Probability of getting a 'better than average' tame
I'm trying to ascertain the probability of getting an 'above average' tame. For this question, I will use an example of taming a level 120 wild sabertooth.
I play on a difficulty 1.0 server, which mean wild dinos have a top wild level of 120. With a perfect tame, these become level 179.
Now as I understand it, there are 7 attributes for a sabertooth :
Heath
Stamina
Oxygen
Food
Weight
Damage
Speed
A freshly tamed 179 sabertooth will have 179 levels randomly assigned to any/all of these attributes. This, on 'average', will be between 25 and 26 levels each. For example :
25 Health + 25 Stamina + 25 Oxygen + 26 Food + 26 Weight + 26 Damage + 26 Speed = 179 points in total
However it is possible that a 179 tame will come out with random attributes such as :
32 Health + 20 Stamina + 12 Oxygen + 17 Food + 28 Weight + 35 Damage + 35 Speed = 179 points in total
A rather nice tame, with 4 attributes above average.
My question
What is the probability that a single attribute (e.g. Health) will be better than average? Specifically, for this question, that a single attribute will have 32 or more points assigned to it after a tame?
Aside note : The attributes Food, Oxygen and Speed are typically useless for a tame except in special circumstances (also Speed will always come out at a fixed number, irrespective of how many points are assigned to it immediately after taming). So :
Bonus question
What is the probability that a single attribute of either Health, Stamina, Weight and Damage will have 32 or more points assigned to it immediately after a tame?
Information for finding wild levels
For those people that do not know how to work out what points have been assigned to a freshly tamed dino, please look at the calculator on the wiki page here : Sabertooth Wild Stats Level-up
For example a freshly tamed 179 sabertooth with the follow stats have the following wild levels :
Health : 1815.1 = 28 levels
Stamina : 540 = 17 levels
Oxygen : 585 = 29 levels
Food : 4687.1 = 29 levels
Weight : 292 = 23 levels
Damage : 242.8 = 29 levels
Speed : Always 130 = 23 levels (whatever remaining levels are left over)
First, like players, an animal in Ark does not get a stat point at level 1, so a level 179 will have 178 stat points.
What is the probability that a single attribute (e.g. Health) will be better than average?
If you are looking for the chance that a specific attribute meets a certain threshold this can be calculated using the Binomial Distribution. The formula is available in many spreadsheets or math libraries and I have included the results of that below. For completeness I will explain the Binomial Distribution and how to use it. Skip the math break if you are not interested in the detailed calculations.
MATH BREAK
These formulas for calculation would use an animal with m levels, and we need to know the odds that it levels a specific stat at least j times. Consider if we had a level 6 (m = 6) Dodo, and wanted to the chance that it put at least 3 level-ups in weight (j = 3). Now there a 7 stats the Dodo could level up, which means the probability, p, that one is picked to be 1/7, or about 0.143 (or 14.3%). When leveling up, the dodo must have either leveled weight 0,1,2,3,4 or 5 times (*remember a level 6 creature leveled up 5 times). So the odds of it leveling weight at least 3 times is equal to 1 - ( (the odds of leveling it zero times) + (the odds of leveling it one time) + (the odds of leveling it two times)). So we need the odds of leveling a stat exactly k times, which is what the binomial distribution really is. So what are the odds the dodo leveled weight twice? It could have upped weight when it reached level 2,3,4,5 or 6. How many ways could it have done 2 weight level ups with these five options? The general solution is to use the Combination formula, nCr (read as n choose r). We have n = 5 levelups and want to r = 2 of them. The formula is n!/(r! * (n-r)!) where ! denotes the Factorial function. So 5C2 = 5! /(2! * 3!) = 120 / (2 * 6) = 10. Now we need the probability of two weight levelups, which is just p^j = (1/7)^2 = (1/49) ~= 0.0204 = 2.04%. And we need the probability of the other level-ups not being weight. The general form is (1 - p)^(n - j) = (1 - 1/7)^(5 - 2) = (6/7)^3 = 216/343 ~= 0.630 = 63%. We combine all of this and get the chance of exactly 2 weight levelups is 10*(1/49)*(216/343) = 2160/16,807 ~= 0.129 = 12.9%. There is a 12.9% chance that the dodo had exactly 2 weight levelups. But we still need to the values for 0 and 1 to get our final answer. Using the binomial formula nCj * p^j * (1 - p)^(n - j), with n = m - 1 = 5, p = 1/7, and vary j as necessary. For j = 0 we get 7776/16807 ~= 0.463 = 46.3%, and for j = 1 we have 6480/16807 ~= 0.386 = 38.6%. When we add the values for j = 0 to 2 we get 16416/16807 ~= 0.977. The value for at least 3 is one minus this so we get the a final answer of 391/16807 which is about 2.33%.
END OF MATH BREAK
Unfortunately I have been unable to create a formula for the other parts of your question, a bit too rusty on conditional probabilities. For completeness I used your method of a million animal simulation and calculated the summary statistics from the population. Our Any Attribute results are off by one (looking at level vs percentage), likely do to you giving animals 179 levelups rather than 178. Our Good Attribute results appear to be off by 2. I have results for 1 to 178, but have limited the results to 20 to 50, where it is more interesting.
For the table, The first column (Lv) is the minimum value you want for the attribute (number of level-ups), and the right is the probabilities. Column 2 (SpecificAttr) is the chance that a specific attribute has at least n level-ups. Column 3 (AnyAttr) is the chance that at least one attribute has at least n level-ups. Column 4 (GoodAttr) is the chance that at least one of Health, Stamina, Weight, and Melee have at least n level-ups.
The results show that a random post-tame 179 would barely have a 1% chance of having 37 or more health level-ups.
Lv- SpecificAttr AnyAttr GoodAttr
20- 90.124380% 100.0% 99.9993%
21- 85.540549% 100.0% 99.99329%
22- 79.792571% 100.0% 99.959%
23- 72.955960% 100.0% 99.7772%
24- 65.227618% 100.0% 99.1432%
25- 56.908916% 100.0% 97.33311%
26- 48.368382% 99.996% 93.4041%
27- 39.992090% 99.6484% 86.510704%
28- 32.132852% 97.06529% 76.4804%
29- 25.068894% 89.6619% 64.2297%
30- 18.979275% 77.1647% 51.0745%
31- 13.938424% 61.6378% 38.582%
32- 9.927425% 46.097% 27.7682%
33- 6.856503% 32.5225% 19.126%
34- 4.592086% 21.8371% 12.6564%
35- 2.982574% 14.0889% 8.1014%
36- 1.878908% 8.7513% 5.0086%
37- 1.148241% 5.2246% 2.9788%
38- 0.680877% 3.0343% 1.7251%
39- 0.391850% 1.7126% 0.9738%
40- 0.218927% 0.95% 0.543%
41- 0.118776% 0.503% 0.2906%
42- 0.062594% 0.2548% 0.1509%
43- 0.032050% 0.1302% 0.0778%
44- 0.015950% 0.062% 0.038%
45- 0.007717% 0.0299% 0.0184%
46- 0.003631% 0.0137% 0.0087%
47- 0.001662% 0.0056% 0.0031%
48- 0.000740% 0.0028% 0.0015%
49- 0.000321% 0.0013% 0.0009%
50- 0.000135% 0.0007% 0.0006%