[firmit]

Playing around with some numbers. Some calculation problems. I have a set of distribution, but how to calculate the probability with hp?

Example:
The balanced hands
4333 - 10,54%
4432 - 21,55%
5332 - 15,52%
Bal. = 47,60%

12 hp with avg. control ( A=2,K=1) 3,75, is 8,03% of all hands.

Now, how to calculate a 4432 hand with 12 hp? Is it as simple as 21,55 * 8,03 / 100 = 1,73%?

Edited: Appr: of the 1/5 4432 hands, 1/12 of them yields the given result -> 1/60= 1,67 %?

If this is incorrect, please give me the correct formula.

[gwnn]

While that isn't the 100% correct formula, it is certainly very close to the correct one.

For example, a 4333 hand can reach the maximum possible of 37 hcp and a 4432 only 36. Skipping a few logical steps, it can be indeed said that the more balanced the hand, the more likely it is that it has much hcp. However, the difference between these balanced hands is very very small.

The average hcp, or any other data you want from an unbalanced hand type (probably from 5422 onwards) is probably best calculated from a simple simulation program.

[Free]

Yes, this is accurately enough. It's not 100% correct however, just to show you a simple example: if you want to know the frequency of a 13-0-0-0 hand with 11hcp, you'll get a number which is higher than 0%. However, the probability is 0% since you always have 10hcp exactly. So there's a very small error, but it's soooo small that you shouldn't take it into account.

[FrancesHinden]

There's no straightforward way to calculate the table you want, but you certainly don't need to do a simulation: it is all calculable precisely.

I have a spreadsheet giving the following tables:

- HCP by distribution (or distribution by HCP)

- Combined HCP between two hands (e.g. if I have 12 points what is the probability partner has 13)

- Combined distribution between two hands (e.g. if I have a 4432 what is the probability that partner has a 2335)

Happy to make them generally available if anyone can tell me a good way to do so!

[Free]

Use a blog!

[kenberg]

To vary Free's example a little:

If your 12 points consists of three aces than obviously the three aces are in different suits. If the 12 points consist of two aces, a J and a K, then these four cards may lie in two suit, or three suits, or four suits. This affects the calculation and presumably the result. It also indicates why working out the exact answer by hand is apt to be something of a pain. Using, say, Mathematica it should be within reach to write a program that would work this out exactly. You need to go through a lot of cases, but that's what computers are happy to do, and Mathematica doesn't mind at all working with exact numbers with a large numbers of digits. It may not give you a lot of insight, but it will give you numbers.

Simply multiplying the probabilities would be correct if the distribution of high cards and the shape of the hands had no influence on each other, but as Free's example makes clear, this isn't true. But for questions such as 12 hcps and 4-3-3-3 distribution my guess is that it is very close to true.

I hope Frances can make her spreadsheet (You wrote this, Frances? Seems like real work.) available somewhere/how. Or lacking that, maybe just a few sample numbers to indicate the typical deviation of the true answer from the approximation obtained by simply multiplying the probabilities. With all the effort that must have gone into this calculatin, it should see the light of day.

[FrancesHinden]

I wrote it in Excel. I'm not a very efficient writer/programmer so the resulting spreadsheet was huge, but the results aren't that big.

I'll email to anyone who wants, but I don't have any great desire to start a blog or my own webpage.

For a 4432, simply multiplying the percentages together gives you a pretty good answer - for a 10-count multiplying together gives you 2.03% while the 'real' answer is 1.98%.

[hotShot]

Based on a simulation of 1 million hands:

21,57% have 4432 shape and 1,68% have 4432 shape and exactly12 HCP

probability    HCP
0.0922          0
0.1820          1
0.3170          2
0.5518          3
0.8770          4
1.1157          5
1.4093          6
1.7255          7
1.8996          8
1.9728          9
1.9969          10
1.9010          11
1.6842          12
1.4693          13
1.2216          14
0.9593          15
0.7153          16
0.5224          17
0.3581          18
0.2390          19
0.1538          20
0.0902          21
0.0537          22
0.0310          23
0.0183          24
0.0083          25
0.0030          26
0.0013          27
0.0006          28
0.0005          29
0.0002          30

[FrancesHinden]

hotShot, on May 1 2007, 02:07 PM, said:

Based on a simulation of 1 million hands:

21,57% have 4432 shape and 1,68% have 4432 shape and exactly12 HCP

I'm interested to see how far out (in context) even a simulation of 1 million hands is:

The real answer is 21.5512% and 1.6996%

[firmit]

I am somewhat of a competent programmer. And all my probabilities so far is not based upon a sample of a 1 million hands; I am working with them all; that's a "sample" of 635 013 559 600 hands. I don't know if this is pratical or suitable, but the results should not lie... I hope!

Is there a way of posting html? In that way I may show you, if ANY interest at all, what I have been working on. It consists of tables and probabilities for example different 1NT ranges, weak 2 openings and so forth. I may easily query the results I want.

The actuall reason for this work is that I have been reading about Zar points, and about inquiry2over1 system, where the Zar points is the base handevaluation system ( I think ). I wanted to legitimate the different bids and find out how frequent the different bids do occur. However, this is somewhat of a huge task...

I don't know if can me used to anything rational, other that confuse other people with big, boring tables of probability....

(I am writing it in php)

[hotShot]

FrancesHinden, on May 1 2007, 03:15 PM, said:

hotShot, on May 1 2007, 02:07 PM, said:

Based on a simulation of 1 million hands:

21,57% have 4432 shape and 1,68%  have 4432 shape and exactly12 HCP

I'm interested to see how far out (in context) even a simulation of 1 million hands is:

The real answer is 21.5512% and 1.6996%

Well a sample of 1 000 000 hands of 635 013 559 600 means that 1 out of 635 013 deals is picked.

This is like making a poll in Great Britain asking about 95 of it's about 60 610 000 inhabitants. One would ask about 1000 people to make a quick poll.

Actually the results are quite good. Both values have an error of less than +/-0.02 and I doubt this makes a difference at the table.

For big numbers (more than 10%) 100 000 deals are good enough, but for numbers less than 10% you need more, if you want to use numbers less than 1% a sample size of of 10^7 or more should be used.

Simulation gets more interesting when the calculation gets much harder than this.
e.g. How often do I have a Cappeletti 2♣ overcall in direct seat over a strong NT.

[helene_t]

FrancesHinden, on May 1 2007, 03:15 PM, said:

hotShot, on May 1 2007, 02:07 PM, said:

Based on a simulation of 1 million hands:

21,57% have 4432 shape and 1,68%  have 4432 shape and exactly12 HCP

I'm interested to see how far out (in context) even a simulation of 1 million hands is:

The real answer is 21.5512% and 1.6996%

The standard deviation of the binomial distribution is
sqrt( N p (1-p))

To get the stard deviation of the estimated proportion, you must devide by N, i.e.
sqrt( p (1-p) / N)

Here, N= 1000000 and p=0.02. For a quick calculation without a pocket calculator you can assume (1-p) = 1 so the stardard deviation on the estimated proportion of 1000000 hands, when the true proportion is 0.02, is

sqrt ( 0.02 /1000000) = 0.00014

A popular confidence bound is +/- 2 SE (sorry for confusing SE with SD but in this case it doesn't matter), i.e.

0.02 +/- 2*0.00014

or

[ 0.01972 ; 0.02028 ]

[Echognome]

I believe Richard Pavlicek has these conditional probabilities on his website. That is hcp by shape.

[FrancesHinden]

hotShot, on May 1 2007, 04:07 PM, said:

Simulation gets more interesting when the calculation gets much harder than this.
e.g. How often do I have a Cappeletti 2♣ overcall in direct seat over a strong NT.

Oh yes, I'm a fan of using simulations when useful. Just not a fan of using them when the exact probability can be calculated without too much difficulty.

The main problem with simulations such as "how often do I have a cappeletti overcall" is that when I look through the simulation results I discover I have to throw out about 10% of the hands generated!

[hotShot]

FrancesHinden, on May 1 2007, 06:52 PM, said:

Oh yes, I'm a fan of using simulations when useful. Just not a fan of using them when the exact probability can be calculated without too much difficulty.

The main problem with simulations such as "how often do I have a cappeletti overcall" is that when I look through the simulation results I discover I have to throw out about 10% of the hands generated!

By using a modified approach I can get the result of a "simulation" in somewhere between 5-60 seconds.
So its easy for me, to optimize the conditions to get the hands I want.

[bid_em_up]

helene_t, on May 1 2007, 10:19 AM, said:

FrancesHinden, on May 1 2007, 03:15 PM, said:

hotShot, on May 1 2007, 02:07 PM, said:

Based on a simulation of 1 million hands:

21,57% have 4432 shape and 1,68%  have 4432 shape and exactly12 HCP

I'm interested to see how far out (in context) even a simulation of 1 million hands is:

The real answer is 21.5512% and 1.6996%

The standard deviation of the binomial distribution is
sqrt( N p (1-p))

To get the stard deviation of the estimated proportion, you must devide by N, i.e.
sqrt( p (1-p) / N)

Here, N= 1000000 and p=0.02. For a quick calculation without a pocket calculator you can assume (1-p) = 1 so the stardard deviation on the estimated proportion of 1000000 hands, when the true proportion is 0.02, is

sqrt ( 0.02 /1000000) = 0.00014

A popular confidence bound is +/- 2 SE (sorry for confusing SE with SD but in this case it doesn't matter), i.e.

0.02 +/- 2*0.00014

or

[ 0.01972 ; 0.02028 ]

You mean, some people actually know this stuff?

Makes my head hurt just looking at it.

[Mr. Dodgy]

4432&12HCP = 1.7% according to my spreadsheet

[Echognome]

I guess no one believed me. Pavlicek calculates the odds using combinatorics. So he finds Pr(4432 and 12hcp) is 1.699564912409. I'll give Mr. Dodgy that 1.7 is close enough.

You can find all the info you'll need on it here:

http://www.rpbridge.net/7z76.htm

[firmit]

Richard Pavlicek sure knows his numbers!

The different tables are not so difficult to make - I have already done this. I make queries based upon the wanted hp range and handpatterns - thus I can get the cumulative sum of percentage of the wanted hands.

I am currently making the Query function regards to specific handpatterns with suites, like having a specific hand after opps opening 1NT 15-17. This may also have been done before, but I think it is quite interesting to work it out myself. Maybe I can make a conclusion of what NT defence that has the highest frequency of occuring - who knows!

Example the 15-17 range for 4432 ( note the relative % is not in this table ), with ZarPoints. The A.Cp=Average Control points.

4-4-3-2 21.55%, 10 ZD
Hp	A.Cp	-ZP-	--%-	-Cum
15	4.99	30.0	0.95	0.95
16	5.39	31.4	0.71	1.67
17	5.81	32.8	0.51	2.18

Maybe not so many cares about this stuff, but I find it interesting.