This may have been added to the linked page after this question was asked, but the wiki page for the oviraptor linked to in the question does state that it increases feces production, the very beginning of the "Utility" section:

"nce tamed, setting the Oviraptor on wander will increase the production speed of eggs in nearby tamed creatures (including mammals), indicated by an egg icon above all affected animals. **This effect will also increase the feces laying rate on all dinos nearby.**"

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 *n*C*j* * *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%

## Best Answer