What are the odds that Sir Robin will appear naturally?

[Rhombic]

It seems extremely unlikely for Sir Robin to appear naturally... but there are 2^(16^2) soups out there!
So exactly how likely is it that Sir Robin would appear naturally in one soup?

i.e. if the probability was 2^-150 that would be extremely unlikely but odds are there are quadrillions of permegillions of soups with the Sir Robin anyway, it would just be extremely unlikely that we find them. Hoswever, the odds might be closer to 2^-500 in which case there is pretty much no chance a single 16x16 soup exists with Sir Robin.


on another note, should it be standardised to, for instance, "synthesis of the Sir Robin" or "synthesis of Sir Robin"?

[Macbi]

Well there are about 643 on and off cells in Sir Robin's "area".

Code: Select all

x = 63, y = 79, rule = B3/S23
4b2o30b2o$4bo2bo28b4o$4bo3bo27b5o$6b3o26b7o$2b2o6b4o20b12o$2bob2o4b4o
20b13o$bo4bo6b3o17b15o$2b4o4b2o3bo17b15o$o9b2o20b15o$bo3bo27b14o$6b3o
2b2o2bo18b14o$2b2o7bo4bo17b2o5b8o$13bob2o25b8o$10b2o6bo23b9o$11b2ob3ob
o23b9o$10b2o3bo2bo23b9o$10bobo2b2o25b9o$10bo2bobobo24b9o$10b3o6bo22b
10o$11bobobo3bo23b9o$14b2obobo23b9o$11bo6b3o22b10o$43b10o$11bo9bo21b
11o$11bo3bo6bo20b12o$12bo5b5o21b11o$12b3o29b8o$16b2o26b7o$13b3o2bo25b
7o$11bob3obo25b8o$10bo3bo2bo24b10o$11bo4b2ob3o21b11o$13b4obo4b2o19b13o
$13bob4o4b2o20b12o$19bo31b6o$20bo2b2o27b5o$20b2o30b5o$21b5o26b6o$25b2o
25b7o$19b3o6bo22b10o$20bobo3bobo22b10o$19bo3bo3bo23b10o$19bo3b2o26b10o
$18bo6bob3o20b12o$19b2o3bo3b2o21b11o$20b4o2bo2bo22b10o$22b2o3bo25b7o$
21bo31b5o$21b2obo28b4o$20bo31b4o$19b5o27b5o$19bo4bo26b6o$18b3ob3o25b7o
$18bob5o25b7o$18bo31b6o$20bo28b7o$16bo4b4o23b9o$20b4ob2o22b10o$17b3o4b
o24b3o3b4o$24bobo29b4o$28bo27b5o$24bo2b2o27b5o$25b3o27b6o$22b2o30b7o$
21b3o5bo23b9o$24b2o2bobo22b10o$21bo2b3obobo22b10o$22b2obo2bo25b9o$24bo
bo2b2o24b8o$26b2o27b7o$22b3o4bo24b8o$22b3o4bo24b8o$23b2o3b3o24b8o$24b
2ob2o27b5o$25b2o30b3o$25bo31b2o$57b2o$24b2o30b3o$26bo31bo!

So we could guess that the probability of him appearing in any given soup is about 2^-643. So we need to try about 2^643 soups before he appears. Since there are only 16*16 = 256 cells in an apgsoup the chance of him appearing if we tried all of them is only 2^(256-643) = 2^-387 = 10^-116.

[dvgrn]

Since there are only 16*16 = 256 cells in an apgsoup the chance of him appearing if we tried all of them is only 2^(256-643) = 2^-387 = 10^-116.

Sir Robin has six phases, though, so that should improve the probability to ... um ... about the same as before, really.

[muzik]

What's the smallest possible predecessor for it, in terms of bounding box? That way, we can at least limit our theoretical soup search to a specific bounding box which will definitely eventually release one.

[Apple Bottom]

on another note, should it be standardised to, for instance, "synthesis of the Sir Robin" or "synthesis of Sir Robin"?

The latter, definitely; the former just doesn't parse as a proper English sentence.

[77topaz]

The latter, definitely; the former just doesn't parse as a proper English sentence.

I think it's a bit weird that some patterns are proper nouns and others common nouns, but this may not be the first example of the former (e.g. "the Gemini" isn't commonly used, I think, and it's always capitalised).

[Rhombic]

What's the smallest possible predecessor for it, in terms of bounding box? That way, we can at least limit our theoretical soup search to a specific bounding box which will definitely eventually release one.

This is precisely what I was looking for.
From all the combined probabilities for each possible predecessor to appear naturally, how would the probability be determined? Otherwise you're assuming a set of cells in potential on/off candidates and mapping it to an unrelated 16x16, which is not the same I guess, because if a predecessor is smaller that would mean that there is a 100% (1) probability that Sir Robin appears, giving an immensely higher chance of appearing naturally.
So the real chance of it appearing naturally is the sum of the probabilities of every predecessor appearing naturally.