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Messages - Carca_Maker

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Muy buenas noches, gente carcasonera!...
Aprovecho para comunicaros que el compendio de mi variante base "La Reserva Etc v1.pdf" para futuros lanzamientos está disponible en el siguiente enlace:

Son 8 páginas completitas de estructura "cristalina" y enriquecida con subvariantes...

Estaría encantado si hubiera algún proceso de traducción al inglés, con "+1" merits y reseña en los créditos finales de agradecimientos.

En inglés sería "The Backup Etc". Espero que la disfrutéis!. Gracias!.

General / Re: Capitain Kirk
« on: July 18, 2019, 06:24:25 AM »
Oldbonz, how much tiles back do you need? One? An A4 full?

Enviado desde mi H3311 mediante Tapatalk

A mí me gusta la modularidad, el juego que da, con una base tan simple. Cuando iba al instituto, o más tarde vi, veía mapas de ciudades antiguas medievales, o renacentistas, en estrella con sus murallas y baluartes, me encantaba. "Algún día saldrá algo de aquí para hacer mapas con caminos, ciudades y bosques".Que fuera un mapa aleatorio, incluso aprovechable para wargames, etc.
Bueno, el colmo de mis mapas favoritos son los holandeses del siglo XVII, la cartografia, libros de viajes y expediciones, descubrir nuevos mundos (por ejemplo el Colonization de Ms dos de PC, lo flipas). Y cómo no, ¡una versión analógica en la que puedes aportar ideas en compañía de los otros aficionados!...
Ah... Yo prefiero el C1 ;-) 

No, DrMeeple but that question about to feed the Constructor it's contemplated, at my Jewels future expansion.

I thank your help, Meepledrone. I give you +1 merit. Glad to meet you!.

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.

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.

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.

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.

In my variant about "Jewels" there are reserve spaces, best scores, and "doomed Jewels", yeah!.

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.

I will, of course!, no doubt.

Ha, ha, ha! The system says that I have to wait one hour for give you another "karma" action, "+1 merit" more.

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!.

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.

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