How many fish should I put in a tank in megaquarium to get optimal prestige?

In Megaquarium, each species of animal added to a tank increases the ecology and science bonuses of a tank. Each tank also has a percentage chance for each visitor that sees it to increase your prestige by the indicated amount.

For example, Azure Demoiselle is worth 1 prestige point. Let's say our tank currently has one fish in it, I believe that gives a 30% chance off the top of my head, then every 100 visitors seeing the tank would give us about 30 points.

Adding more and more fish of the same species gives you more and more points, but there is a limit: There are diminishing returns and adding more than 13 Azure Demoiselles seems to give you no higher probability. But the number is different for other species of fish. How can I figure out how many fish I need to add beforehand, so I can plan the size of my tanks?

Also, what exactly happens when I start mixing multiple species in the same tank? Is it possible to use 13 different small species of fish and get the maxed out probability with a much higher value? Or is it better to give each species their own tank? Or is it linear, meaning as long as I have the required amount of each species, a tank with the sum of 2 other tanks' fish in it will give me the sum of their prestige?

First, let's consider only a single species in a tank. The amount of fish you need for maximum (100%) prestige chance in one go depends on the size of the fish. Consult this table:

+------+----------+------+
| Size | Required | Tank |
+------+----------+------+
|    2 |       14 |   28 |
|    3 |       12 |   36 |
|    4 |       11 |   44 |
|    5 |       11 |   55 |
|    6 |       10 |   60 |
|    7 |        9 |   63 |
|    8 |        9 |   72 |
|    9 |        9 |   81 |
|   10 |        9 |   90 |
|   11 |        8 |   88 |
|   12 |        8 |   96 |
|   13 |        8 |  104 |
|   14 |        8 |  112 |
|   16 |        8 |  128 |
|   18 |        7 |  126 |
|   26 |        6 |  156 |
|   60 |        4 |  240 |
|  180 |        2 |  360 |
+------+----------+------+

When trying to combine multiple species; if you want 100% prestige chance on every species, you'll need as much as the table indicate for each species. For example, when combining a size 2 and size 3 species, you'll need 28 + 36 = 64 tiles for maximum prestige.

A guest can only give prestige once per species. If a guest would give prestige twice (or 3 times or more) instead you gain nothing. Therefore, it's important not to repeat any fish in your aquarium too much in separate tanks, as even a single fish can already give you over 50% chance of getting the prestige, and at 50% or above, having 2 tanks with the same fish actually gains you less prestige points than you'd think (only 75%), while 3 tanks would be 87.5%, and so on, eventually having them all in the same tank gains you more.

Worse, with 3 or more viewings, guests will lose some prestige instead. I haven't worked out how many tanks is optimal for species that dislike conspecifics yet, more to come about that.

Essentially, this boils down to three points to maximize gains, when you are required to have one species in multiple tanks (for example, the fish that don't like conspecifics):

1. Use a minimum amount of tanks per species.

2. Don't let the total prestige percentage go over 100%.

3. Evenly distribute the species over the tanks.

At x chance, with a P prestige fish:

With one tank: xP prestige

Using basic probability; where the chance of seeing K out of N tanks is (N over K) * P^K * (1-P)^(N-K);

With two tanks: x(1-x) * 2P + x^2 * P + (1-x)^2 * P = x(1-x) * 2P. Equating this to xP has a solution x = 0.5.