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Frequency

#1 User is offline   Wiste1 

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Posted 2004-July-02, 15:52

Hi can anyone tell me the frequency(%) of :

8-10 hcp with 5+card any suit, not 5332 but all unbalanced hands 5+-4,
5-5, 6+ etc.





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#2 User is offline   pbleighton 

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Posted 2004-July-03, 13:34

13.7%

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#3 User is offline   Wiste1 

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Posted 2004-July-03, 15:28

Thx pbleighton :blink:







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#4 User is offline   Trpltrbl 

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Posted 2004-July-04, 12:05

My question is, how do find that out?
Is there a program for that ?

Mike ;)
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#5 User is offline   hrothgar 

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Posted 2004-July-04, 12:08

Trpltrbl, on Jul 4 2004, 09:05 PM, said:

My question is, how do find that out?
Is there a program for that ?

Mike ;)

Pretty easy to simulate with a program like Dealer or Borel.

With this said and done, HCP and shape are independant of one another, so its pretty east to multiple the percentage chance that opener has shape ABC by the chance that opener has HCP XZY...
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#6 User is offline   pbleighton 

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Posted 2004-July-04, 12:30

"With this said and done, HCP and shape are independant of one another, so its pretty east to multiple the percentage chance that opener has shape ABC by the chance that opener has HCP XZY... "

http://www.bridgeguy...MathTables.html

or Mathematical Tables in the Encycolopedia Of Bridge.

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#7 User is offline   Trpltrbl 

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Posted 2004-July-04, 14:04

Thx, Peter.

Mike ;)
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#8 User is offline   blahonga 

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Posted 2004-July-05, 04:00

hrothgar, on Jul 4 2004, 01:08 PM, said:

With this said and done, HCP and shape are independant of one another, so its pretty east to multiple the percentage chance that opener has shape ABC by the chance that opener has HCP XZY...

They are not independent of each other. A 13-0-0-0 will always have 10 hcp. The flatter the hand the higher the chance for more hcp i think.

/Mattias
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#9 User is offline   Fluffy 

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Posted 2004-July-05, 05:46

Good point ;) they are considered indedpendant because elsewhere it would become imposible to calculate.


I once tried to find the % of finding a ♠K on a 4♥ contract where you know west has 2♠+ 19/20 HCP against west who is known to hold 4-5 HCP +5♠, the result was so disapointing: it depends on wich exactly 24 HCP you are missing, I mean, depends on wihch honnors you miss exactly and in wich suits... imposible to calculate in time.
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#10 User is offline   hrothgar 

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Posted 2004-July-05, 05:52

blahonga, on Jul 5 2004, 01:00 PM, said:

hrothgar, on Jul 4 2004, 01:08 PM, said:

With this said and done, HCP and shape are independant of one another, so its pretty east to multiple the percentage chance that opener has shape ABC by the chance that opener has HCP XZY...

They are not independent of each other. A 13-0-0-0 will always have 10 hcp. The flatter the hand the higher the chance for more hcp i think.

/Mattias

URK. I standard corrected. Mattias is completely correct. If we start looking at extreme shapes, the two dimensions are no longer independant. With this said and done, I don't think that this would matter much for MOST calculations.

In general, I prefer to use Monte Carlo simulation anyway.
MUCH easier...
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#11 User is offline   Trpltrbl 

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Posted 2004-July-05, 07:13

Where can I find the Monte Carlo simulation, Richard ?

Mike ;)
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#12 User is offline   hrothgar 

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Posted 2004-July-05, 07:25

Trpltrbl, on Jul 5 2004, 04:13 PM, said:

Where can I find the Monte Carlo simulation, Richard ?

Mike ;)

Monte Carlo simulation is a technique in which you using a constrained random simulation to model a complex problem.

As I noted earlier, Dealer is a great freeware program that can be used for this type of analysis.
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#13 User is offline   Trpltrbl 

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Posted 2004-July-05, 07:56

You happen to have a link to this ?

Mike ;)
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#14 User is offline   hrothgar 

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Posted 2004-July-05, 08:01

Trpltrbl, on Jul 5 2004, 04:56 PM, said:

You happen to have a link to this ?

Mike ;)

http://www.dombo.org/henk/dealer.html
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