[hrothgar]

 AL78, on 2021-February-20, 04:40, said:

Surely it is 0.732097^25, the probability of the event happening 25 consecutive times. What you have is the probability of getting more than 12 HCP 25 consecutive times.

Yeap. I shouldn't answer stuff like this right before bedtime

[thepossum]

I feel people are over-simplifying this issue

Firstly can we clarify are we discussing the probability of the OP getting this outcome in any particular set of 25 hands. How many sets of 25 does the OP intend to play etc
Are we talking about someone like me who may never play 25 hands in one session?

What kind of person are we talking about. Do we have to work out the distribution for all players over 1 year, over their lives, over all players over a finite or potentially indefinite time period etc

Plus obviously its a complex combinatorial calculation. The probability goes up the more sets of 25 hands you play etc. Chance it happens in set 1 + (1-that chance)* (chance in set 2) etc

Problem for me is that P I am using is being estimated above using the CLT and so it goes

And since its already happened once that complicates the calculation even more

For me to contribute I think I need more clarity on the specific probability we are looking for

PS And I have a very vague recollection over how long things like the CLT can take to kick in with many distributions etc

But I also appreciate these days with economics of time and limited resources only a small number of replications (if any sometimes) are deemed necessary so maybe a small sample from a sim will get close

But I did a little sim just looking at South's hand (too complicated otherwise) and I estimated around 28% of South hands had less than 8 points over about 10 sims of 4000 hands each

That gave me a probability of around 0.007 of South getting 25 hands of less than 8 points (ignoring many other factors and complexities of course)

Oops wrong calculation (thought it looked wrong, thats e^0.28) of course its 2.5 x 10^-14 (approximately)

[smerriman]

 thepossum, on 2021-February-20, 16:27, said:

That gave me a probability of around 0.007 of South getting 25 hands of less than 8 points (ignoring many other factors and complexities of course)

Oops wrong calculation (thought it looked wrong, thats e^0.28) of course its 2.5 x 10^-14 (approximately)

You've misread the question(s) - the first was *averaging* 7.5, not getting less than 8 every hand. The second was getting less than 13 every hand.

[thepossum]

 smerriman, on 2021-February-20, 20:17, said:

You've misread the question(s) - the first was *averaging* 7.5, not getting less than 8 every hand. The second was getting less than 13 every hand.

I'm sure I did. I don't even know if I can calculate what I want simply with the tools I have available
-rephrasing I could but I'm trying to do it with minimum effort and not sure I can - my coding days are many years ago - I need to do stuff with simple tools these days or I get bored very easily

back to reading reports on mental health and substance use - strangely boredom is mentioned a fair bit

[PeterAlan]

Possum, you can think about the original problem in this way:

Consider the formal polynomial p(X) = a(0) + a(1).x + a(2).x^2 + ... + a(37).x^37

where the coefficients a(i) = probability of i HCP in a single hand (a(0) = 0.003639, a(1) = 0.007884 etc).

Essentially, the power of x in the formal polynomial is used to label the number of HCP.

For N hands (here N=25) calculate p(x) ^ N = b(0) + b(1).x + ... b(37.N).x^(37.N); the coefficients b(i) then represent the probability of holding exactly i HCP in N hands.

For the probability of holding <= M HCP in total (here M=187), sum b(i) for 0 <= i <= M.

Calculating the coefficients b(i) of p(x)^N is quite straightforward - you can do it in a spreadsheet from the raw a(i) data (which obviously you need). I wrote a program instead: it accepts N & M as its input and does just that calculation.

[hrothgar]

Or, alterative, you can use the code that I posted earlier to generate a million or 10 million or 100 million hands and approximate things...

[PeterAlan]

 hrothgar, on 2021-February-21, 10:59, said:

Or, alterative, you can use the code that I posted earlier to generate a million or 10 million or 100 million hands and approximate things...

Indeed Richard. Or, if looking for an approximate answer with a quick calculation, you can follow smerriman's and ALT78's CLT route, though that needs some technical knowledge (and the standard deviation in question).

But thepossum seemed to be looking for some way to think about the problem, and I was trying to give that. Simulations have their place, but they do nothing to help you understand the underlying problem and its possible solutions.

[thepossum]

 PeterAlan, on 2021-February-21, 06:22, said:

Possum, you can think about the original problem in this way:

Consider the formal polynomial p(X) = a(0) + a(1).x + a(2).x^2 + ... + a(37).x^37

where the coefficients a(i) = probability of i HCP in a single hand (a(0) = 0.003639, a(1) = 0.007884 etc).

Essentially, the power of x in the formal polynomial is used to label the number of HCP.

For N hands (here N=25) calculate p(x) ^ N = b(0) + b(1).x + ... b(37.N).x^(37.N); the coefficients b(i) then represent the probability of holding exactly i HCP in N hands.

For the probability of holding <= M HCP in total (here M=187), sum b(i) for 0 <= i <= M.

Calculating the coefficients b(i) of p(x)^N is quite straightforward - you can do it in a spreadsheet from the raw a(i) data (which obviously you need). I wrote a program instead: it accepts N & M as its input and does just that calculation.

Hi Peter

I dont need a lesson in anything but thanks for posting it for other readers

You mistook a certain lack of interest, motivation or carelessness in my response - basically not giving a *** - as indicative of not understanding something

Can I ask (you and others) did you actually understand some of the considerations I mentioned

Of course we all know there is a very simple way of looking at it using sample theory and the CLT. But I was trying to think around the problem a bit and ended up answering the wrong question

I was trying to get an estimate on the distribution of the mean from a particular package I was using but couldn't actually script it up or even think about it in the few minutes I allocated. I ended up with a much simpler problem

The one thing I do find interesting is trying to think about the different approaches to conceptualising Bridge statistics and distributions. The chance of an individual picking up 25 hands in a row with that average is a very challenging distribution to think about. It depends on thinking about the individuals involved. For most of the world the probability is zero etc

I was thinking about the original poster and looking at it this way. Whats the chance of it happening twice. Very small. Since its already happened once the chance of it happening again is therefore very small

But more seriously I actually read the problem as getting that exact mean/sum HCPs in 25 hands, not less than or equal which is what most people calculated

Also I'm curious that you have .0078 as one of your coefficients. I couldnt for the life of me work out where I had that number from. I had a number approximately .007 from somewhere but wasnt sure what it was

I do like thinking about the number of possible Bridge hands and how often we try to estimate things/differences in often extremely small probabilities, often without the precision to do so

[PeterAlan]

I can't tell whether or not you're interested in the answer, possum, but the probability of exactly a total of 187 HCP in 25 independent hands is 0.0001538068 (1 in 6502).

[thepossum]

 PeterAlan, on 2021-February-21, 18:54, said:

I can't tell whether or not you're interested in the answer, possum, but the probability of exactly a total of 187 HCP in 25 independent hands is 0.0001538068 (1 in 6502).

I'm not really interested in the problem or the answer at all. And certainly not in a lesson or being patronised and singled out.

I don't know if your a teacher for example but I am not a child in your class

Singling an individual out in this way as has happened regularly on this forum is a way of undermining and disempowering an individual. I certainly need no lessons here

I regard the way some people talk to others on here as very rude and insulting and for some reason I seem to be one of the chosen recipients of such behaviour

It feels as if people are cruising around looking for someone to give an unwanted lesson to. The thread was not singling out individuals. But seemingly I needed to be singled out

EDIT I'm out of here again for a while before anyone else weighs in. Its like a subtle form of trolling (maybe unintentional). I wish I had never even commented on the thread. I certainly should not have bitten the first time and certainly not the second time. I even came back to delete my first bite but was too late. See you later everyone, especially before any of the more usual suspects show up and weigh in

I will say finally that kind of language used (to single out and patronise), whether intentional or not has the effect of an ad hominem attack or troll to cause a response

[PeterAlan]

 thepossum, on 2021-February-21, 18:07, said:

I was thinking about the original poster and looking at it this way. Whats the chance of it happening twice. Very small. Since its already happened once the chance of it happening again is therefore very small

Let's say that the probability of some outcome (total HCP <= 187, exactly 187, whatever) over a set of 25 boards is p. Then the a priori chance of it happening twice in two sets of 25 boards (assuming independence throughout) is (p^2). But once it has happened in the first such set, the chance of it happening in the second set is still p, and not something less.

[PeterAlan]

Quite frankly, you single yourself out possum. You pose a bunch of questions, some clearly expressed, others less so, but if someone takes you seriously and engages with you, you then accuse them of being patronising, rude, etc. As far as I am concerned, all the rudeness is coming from your side. You can engage in civil debate, or be ignored: which would you prefer?

[hrothgar]

 PeterAlan, on 2021-February-21, 19:08, said:

Let's say that the probability of some outcome (total HCP <= 187, exactly 187, whatever) over a set of 25 boards is p. Then the a priori chance of it happening twice in two sets of 25 boards (assuming independence throughout) is (p^2). But once it has happened in the first such set, the chance of it happening in the second set is still p, and not something less.

You are framing this issue in a quite specific manner (note: this is the same way that I approached the original question) It's not the only way.

Let's assume a more Bayesian world

• I am playing in a bridge tournament, sitting North
  
• I get dealt a hand with less than 12 HCP on the first deal
  
• The same happens on the second deal
  
• ...

Let's reframe the question such that we are adjusting our priors around two different hypotheses

1. The dealing process is fair, but the card gods have it in for North
2. The dealing process is not fair

I am guessing that ThePossum was thinking about a problem that is more in this vein

[PeterAlan]

 hrothgar, on 2021-February-22, 07:20, said:

You are framing this issue in a quite specific manner (note: this is the same way that I approached the original question) It's not the only way.

Guilty, and I'm glad you pointed this out.

In your Bayesian realm, I presume that one would start with a null hypothesis concerning the fairness of the dealing algorithm. Making this well-defined may not be straightforward: see below.

I probably have much more interest in these questions than most; if you're not interested in further discussion I quite understand. However, and at the risk of digging a still deeper hole, there are a couple of things still to be said about implicit hypotheses in everyone's answers to the original question (independence has already been mentioned):

• I and the CLT folk are giving probabilistic answers for the mathematical universe of all deals and its HCP distribution, or models of it. The BBO dealing software in question might not conform to that, and to give a better answer to the question originally posed (what's the probability in BBO etc) we'd really need to know the single-hand HCP distribution that results from it - of course, we don't have that information. There's also the possibility that the BBO software could have precisely the correct overall HCP distribution for single hands but nevertheless have bias as to which hands get which points; we'd need information about the full deals to get a better handle on that; and
  
  
• A simulation approach doesn't actually answer the question posed: what it gives is answers about a third HCP distribution, namely that of the dealing algorithm underlying the simulation software. This seems to be important.

Of course, if all the dealing algorithms are pretty close one-to-another and to the theoretical universe then the results are sufficiently good for the everyday question.

All this is really saying is that a complete answer to just the original question is not straightforward; we all simplify the problem by adopting one assumption or another in order to come up with an answer, and in doing so pick up our calculation tool of choice.

[hrothgar]

 PeterAlan, on 2021-February-23, 06:37, said:

Guilty, and I'm glad you pointed this out.

In your Bayesian realm, I presume that one would start with a null hypothesis concerning the fairness of the dealing algorithm. Making this well-defined may not be straightforward: see below.

I probably have much more interest in these questions than most; if you're not interested in further discussion I quite understand. However, and at the risk of digging a still deeper hole, there are a couple of things still to be said about implicit hypotheses in everyone's answers to the original question (independence has already been mentioned):

• I and the CLT folk are giving probabilistic answers for the mathematical universe of all deals and its HCP distribution, or models of it. The BBO dealing software in question might not conform to that, and to give a better answer to the question originally posed (what's the probability in BBO etc) we'd really need to know the single-hand HCP distribution that results from it - of course, we don't have that information. There's also the possibility that the BBO software could have precisely the correct overall HCP distribution for single hands but nevertheless have bias as to which hands get which points; we'd need information about the full deals to get a better handle on that; and
  
  
• A simulation approach doesn't actually answer the question posed: what it gives is answers about a third HCP distribution, namely that of the dealing algorithm underlying the simulation software. This seems to be important.

Of course, if all the dealing algorithms are pretty close one-to-another and to the theoretical universe then the results are sufficiently good for the everyday question.

All this is really saying is that a complete answer to just the original question is not straightforward; we all simplify the problem by adopting one assumption or another in order to come up with an answer, and in doing so pick up our calculation tool of choice.

FWIW, I agree completely with what you are claiming here

The one area where I might quibble is the amount of information that we have about the BBO dealing algorithm. We do have a couple data points about this

1. BBO had the algorithm tested years back (and they claim that they have no touched the code since then)
2. Lots of people have raised lots of claims about the BBO dealing algorithm (and too my knowledge no one has ever found a problem with it)

Nicolas Hammond has accumulated an impressive corpus of hands from various types of ACBL tournaments. He's probably in a position to analyze the output of the deal generator in a more systemic manner.

[Vinish]

Wow! There are some impressively knowledgeable people on here about statistics. I certainly in my original question and in the follow-up question wanted people offering help to assume that the BBO dealing was random and that the assignment of hands to a seat position was also random. Sure, this might not be true but I am glad that post #35 suggests there is some evidence that it is true.

I am sorry that poster thepossum got upset but I saw nothing posted that was insulting or patronising.

Thanks for the help on my two questions.