 33%?

[mike777]

Dealer: North Vul: Both Scoring: IMP

♠ Jxxx ♥ AT9xxx ♦ [space] ♣ AJx

 

2Nt-------3D

3H--------3S

3NT-------4C

4D--------???

 

2nt=3d

3h(no super accept)=4c

?

 

 

3S was slammish at least 4 card.

4D was 1st round.

too hard:

 

 

prefer 3d and then 4c

[whereagles]

I have 7 losers and a 2NT opener usually has 6 cover cards, so 6 is in order.

 

However, this auction is very messy. I have no clue what's happening, except that it seems that the hands don't fit well. I'll just play safe and bid 4♥ now. If pard can make a move over this, I'll oblige.

[benlessard]

how is it 20%? isnt it supposed to be 33% or very close to 33%

Dont tell me you really check his post when he put tons of numbers I always skip them, just a quick look a them and i laugh.

 

I mean

 

QH 61%

QC 56%

QS 52%

QD 46%

 

How is it possible that partner is more likely to have the Q of H than the Q of D ?

 

If you have a void in D and 6H it doesnt take a genius to know that partner is more likely to be long in D and short in H. So its simply not possible to have the HQ more often than any other queens....

 

 

 

Just kidding... i believe your number are usually write but youve surely done something wrong this time.

[hotShot]

I mean

 

QH 61%

QC 56%

QS 52%

QD 46%

 

How is it possible that partner is more likely to have the Q of H than the Q of D ?

 

If you have a void in D and 6H it doesnt take a genius to know that partner is more likely to be long in D and short in H. So its simply not possible to have the HQ more often than any other queens....

The 7 missing ♥ are KQJxxxx so I estimate that holding the average 2.33 cards in ♥ give you a close to 50% chance to have one of the honors.

 

Additionally if 2NT shows 20-21 and my hand contains 10 than partner has (more) than 66.6% of the HCP.

 

If you combine these it is not unlikely that the chance to find QH with partner is as big as suggested.

[JLOL]

too hard:

 

 

prefer 3d and then 4c

LOL

[benlessard]

If you combine these it is not unlikely that the chance to find QH with partner is as big as suggested.

What im saying is that the Q of D is more likely than the Q of H and its not close.

[maggieb]

I see I already posted 4H is best, I still think so!

[Cascade]

I would be interested in the simulations %s for the heart and spade pips also.

I am not sure exactly what you mean since we have the heart 10 and 9.

 

♠ 10 - 20%

♠ 9 - 20% - all spots should be the same in each suit since they don't count towards the hcp total

 

♥ J - 43%

♥ 8 - 25%

how is it 20%? isnt it supposed to be 33% or very close to 33%?

because a 20-22 count has fewer spot cards and more pictures than an average hand.

[Cascade]

how is it 20%? isnt it supposed to be 33% or very close to 33%

Dont tell me you really check his post when he put tons of numbers I always skip them, just a quick look a them and i laugh.

 

I mean

 

QH 61%

QC 56%

QS 52%

QD 46%

 

How is it possible that partner is more likely to have the Q of H than the Q of D ?

 

If you have a void in D and 6H it doesnt take a genius to know that partner is more likely to be long in D and short in H. So its simply not possible to have the HQ more often than any other queens....

 

 

 

Just kidding... i believe your number are usually write but youve surely done something wrong this time.

So we are missing 7 hearts and 13 diamonds - lets look at some probabilities for the queens in each of these suit:

 

We assume partner has a balanced hand ...

 

If partner has two hearts:

 

P(Q) = 1 queen x 6 other cards out of 7 choose 2 possible doubletons = 2/7

 

If partner has three hearts:

 

P(Q) = 1 queen x 6 choose 2 other cards out of 7 choose 3 = 6C2 / 7C3 = 3/7

 

If partner has four hearts:

 

P(Q) = 1 queen x 6C3 / 7C4 = 4/7

 

If partner has five hearts:

 

P(Q) = 1 queen x 6C4 / 7C5 = 5/7

 

If partner has two diamonds:

 

P(DQ) = 1 queen x 12C1 / 13C2 = 2/13

 

If partner has three diamonds

 

P(DQ) = 12C2/13C3 = 3/13

 

If partner has four diamonds

 

P(DQ) = 12C3/13C4 = 4/13

 

If partner has five diamonds

 

P(DQ) = 5/13

 

Summarizing:

 

With a doubleton heart partner is more likely to have the heart queen than the diamond queen if he has two or three diamonds

 

With a tripleton or longer heart partner is more likely to have the heart queen than the diamond queen if he has five or fewer diamonds

 

These probabilities did not take into account that partner is 20-22 hcp. The calculations are more complex if you take that into account but the probabilities nevertheless favour partner having fitting honours.

[benlessard]

Youre a not using the right approach to calculate.

 

You cannot visualize a distribution and put HCP in it. The distribution is a restriction for the the hcp and the hcp is a restriction for the distribution you have to sims them together not 1 after the other.

 

Another way to see it...

 

Partner has 2H. Only time hes got 3 hearts is if slow values in the minors.

 

If hes has 2D. It means 4-5 in the blacks

Holding AJ of clubs make it higly unlikely that OP has 5 clubs because the missing HCP is a restriction on his distribution. So therfore the % of having short D drops proportionally to the hcp restriction.

 

Another example

 

Q

xxxx

xxxx

AKQJ

 

 

You know that partner got a strong NT. Because hes got 0 HCP in clubs hes more likely to be long in D/H than in clubs.

 

So when holding

 

J

Axxx

xxxx

AKxx

 

The odds of having the Q of clubs isnt really higher than having the Q of D.

 

 

Its not that easy to understand or to explain but i can assure you that im 100% sure that your numbers are wrong.

[jdonn]

Wayne, I know you admitted you weren't taking into account that partner has already denied heart support, but the problem is you also aren't taking into account our hand to determine the length partner tends to have in each suit. Your conclusion is essentially saying "he is moremuch likely to have the heart queen in any given-length heart holding than the diamond queen in the same length diamond holding because the queen is 1/7 of the missing hearts and 1/13 of the missing diamonds." But that completely ignores that partner's diamonds are going to be on average (much?) longer than his hearts since we are 6-0 in those suits.

[hotShot]

I hope that it is save to assume that partner did not open 2NT with a single or void in ♥.

[Cascade]

Wayne, I know you admitted you weren't taking into account that partner has already denied heart support, but the problem is you also aren't taking into account our hand to determine the length partner tends to have in each suit. Your conclusion is essentially saying "he is moremuch likely to have the heart queen in any given-length heart holding than the diamond queen in the same length diamond holding because the queen is 1/7 of the missing hearts and 1/13 of the missing diamonds." But that completely ignores that partner's diamonds are going to be on average (much?) longer than his hearts since we are 6-0 in those suits.

The initial simulation did exactly what you are saying.

 

I forced north to have the actual hand and then constrained south to have any 20-22 balanced and only then counted up how many times south had the various missing honour cards.

 

It is a little known principle - I have never seen it discussed :

 

If you know that partner has a "balanced" hand then partner is more likely to have fitting honours than misfitting honours - well a particular fitting honour than a particular non-fitting honour.

 

This can be shown when you do not assume anything about partner's strength.

 

When you add in that partner has some known (usually strong) hand then the affect is even greater.

[Cascade]

Youre a not using the right approach to calculate.

 

You cannot visualize a distribution and put HCP in it. The distribution is a restriction for the the hcp and the hcp is a restriction for the distribution you have to sims them together not 1 after the other.

I did that and you do not believe the numbers.

[Cascade]

The code is very simple in dealer:

 

predeal north SJ432, HAT9432, CAJ2

 

I predealt the exact hand (replacing x's with small spots)

 

twonotrump =

 

hcp(south)>=20 and

hcp(south)<=22 and

shape(south,any 4333 + any 4432 + any 5332)

 

I constrained the hand opposite to be in the range 20-22 hcp and required that it was a traditionally balanced distribution.

 

condition twonotrump

 

I forced the simulator to only use hands that met the above constraint.

 

SsA = hascard(south,AS)?1:0

 

I created a 0-1 variable that tood the value one only when the strong 2NT hand had the ♠A.

 

SsK = hascard(south,KS)?1:0

SsQ = hascard(south,QS)?1:0

HhK = hascard(south,KH)?1:0

HhQ = hascard(south,QH)?1:0

DdA = hascard(south,AD)?1:0

DdK = hascard(south,KD)?1:0

DdQ = hascard(south,QD)?1:0

CcK = hascard(south,KC)?1:0

CcQ = hascard(south,QC)?1:0

 

SsT = hascard(south,TS)?1:0

Ss9 = hascard(south,9S)?1:0

 

HhJ = hascard(south,JH)?1:0

Hh8 = hascard(south,8H)?1:0

 

I created similar variables for all other significant honours and some spot cards.

 

generate

10000000000

 

produce 1000000

 

I restricted the simulation to some gaziillion trials (10^10) this meant the simulation would terminate after 10^10 trials even if it did not find the million (10^6) examples I was looking for.

 

action

 

frequency "AS" (SsA,0,1),

frequency "KS" (SsK,0,1),

frequency "QS" (SsQ,0,1),

frequency "KH" (HhK,0,1),

frequency "QH" (HhQ,0,1),

frequency "AD" (DdA,0,1),

frequency "KD" (DdK,0,1),

frequency "QD" (DdQ,0,1),

frequency "KC" (CcK,0,1),

frequency "QC" (CcQ,0,1),

frequency "TS" (SsT,0,1),

frequency "9S" (Ss9,0,1),

frequency "8H" (Hh8,0,1),

frequency "JH" (HhJ,0,1),

 

Finally i printed out some statistics on the variables I had created.

 

Here are the raw results from such a run:

 

Frequency AS:

    0    135133

    1    864867

Frequency KS:

    0    282978

    1    717022

Frequency QS:

    0    476215

    1    523785

Frequency KH:

    0    209175

    1    790825

Frequency QH:

    0    387077

    1    612923

Frequency AD:

    0    170849

    1    829151

Frequency KD:

    0    343989

    1    656011

Frequency QD:

    0    537129

    1    462871

Frequency KC:

    0    246988

    1    753012

Frequency QC:

    0    438054

    1    561946

Frequency TS:

    0    802461

    1    197539

Frequency 9S:

    0    802664

    1    197336

Frequency 8H:

    0    745779

    1    254221

Frequency JH:

    0    570543

    1    429457

6.50915

Generated 308757459 han

Produced 1000000 hands

Initial random seed 122

Time needed 358.00 sec

 

You can believe the results or not - your choice.

 

But I am convinced that by these numbers for example you are about a 4:3 favourite to have the ♥Q over the ♦Q.

[Cascade]

So when holding

 

J

Axxx

xxxx

AKxx

 

The odds of having the Q of clubs isnt really higher than having the Q of D.

This is a different problem since you have equal length in clubs and diamonds.

 

Its not that easy to understand or to explain but i can assure you that im 100% sure that your numbers are wrong.

 

I am pretty confident they are right. I hope the previous post convinces you.