Author Topic: TREES question for my next variant.  (Read 862 times)

Offline Meepledrone

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Re: TREES question for my next variant.
« Reply #15 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!
« Last Edit: July 10, 2019, 04:36:08 PM by Meepledrone »

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #16 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!.

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #17 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.

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #18 on: July 11, 2019, 06:25:42 AM »
I will, of course!, no doubt.

Offline Meepledrone

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Re: TREES question for my next variant.
« Reply #19 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...
« Last Edit: July 11, 2019, 09:12:43 AM by Meepledrone »

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #20 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.

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #21 on: July 11, 2019, 10:03:27 AM »
In my variant about "Jewels" there are reserve spaces, best scores, and "doomed Jewels", yeah!.

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Re: TREES question for my next variant.
« Reply #22 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!
« Last Edit: July 11, 2019, 12:09:35 PM by Meepledrone »

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #23 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.

Offline Meepledrone

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Re: TREES question for my next variant.
« Reply #24 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!

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #25 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.


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Re: TREES question for my next variant.
« Reply #26 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

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!

Offline Carca_Maker

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Re: TREES question for my next variant. First results!.
« Reply #27 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.

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Re: TREES question for my next variant. First results!.
« Reply #28 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!

Offline Carca_Maker

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Re: TREES question for my next variant.
« Reply #29 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.



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