[hotShot]

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

[Cascade]

jdonn, on Dec 7 2008, 05:56 AM, said:

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]

benlessard, on Dec 7 2008, 04:27 AM, said:

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:

Quote

predeal north SJ432, HAT9432, CAJ2

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

Quote

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.

Quote

condition twonotrump

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

Quote

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.

Quote

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.

Quote

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.

Quote

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:

Quote

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]

benlessard, on Dec 7 2008, 04:27 AM, said:

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.

Quote

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.

[jdonn]

Cascade, on Dec 6 2008, 01:53 PM, said:

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.

Could you please explain why?

Edit: Having thought about it more...

Since 4/13 hearts are honors, and we have 1 honor and 5 small hearts, any heart partner has is (4-1)/(13-6) = 42% to be an honor, whereas any diamond he has is 4/13 = 31% to be an honor. So it seems your claim is true as long as on average at least (4/13)/(4/13+3/7) = 42% of partner's red cards are hearts (not the exact same as the above 42%, just a coincidence.)

If we apply that logic, on a random hand (opposite the hand we hold) we are missing 7 hearts and 13 diamonds, so 7/(7+13) = 35% of his red cards on average are hearts and partner would not have proportionally more fitting than nonfitting honors. However when we constrain partner to be balanced to match your claim, his average heart length certainly increases relative to his average diamond length.*** So I guess what you are saying could very well be true. Interesting, I would never have thought that.

It doesn't change my opinion of what to bid on the actual hand (our hand is shown, partner has denied heart support and can judge for himself if he has fitting heart honors, he has shown a potentially wasted diamond honor, he can bid over 4♥ if he wants, etc.) However, thanks for pointing out that interesting principle.

*** Essentially neither suit can be very long or short any more, but it was already extremely unlikely that he was long in hearts or short in diamonds, so we are just eliminating hands where he was very short in hearts or long in diamonds. Therefore, his heart length increases relative to his diamond length.

[cherdano]

jdonn, on Dec 6 2008, 04:59 PM, said:

Cascade, on Dec 6 2008, 01:53 PM, said:

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.

Could you please explain why?

I think the principle is actually simpler than that:
If partner has a balanced hand, he is more likely to have a fit for us. I think we are all intuitively familiar with that...
(Now, if our long suit has missing many honors, then he is also more likely to have fitting honors, but that is really a side effect.)

[jdonn]

cherdano, on Dec 6 2008, 07:38 PM, said:

jdonn, on Dec 6 2008, 04:59 PM, said:

Cascade, on Dec 6 2008, 01:53 PM, said:

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.

Could you please explain why?

I think the principle is actually simpler than that:
If partner has a balanced hand, he is more likely to have a fit for us. I think we are all intuitively familiar with that...
(Now, if our long suit has missing many honors, then he is also more likely to have fitting honors, but that is really a side effect.)

It looks to me like you are making a different claim than the one stated. You are saying if partner is balanced he is more likely to have fitting honors for us than if he is not balanced. This is not the same as saying if partner is balanced he is more likely to have fitting honors for us than nonfitting honors.

[cherdano]

jdonn, on Dec 6 2008, 08:11 PM, said:

It looks to me like you are making a different claim than the one stated. You are saying if partner is balanced he is more likely to have fitting honors for us than if he is not balanced. This is not the same as saying if partner is balanced he is more likely to have fitting honors for us than nonfitting honors.

As it happens, the two claims are the same after all. If you have neither ♥Q nor ♦Q, than an average hand for partner will contain either honor exactly 33.333333% of the time. (If you specify a hcp range, he will still have either honor equally often.)

[jdonn]

cherdano, on Dec 7 2008, 12:35 AM, said:

jdonn, on Dec 6 2008, 08:11 PM, said:

It looks to me like you are making a different claim than the one stated. You are saying if partner is balanced he is more likely to have fitting honors for us than if he is not balanced. This is not the same as saying if partner is balanced he is more likely to have fitting honors for us than nonfitting honors.

As it happens, the two claims are the same after all. If you have neither ♥Q nor ♦Q, than an average hand for partner will contain either honor exactly 33.333333% of the time. (If you specify a hcp range, he will still have either honor equally often.)

They still aren't the same. What if making partner balanced would mean Q♥ = 40%, Q♦ = 50%?

[Fluffy]

I didn't follow up everything, but just wanna point out that the 3NT bid means that most of the strenght, lenght and spots are in the minors, not in the majors.

Our chances for slam are that we can handle enough ♠ discards from the minors I think, All of a sudden, our 6430 now looks the same as a 6322 (even worse because void in diamonds won't help us develop that suit). Really it is up to partner to make next move.

[benlessard]

Having a balanced hands tend to bonify the short suits. So if my hand is

4603
and no restrictions are put.
partner chance of having the 2 of clubs VS the 2 of D are the same & every cards that i dont hold partner will have the same prob of getting them.

But if hes got a balanced hand. Then all the hands where he hold with a stiff H/S or with 6D/6C are discarded. These will be much more frequent than the opposites hands that are also discarded (stiff in a m or 6M suits.) This created an inflation on the numbers of cards partner hold in my long suits. But in no way this increased the likeliness that each of partner H card will be the Q vs the ratio of D for being a Q, it just mean that his average H lenght is longer than expected because he is biased by having a balanced hand.

But these are completly irrelevant here and in most bridge hands.

edited ive understand that youve put no restriction for the bidding. I dont really understand why you did that but other than that your sims are surely good.


Here partner has exactly 2H and cannot have 4S. So partner chances of holding the HQ and not holding the Q of D vs holding the DQ and not the HQ are directly proportionnal to the average expected lenght that we give for the diamonds if we use a conservative lenght of 4.5 diamonds

2♥/4.5 ♦

Partner will hold the Q of ♥ without the Q of ♦ 3 times and will hold the Q of ♦ without the ♥Q 7 times. So partner is at least twice as likely to hold a D honor without the counterpart in H then holding the H honnor without the equal honnor in Diamonds.

[655321]

benlessard, on Dec 7 2008, 05:28 AM, said:

Here partner has exactly 2H and cannot have 4S. So partner chances of holding the HQ and not holding the Q of D vs holding the DQ and not the HQ are directly proportionnal to the average expected lenght that we give for the diamonds if we use a conservative lenght of 4.5 diamonds

2♥/4.5 ♦   

Partner will hold the Q of ♥ without the Q of ♦ 3 times and will hold the Q of ♦ without the ♥Q 7 times.  So partner is at least twice as likely to hold a D honor without the counterpart in H then holding the H honnor without the equal honnor in Diamonds.

Ah, a graduate of the whereagles School of Mathematics.

We are talking about responder's original 4603 hand, right?

On your numbers, giving Opener 2 hearts and 4.5 diamonds:

• Opener has 2 of the missing 7 hearts, so will have the ♥Q 2/7 times (28.57%)
  
• Opener will have the ♦Q 4.5/13 = 34.62% of the time.
  
• Opener will have only the Heart Queen 18.68% of the time.
  
• Opener will have only the Diamond Queen 24.72% of the time.

Probably there are real mathematicians who know how to adjust these numbers to allow for Opener's 20-22 HCP, I don't know how to do that.

[benlessard]

Yep my bad.

My number arent good because we have to take into account that some H cards are gone while none of the D cards are gone.

So with 2H and 4.5 diamonds.

1/7 x 2 vs 1/13 x 4.5 diamonds. Will give the right proportion. As the points go up the proportions should stay the same unless you are very near to the limit.


But my main point was that if i hold.

JT98765432
2
2
2

Without restrictions every honnor has equal prob.

Now if parnter got a balanced hand its obvious that hes going to have a S honnor more often than any other honnors. But its pertinent to a much lower degree to the cases where getting card is proportionnal to the expected length.

[whereagles]

655321, on Dec 7 2008, 11:20 AM, said:

Probably there are real mathematicians who know how to adjust these numbers to allow for Opener's 20-22 HCP, I don't know how to do that.

Best is probably run a simulation and get empirical odds.