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Carc Central Community => Quizzes, Puzzles and Challenges => Topic started by: Carca_Maker on July 08, 2019, 02:06:07 PM

Title: TREES question for my next variant.
Post by: Carca_Maker on July 08, 2019, 02:06:07 PM
Hi everybody!
I'm working on my Carcassonne variant about some "elements", but I'm stopped at this point:

To build a sets of 3, (and in the other side) , sets of 4 of this “elements". ("Triple",  and "Quaterna" respectively?).  Sets of one, and two elements are trivial.

So we are at this starting point:

*We have got an amount of 10 elements: one "1", two "2", three "3", and finally four "4“.

A) If we extract randomly  THREE of them without replacement, how many cases are at all at every single branch?. (See my accompanying trees of three extracted elements).

I think that repeated sets of this  three elements are considered as "different" cases, although they have got the same composition.

So it looks that it doesn't matter the way, the order in that this elements are considered.

Every branch of this could be ordered in any order. It only matters its composition and how many repeated times appears every case inside its own branch.

Very important: although for example, we have extracted one "2", this is DIFFERENT  from the other existing "2", this means that at the end we could have got apparently some "repeated cases", but they have to be seen as "DIFFERENT", although belonging to the same  branch, of course.

I estimated approximately among 720 cases at all, at all the "triplets" trees, but I'm not sure if it's true.

I followed my own describing graphical order, where I began by the most unusual, uncommon element to the most frequent elements, in my trees.

B)The same for the sets, branches,  of FOUR of this extracted elements from the total amount of this 10 possible elements, (without replacement, etc).

Finally, a basic question:
It would be a  clarification for me to knowing how many cases are the most representative branches "1234", and "4444“,  separately, for example. Thanks.

I would include the helpers and solvers of this questions at the acknowledgements of my variant, and I would give a "+1“s thanks to them.

POSDATA: I wonder also if anyone can tell me if I could write all this in Spanish, my usual language. Is it possible too?. Greetings from Carca_Maker.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 08, 2019, 02:11:37 PM
Sorry... Maybe the two images that I accompanied are too big for a smartphone screen? Tell me if it does. Carca_Maker.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 08, 2019, 02:22:23 PM
The two images you uploaded are huge.

However, something weird is happening with images. I just uploaded one and the Forum is displaying it full size instead of a thumbnail you can click and magnify.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 08, 2019, 02:31:38 PM
Meepledrone, I sized both two images when I uploaded them, to a A4 printable size. I don't know if that is the matter of the trouble. I'm going to souper, it's late, and disconnect. Bye.
Title: Re: TREES question for my next variant.
Post by: wolnic on July 08, 2019, 03:04:12 PM
The two images you uploaded are huge.

However, something weird is happening with images. I just uploaded one and the Forum is displaying it full size instead of a thumbnail you can click and magnify.

I noticed that last night, too, when I uploaded the sample Cliffs tile ...
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 08, 2019, 04:15:10 PM
Hi Carca_Maker,

The size of the images should be displayed smaller, no matter the real size.

In any case, I wanted to verify if I understood your question correctly.

1. The elements: As far as I understood your elements belong to 4 categories but they are different. If so, may I call them univocally with an ID the represents the category plus a letter to be able to tell them apart:
    - one "1": 1A
    - two "2": 2A, 2B
    - three "3": 3A, 3B, 3C
    - four "4“: 4A, 4B, 4C, 4D

2. Extractions: I'm assuming the order of the elements is irrelevant and you are only interested in the elements of the subsets. In this case, you are representing each extraction and sorting the elements by ID. This would lead to a series of sorted extractions that could be grouped and represented as tree branches.

If so all your elements are unique, the number triplets and quadruplas can be computed with binomial coefficients if the order of the elements in one extraction is irrelevant:

(https://wikimedia.org/api/rest_v1/media/math/render/svg/08bdf0fff474c26293414f9eb01ab4bc73ef941f)

For triplets:
n = 10 and k = 3 --> 120 extractions

For quadruplas:
n = 10 and k = 4 --> 210 extractions

Let me know if this is what you were looking for.

Cheers!
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 09, 2019, 12:44:48 AM
The two images you uploaded are huge.

However, something weird is happening with images. I just uploaded one and the Forum is displaying it full size instead of a thumbnail you can click and magnify.

I noticed that last night, too, when I uploaded the sample Cliffs tile ...

Hi wolnic,

I reported this issue to gantry along with the Download area issue just in case they are connected (although it seems unlikely to me.)

http://www.carcassonnecentral.com/community/index.php?topic=4299.msg63615#msg63615 (http://www.carcassonnecentral.com/community/index.php?topic=4299.msg63615#msg63615)

Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 09, 2019, 10:46:31 AM
Hello Meepledrone!
I'm not still familiarized with "quote" action, though my smartphone, it's difficult seeing from it. Sorry. I'm going to talk from "recently" memory.

That's a good idea to identify the four kinds of elements with A, B, C, D.

I can say that your two calculations, look like, seem to "unify" some cases belonging to a "upper", category of cases. "120“ cases, and "210“ approximately in the other hand, it seems to me a nice approximation to the goal, because it's not a huge amount of rather cases. But I don't still know what does it means, so brief numbers. 

Now, the question is how to FIT  your first calculation,  to the most simple cases, that I know and I calculated too: the PAIRS of elements. (And the trivial singles cases too in one table).

I would be glad if your mathematical model can get to, describe the same as my table of pairs.
This is, how many cells are belonging to each pair, etc?. That's the question that I need for first approximation, for every trees and its branches.

Posdata: The cells of this pairs table are empty, because are an example.Thanks.

Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 09, 2019, 01:42:20 PM
Hello Meepledrone!
I'm not still familiarized with "quote" action, though my smartphone, it's difficult seeing from it. Sorry. I'm going to talk from "recently" memory.

Hi there, Carca_Maker!

I've been using the "quote" action in several ways:

1. Quoting a previous message when posting a new message: You can do this by clicking on the "Quote" button on the right side of the header of the post you want to quote.

2. Quoting a previous message in the thread while you are editing: Just position your cursor where you want to insert the quote, scroll down and click on the "Insert Quote" button on the right side of the header of the post you want to insert.

3. Inserting a dummy quote in a message: click the "insert quote"  button on the toll bar represented by a text baloon ((http://www.carcassonnecentral.com/community/Themes/default/images/bbc/quote.gif)) or type the following:

Code: [Select]
[quote]Blah blah blah [/quote]
You can edit the text of the quoted post to keep the parts you need or split it in sections by closing and opening "quote" tags to the text, as I'm doing in this post to reply inline.

Example:

Code: [Select]
[quote author=Carca_Maker link=topic=4332.msg63629#msg63629 date=1562694391]
Hello Meepledrone!
I'm not still familiarized with "quote" action, though my smartphone, it's difficult seeing from it. Sorry. I'm going to talk from "recently" memory.
[/quote]

My comments... before repeating the header and continuing with the quoted post

[quote author=Carca_Maker link=topic=4332.msg63629#msg63629 date=1562694391]
That's a good idea to identify the four kinds of elements with A, B, C, D.
....
[/quote]

Hope this helps.  :D

That's a good idea to identify the four kinds of elements with A, B, C, D.

Keep this in mind, I'm going to use it below.

I can say that your two calculations, look like, seem to "unify" some cases belonging to a "upper", category of cases. "120“ cases, and "210“ approximately in the other hand, it seems to me a nice approximation to the goal, because it's not a huge amount of rather cases. But I don't still know what does it means, so brief numbers. 

Now, the question is how to FIT  your first calculation,  to the most simple cases, that I know and I calculated too: the PAIRS of elements. (And the trivial singles cases too in one table).

We are closer than you think. The key point here is if you are considering permutations or not in your pairs: For example: In the "4,4" pairs you may have 4A-4B and 4B-4A... Are you considering them to be the same pair or not?

I was assuming they are, but your example is counting them as different pairs. If so, we have to factor back in the possible permutations of the elements in the triplets and quadruplas.

I would be glad if your mathematical model can get to, describe the same as my table of pairs.
This is, how many cells are belonging to each pair, etc?. That's the question that I need for first approximation, for every trees and its branches.

Posdata: The cells of this pairs table are empty, because are an example.Thanks.


Let me elaborate how you can simplify your computation if using pairs...

First, back to the binomial coefficient. This formula defines the number of subsets of size "k" you can make with "n" elements, where the order of the elements in a subset is unimportant (for example, how many different teams with 3 (k) people could you create with a total 10 people (n)?): 

(https://wikimedia.org/api/rest_v1/media/math/render/svg/08bdf0fff474c26293414f9eb01ab4bc73ef941f)

The numerator N = n * (n - 1) * ... * (n - k + 1) calculates the number of subsets.

The denominator D = k * ( k - 1 ) *... * 2 * 1  calculates the number or permutations you have within a set.

In your example, n = 10 and k = 2. Thus,

N = 10 * ... * (10 - 2 + 1 ) = 10 * 9 = 90

D = 2 * 1 = 2

There is no magic here: You have 10 elements, you pick 1 out of 10 and then you have 9 left to choose from. So you have 10 for your first pick and 9 left for your second pick. That is, 10 * 9 = 90 combinations in total. So here you stop after two picks.

Analogously, the logic for the number of permutations with two elements is similar: you have 2 elements, you pick 1 out of 2 and then you have only 1. So you have 2 for your first pick and only 1 left for your second pick. That is, 2 * 1 = 2 permutations in total. For example, if you have a set with two elements 4A and 4B, there are two possible permutations (4A,4B) and (4B,4A.) Top notch?

So getting back to you example, we have the following two scenarios.

1. Element order is considered for pairs: the same elements in different order represent different pairs (as per your example)

Here the result is N, as per the equations above. We want to calculate all the possible subsets without removing the possible permutations within each set.

So the result is N = 90, that matches your result for pairs with 10 elements. By the way, this result matches as well the number of cells in your grid if you remove the cells in the diagonal. Makes sense, right?  :o

2. Element order is NOT considered for pairs: the same elements in different order represent different pairs (as per my calculations)

Here the result  is N/D = 90 / 2 = 45, This would match the number of cells in your grid above (or below) the diagonal. This reminds me s the distance charts you may find around:

(https://antihrap.com/antihrap/mileage-charts-maths-worksheets/700-11_mileage-charts-maths-worksheets_question-1-of-10-look-at-the-following-distance-chart-.png)

You only need a triangle (half of the square grid) as the distance between any two cities is the same no matter which one you pick first.

So, in my previous post I was working with this scenario.



So how do we generalize all this?

Let n be the number of elements and K the size of the subsets you want to create.

Case #1: Elements in different order represent different sets

For this case use:

Combinations = N(n, k) = n * (n - 1) * ... * (n - k + 1)

* Pairs (same value as the example in your previous post):

N(10, 2) = 10 * 9 = 90

* Triplets:

N(10, 3) = 10 * 9 * 8 = 720

* Quadruplas:

N(10, 4) = 10 * 9 * 8 * 7 = 5040


Case #2: Elements in different order represent the same set

For this case use:

Combinations = C(n, k) = ( n * (n - 1) * ... * (n - k + 1) )  / ( k * (k - 1) * ... * 2 * 1 )

that is the same as

(https://wikimedia.org/api/rest_v1/media/math/render/svg/08bdf0fff474c26293414f9eb01ab4bc73ef941f)


* Pairs:

C(10, 2) = ( 10 * 9 ) / ( 2 * 1 ) = 90 / 2 = 45

* Triplets (same value as my previous post):

C(10, 3) = ( 10 * 9 * 8 ) / ( 3 * 2 * 1 ) = 720 / 6 = 120

* Quadruplas (same value as my previous post):

C(10, 4) = ( 10 * 9 * 8 * 7 ) / ( 4 * 3 * 2 * 1 ) = 5040 / 24 = 210




So after all this, which one is your case?


Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 09, 2019, 03:12:47 PM
Good evening, Meepledrone! First of all, I wish to give you a "+1" merit, no doubt, you are a math sage!. Afterwards I will have to read better what you wrote, but in a first approximation I think that "90" cases for pairs, "720" for triplets , etc is the right way.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 09, 2019, 03:26:22 PM
And written in the right way, I add : "Case #1: Elements in different order represent different sets", in the Meepledrone terms.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 09, 2019, 04:01:22 PM
Good evening, Meepledrone! First of all, I wish to give you a "+1" merit, no doubt, you are a math sage!. Afterwards I will have to read better what you wrote, but in a first approximation I think that "90" cases for pairs, "720" for triplets , etc is the right way.

Hi there, Carca_Maker!

Thank you! I just like Combinatorics. Counting things is fun! (Maybe a byproduct of too much Sesame Street?  :o)

So have fun with the light reading. If you have any doubts, just let me know.

Cheers!
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 09, 2019, 04:03:11 PM
And written in the right way, I add : "Case #1: Elements in different order represent different sets", in the Meepledrone terms.

Great! This narrows down the solution space   :))

I'm intrigued with what you will use all this for...

Keep us posted, please.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 10, 2019, 02:58:04 AM
Good morning! As Meepledrone said at his "Case #1: Elements in different order represent different sets" the triplets, 720 cases match with my old solution, here you are a picture.

So, it's only needed by me the solution and solving for Quaternas branches, at the beginning, supposing "Case #1: Elements in different order represent different sets". Ok and thanks.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 10, 2019, 08:35:17 AM
It's important to say that the my triplets notation at the previous picture, means that, for example "432“ means every triplet containing that three elements. Because of it, the total amount of triplets is 720 cases, and no the other smaller result of Meepledrone's supose #2.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 10, 2019, 03:15:10 PM
Hi Carca_Maker,

Based on the 10x10 grid analogy and projecting the exercise to 3 and 4 dimensions, I put together the table below covering the pairs, triplets and quadruplas following your conventions. Please see the image (you may need to scroll right to see the quadruplas part) and the zipped Excel file below.

It computes the number of basic combinations plus the permutations due to symmetries in the corresponding spaces. Reasoning in 4 dimensions was a bit complicated.  :o

Let me know your thoughts.

Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 06:19:01 AM
Congratulations Meepledrone!!
I intuited (intuition) that for the triplets, or quadruplas, the lowest quantity of cases were for 333, and 4444, respectively, but there were some hidden, another not obvious "combinations" which amount, worth the same lowest quantity of cases, maybe I saw, I suspected it at my triplets first trying that thanks to you, you "validated" it to me. Although I completed, I wrote that my triplets table one by one, I weren't sure at all. I'm going to compare my first trying of triplets table with yours, I suppose they match... Since I sent you my pairs table, etc, you understood that, great!.
I only can say that I lift my hat!. You beated the "final big boss monster" of my maths mess!. I give you +2 merits. Now I will apply it to my variant... Thanks Meepledrone!.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 06:23:23 AM
Ha, ha, ha! The system says that I have to wait one hour for give you another "karma" action, "+1 merit" more.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 06:25:42 AM
I will, of course!, no doubt.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 11, 2019, 09:10:06 AM
OMG! You are making me blush! It's okay...  :o

It was fun to do all the calculations... A good mathematical challenge from time to time is always welcome.

Now you have to tell us what all this number crunching is good for...
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 09:54:25 AM
Ooh! Meepledrone...
It looks that there aren't some Quaternas branches, for example your 4221, or 1224, it's the same thing, it's not.

There are an amount of 20 Quaternas branches, remember. Although your computation worths, does the right "5040" cases, so I think that it lacks 4221, 4421, 3221, 4422 branches and their computations.

In the other hand I revised your Triplets computations and they match with mine!.Good job!. I attached here  your table for clarification.

I'm using this calculations for to make an expansion in which appears "Jewels" and its combinations. You will see... Thanks. I'm waiting your wisdom words...greetings.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 10:03:27 AM
In my variant about "Jewels" there are reserve spaces, best scores, and "doomed Jewels", yeah!.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 11, 2019, 12:00:34 PM
Hi, Carca_Maker!

Too beautiful to be fully correct...  :'(

As you requested, some words of wisdom:  :o
1. Never post while watching the finale of Stranger Things Season 3.
2. Never trust the inner calculations are right if the result matches what you expected: there may be errors cancelling each other. In this case:
  - 4 missing quadruplas that added up 600 missing combinations. This is 600 combinations less.
  - An error in the Combinations for quadrupla 4322 and the symmetries for quadruplas 4433 and 3322 that added 600 extra combinations by error.
  - Net sum: 0
3. Contrast your results with an external source: I forgot to do so for the quadruplas as the total number were matching. Thank you for spotting the missing quadruplas and triggering the review.
4. Reasoning in 4D is tricky: Quadruplas such as 4433, 4422 and 3322 present 6 possible symmetries and not 12 as I assumed yesterday because there are 6 possible 4D planes to take into account for the mirroring, and not 12.
5. Reviewing everything after a good sleep is key to find hidden error: And you will do it in a jiffy.

And now, the quadruplas reviewed...

I added the missing quadruples to the listing and marked in pink the offending cells. (Walk of shame for me!)

Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 01:17:35 PM
Ha, ha, ha! Good heavens! The circle seems rounded, and closed. In the beginning it was wrongly "5040", and at last "5040“ rightly ;-). Most probably your calculations now will be right. If you found a mistake, make know it to me, please.
" Whatever it be, you are my Tesserackt hero!"Ha, ha, ha! (because those about the four dimensions). You have a place in the acknowlendgements of my variant. Have a good night.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 11, 2019, 01:39:55 PM
If there are any news about the calculations I will let you know.   O0

I'm eager to see how all this impacts your variant. Have fun with the design!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 11, 2019, 03:08:45 PM
I forgot a last detail, guys!:

As far as I know,...
 Is it true that "it's the SAME  if I do four (for example) MULTIPLE extractions without replacement one by one?" , than "if I do a SINGLE  extraction of four elements in once time?? (From a bag with 10 same balls as I proposed in the starting enunciated), in a probabilistic mode. Greetings.

Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 12, 2019, 03:10:01 PM
Hi Carca_Maker,

I've always seen all probability quizzes related to multiples draws with (or without) replacement been solved by handling one draw at a time. They use probability trees, similar to the trees you use but covering all the cases available. In your trees you are unifying in one case all the possible permutations, that I modeled as symmetries.

Here you can see some examples using probability trees:

http://onemathematicalcat.org/Math/Algebra_II_obj/prob_tree_diagrams.htm (http://onemathematicalcat.org/Math/Algebra_II_obj/prob_tree_diagrams.htm)

So when you say you draw 4444, you would have a probability tree that assumes that you draw 4 (with probability 4/10) followed by another 4 (with probability 3/9 --> three 4 elements left and 9 elements left in total after the first draw), followed by a third 4 (with probability 2/8) and finally followed by the last 4 (with probability 1 / 7.) So the probability of drawing 4444 is:

P(4444) = ( 4 / 10 ) * ( 3 / 9 ) * ( 2 / 8 ) * ( 1 / 7 ) = 24 / 5040  = 1 / 210

Does 24 / 5040 ring a bell? I think so (See the quadruplas table attached.) Well, it is not rocket science: the probability of drawing any quadrupla is equal to the number of combinations for the quadrupla divided by the total of possible quadruplas. Yeahhhh!

As you can imagine, this also works for pairs and triplets.

Hope this responds to your question.



Appendix: On Symmetries

As a side note, symmetries on the table below are equivalent to the number of possible permutations of quadrupla elements (with repetitions if any of the elements is repeated of course), so:

Let me define k! as the factorial of k, where k! = k * (k -1) * ... * 2 * 1

Therefore:
* The number of permutations of k different elements is equal to k!
* The number of permutations of k when several elements are repeated "a", "b",... "m" times is equal to k! / ( a! * b! * ... * m!)

Thus, If we revisit the value given to some quadruplas cases, we can obtain the number of symmetries as follows:

4321 --> 4! symmetries, no repetition = 24
4331 --> 4! / 2! symmetries, with 1 repetition of two elements = 12
4433 --> 4! / ( 2! * 2! ) symmetries,  with 2 repetitions of two elements each = 6
4441 --> 4! / 3! symmetries, with one repetition of 3 elements = 4
4444 --> 4! / 4! symmetries with one repetition of 4 elements = 1

This is applicable to all the quadruplas according to their particular cases. The objective of this additional exercise was to validate the results for the quadruplas.

Moreover, this can be applied to pairs and triplets. And it is valid for higher dimensions too  >:D

Cheers!
Title: Re: TREES question for my next variant. First results!.
Post by: Carca_Maker on July 13, 2019, 07:15:54 AM
Hi people! This is a first approximation to my results. It's about trees branches and its weights at my future variant. It's a "dual" file... Sorry. There are pearls 4, emeralds 3, rubíes (in Spanish, red gems, and a diamond!. Thanks Meepledrone, I'm a little busy... I will read your previous post more slowly, I think that I matched with your post, without being read. Luckily!. Cheers!. JEWELS VALIDATIONS AND ITS WEIGHTS.
Title: Re: TREES question for my next variant. First results!.
Post by: Meepledrone on July 13, 2019, 07:47:48 AM
Hi people! This is a first approximation to my results. It's about trees branches and its weights at my future variant. It's a "dual" file... Sorry. There are pearls 4, emeralds 3, rubíes (in Spanish, red gems, and a diamond!. Thanks Meepledrone, I'm a little busy... I will read your previous post more slowly, I think that I matched with your post, without being read. Luckily!. Cheers!. JEWELS VALIDATIONS AND ITS WEIGHTS.

Hi Carca_Maker,

Take your time. Any other task I can help you with in the meantime?  :)

Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 14, 2019, 04:19:49 AM
Good afternoon, Meepledrone!

After reading your before post, I have a doubt... You wrote
"Does 24 / 5040 ring a bell? I think so (See the quadruplas table attached.) Well, it is not rocket science: the probability of drawing any quadrupla is equal to the number of COMBINATIONS  for the quadrupla divided by the total of possible quadruplas."

As far as I know, probability calculations are based on " possible cases divided by the total amount of cases ".
So for 4444 quadrupla, it's 24 combinations and also 24 possible cases... But in another different case of" not coincidence ", for example 1223, your 3221, there are 6 COMBINATIONS at your table, but I think that its probability of extraction is" 72/5040 ", and not" 6/5040 “, Am I right?.
I can see also that the addition of all of quadruplas COMBINATION  data at your table, column, can't do 5040, so I suppose that you really referred to Total [possible] cases of every single  quaterna, etc.  in your table.

I discovered finally a mistake in my "Weights" column, next to your table :
I should be written "x4" at 433 triplet weight. I don't know now if there are more mistakes.

I obtained my "weights" values as the inverse of the (my right calculation?) probability, "possible cases divided by total cases", (this is, its inverse), knowing that the "unitary weight" should be the single Pearl extraction case, this is a "x1", and continuing proportionatelly with the other all cases. Is this right?.

(I wanted to meliorate the rarity of the extractions).
Attentively, waiting your response.

Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 14, 2019, 06:03:37 AM
Good afternoon, Meepledrone!

After reading your before post, I have a doubt... You wrote
"Does 24 / 5040 ring a bell? I think so (See the quadruplas table attached.) Well, it is not rocket science: the probability of drawing any quadrupla is equal to the number of COMBINATIONS  for the quadrupla divided by the total of possible quadruplas."

As far as I know, probability calculations are based on " possible cases divided by the total amount of cases ".
So for 4444 quadrupla, it's 24 combinations and also 24 possible cases... But in another different case of" not coincidence ", for example 1223, your 3221, there are 6 COMBINATIONS at your table, but I think that its probability of extraction is" 72/5040 ", and not" 6/5040 “, Am I right?.
I can see also that the addition of all of quadruplas COMBINATION  data at your table, column, can't do 5040, so I suppose that you really referred to Total [possible] cases of every single  quaterna, etc.  in your table.

Hi there CarcaMaker!

Yes, you are right. I was talking about total combinations (column Total in the previous tables) the intermediate columns are just to keep track of how the final figure is obtained, but for your objectives are irrelevant.

Please find below a new version of the same Excel and PNG files where I hid everything not useful for you. I also renamed the offending column to avoid ambiguities.

I discovered finally a mistake in my "Weights" column, next to your table :
I should be written "x4" at 433 triplet weight. I don't know now if there are more mistakes.

I obtained my "weights" values as the inverse of the (my right calculation?) probability, "possible cases divided by total cases", (this is, its inverse), knowing that the "unitary weight" should be the single Pearl extraction case, this is a "x1", and continuing proportionatelly with the other all cases. Is this right?.

(I wanted to meliorate the rarity of the extractions).
Attentively, waiting your response.

I'm glad you solved the issue.

Eager to hear from you.

Cheers!
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 14, 2019, 07:32:58 AM
I thank your help, Meepledrone. I give you +1 merit. Glad to meet you!.
Title: Re: TREES question for my next variant.
Post by: Meepledrone on July 14, 2019, 07:39:16 AM
Thank you, Carca_Maker!

It's my pleasure! Happy to help.

See you around with your ideas!  :(y)
Title: Re: TREES question for my next variant. First results!.
Post by: DrMeeple on July 15, 2019, 05:09:25 AM
Hi people! This is a first approximation to my results. It's about trees branches and its weights at my future variant. It's a "dual" file... Sorry. There are pearls 4, emeralds 3, rubíes (in Spanish, red gems, and a diamond!. Thanks Meepledrone, I'm a little busy... I will read your previous post more slowly, I think that I matched with your post, without being read. Luckily!. Cheers!. JEWELS VALIDATIONS AND ITS WEIGHTS.

Like traders and builders mechanism??
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on July 15, 2019, 07:00:07 AM
No, DrMeeple but that question about to feed the Constructor it's contemplated, at my Jewels future expansion.
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on September 09, 2019, 10:25:08 AM
Hello everybody!
I can say today that here is the link to my last expansion "JEWELS - JOYAS", finished thanks to Meepledrone help. The question of this quiz seems  solved at last!.Thanks!. You can find the numerical results included in a single pdf, which is inside this another zip file:

Here is the link to the Spanish zip file "Carcassonne JOYAS expansion v1.zip":

https://drive.google.com/file/d/1AgdfvXhGLZAvGsFMpmWvfb3rGDenICW4/view?usp=drivesdk

Cheers!
Title: Re: TREES question for my next variant.
Post by: Meepledrone on September 11, 2019, 12:45:40 PM
Hi Carca_Maker,

All your documents require a thorough reading... And they are interconnected...

I need to revisit them all. Hope to give some additional feedback after digesting it all.  :o

Cheers!  :D
Title: Re: TREES question for my next variant.
Post by: Carca_Maker on September 11, 2019, 12:59:13 PM
Hiii Meepledrone!!
I'm very glad to read you again!!.
Don't worry, make well the digestion of my readings... As I said you, they come from a past, a past of possibilities and a proof (prueba) and mistake, not in a quickly way as I really could wish, building with the bricks more or less defined... Cheers too!

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