Fitting Honours

[Cascade]

(jdonn)

(Cascade @ Dec 6 2008, 01:53 PM)

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...

 

...

 

So I guess what you are saying could very well be true. Interesting, I would never have thought that.

 

As I said I have never seen this principle expounded but it appears to be true in an simulations I have done.

 

I had been thinking about this for some time when the following dramatic hand came up in a New Zealand Trial:

 

[hv=d=n&v=n&s=shqjt98xxxdcqjt98]133|100|Scoring: IMP

I am not actually 100% sure about the vulnerability and dealer except that this hand did not pass and partner bid first.[/hv]

 

At our table I opened a Mexican 2♦ for us an 18-bad 20 balanced or nearly balanced - note the nearly balanced hands are more likely to have fitting honours in any doubletons (or singletons) otherwise they might be more suited to a suit opening.

 

My partner realizing that she was not at all likely to be able to determine enough useful information to make an informed decision simply took a bash at 6♥ which rolled in when I contributed ♥AK and ♣A.

 

Subsequently I did some simulations on this hand which came out with numbers similar to those below (I have just repeated the simulation):

 

Pr(♠A) = 54%

Pr(♠K) = 44%

Pr(♥A) = 77%

Pr(♥K) = 66%

Pr(♦A) = 54%

Pr(♦K) = 44%

Pr(♣A) = 67%

Pr(♣K) = 56%

 

E(♠ Controls) = 1.53067

E(♥ Controls) = 2.21021

E(♦ Controls) = 1.53273

E(♣ Controls) = 1.90612

 

These are from 1000000 hand simulations (actually based on an 18-19 hcp balanced hand opposite).

 

My intuition at the time told me that partner was more likely to have (high) honours in my short suits but the simulations were telling me very strongly that that intuition was wrong.

 

The simulations strongly suggested that optimism was warranted when partner showed a balanced hands. A corrollory of this is that pessimism is warranted with regard to NTs when you have some distribution (a singleton or void) opposite partner's balanced hand. Partner is much more likely to have a moderate holding in our short suit than a several solid stoppers.

 

As I thought about this some more and did some calculations it became clear that the simulations were right.

 

Opposite the hand in this thread if partner is just known to be balanced then for example:

 

With two hearts:

 

P(♥A) = one ♥A and one other card out of four all out of a total of five choose 2

 

= 4/10 = 2/5

 

Actually this reasoning is too complex - partner has two cards out of five each of which is equally likely to be the ace so the probability of the ♥ A is simply 2/5

 

With three hearts it is 3/5

 

With four hearts it is 4/5

 

With five hearts it is 5/5 = 1 - duh

 

In spades with:

 

Two spades 2/13

Three spades 3/13

Four spades 4/13

Five spades 5/13

 

These numbers are affected by the knowledge that partner has more high-card points. With hands above average in strength there is an obvious tendancy to have a bigger chance of holding the honours. This tendancy is even greater for controls - while a ten count has an average of 3 controls (A=2 K=1) a twenty count has an average of around 7 controls (rule of thumb calculation not the exact number).

 

This however just increases all of these probabilities. It would be relatively difficult to calculate the exact probabilities for a given hcp total or range but simulations suggest that these principles will continue so that partner will have a greater probability of having useful fitting honours than similar honours in our short suits.

[benlessard]

Its not hard to understand.

 

If you have 10 S and patrner show a balanced hands his number of S (2 or 3) is above his normal expectancy of spades (1).

 

If my hand is

 

4441

 

in average he should have a 3334 if he hold a 3334 all the missing cards in the deck have the same probablilty.

 

But if he showed a 5??5.

 

so hes 60% more likely to have a specific S card and 25% more likely to have a specific club card.

[benlessard]

In your case.

 

you have a

 

0805 partner average is

 

4.333 (13 divided by 3 players)

1.666 (5/3)

4.333

2.6666 (8/3)

 

after discarding all hands with a stiff a void or a 6 cards suits. His average hand will be something like

 

3.9

2.2

3.9

3

 

So hes 30% something more favorite to hold any missing H than usual and 15% more likely to have a specific C card.

[Fluffy]

The way I see it is that the fact you hold so much low (non honnor) heart and club cards means that partner's chances of having low cards there is low, so he must have high cards.

 

But if you have AQJ10xx, then chances of fitting won't be as good (althou obviously you only need 1 fitting card here).

 

About calculating odds for certain ranges, it doesn't work very well, I tried with a mathematician friend once, and we concluded that exact percentages would depend upon exactly wich honnor cards were missing, not something you can calculate at a table.

[Cascade]

But if you have AQJ10xx, then chances of fitting won't be as good (althou obviously you only need 1 fitting card here).

Even for this hand with say 3=2=2 distribution outside so:

 

AQJ10xx

xxx

xx

xx

 

opposite a 15-17 1NT

 

The probabilities were

 

♠K 60%

♥K 47%

♦K 46%

♣K 46%

 

opposite a 20-22 1NT

 

The probabilities were

 

♠K 77%

♥K 62%

♦K 60%

♣K 60%

[Fluffy]

Nice indeed. Thanks wayne.

 

this also helps you when you are overcalling against balanced hands, it is likelly the NT hand has your fitting honnors, not your partner.