# 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)

• Oddly this question is merely a mathematical question. I would recommend asking on a math forum. Commented Mar 14, 2016 at 2:52

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%

• Thanks for pointing out that my results slightly off due to skill points assigned to tames. However at the moment I don't believe that this answer is sufficient to award the bounty as it basically just reiterates the existing answer with recalculations. I'm not saying this is a bad answer (upvoted), but a better answer would be a working example of binomial distribution with in-game examples, so those unskilled in advanced mathematics can still understand the working out. I appreciate your being honest in your rustyness, it's been several years for me too! Commented Mar 17, 2016 at 11:59
• @MrDeveloper I've added an explanation for the binomial distribution, If I come up with the general solution I'll add more. Commented Mar 17, 2016 at 15:28
• Excel and Google Spreadsheets won't allow factorials over 170, so I had to do this in a Jupyter notebook. Your post helped immensely. Thanks! Commented Dec 19, 2018 at 12:20

Whilst this is not the answer I'm looking for, as it doesn't include fraction percetiles and the working out of how those probabilities are calculated, I feel that my ongoing investigation is more suited to an answer rather than being as part of the question. It also breaks the question down from being monolithic and provides a quick look up for any players looking for a quick, rough and ready, answer to the question.

Ongoing investigation

Whilst waiting for a more scientific answer, I've put together a small application that calculates the chance that any single attribute point will be above a certain value. Here are the results of 1,000,000 level 179 sabertooths:

Any attribute

Any attribute level 27 or greater : 99.9%

Any attribute level 28 or greater : 99.6%

Any attribute level 29 or greater : 97.1%

Any attribute level 30 or greater : 89.6%

Any attribute level 31 or greater : 77.0%

Any attribute level 32 or greater : 61.5%

Any attribute level 33 or greater : 45.9%

Any attribute level 34 or greater : 32.4%

Any attribute level 35 or greater : 21.7%

So looking here, any single 179 sabertooth tame would have ~61.5% chance of getting a single attribute at level 32 or more.

Here are results from a similar test, but concerned with only attributes Heath, Stamina, Weight and Damage:

Health, Stamina, Weight or Damage

Any attribute level 27 or greater : 97.3%

Any attribute level 28 or greater : 93.4%

Any attribute level 29 or greater : 86.5%

Any attribute level 30 or greater : 76.5%

Any attribute level 31 or greater : 64.2%

Any attribute level 32 or greater : 51.1%

Any attribute level 33 or greater : 38.5%

Any attribute level 34 or greater : 27.6%

Any attribute level 35 or greater : 19.1%

Interestingly, a 179 sabertooth would have a just over 50% chance of getting a worthwhile attribute of 32 levels or more.

Whilst I could go on and expand upon these tests with the chances that Health, Stamina and Damage attributes would be each above a certain number of levels, I wont. That is what the Ark Survival Evolved Breeding mechanic is used for.