I guess this problem is already resolved but here's a fun little matlab script for generating data samples comparable to OP's:
-------
nrolls = 100;
nruns = 1000;
data = randi([1, 14], nrolls, nruns);
mu = mean(data);
sigma = std(mu);
-------
In this case the "variable" is the ensemble mean of a bunch of siege rolls, and then we take the standard deviation of all the means to see what the cross-section of a set of 100 rolls looks like. The standard deviation of the set of means converges on ~0.4 the bigger the data set is.
Okay, so, in English? 100 random integers on 1-14 will have a mean that's up to 0.4 off of the expected mean (7.5) 68% of the time. They'll be even further from the expected mean 32% of the time! Which means you'll have an average of 7 or below or 8 or above 32% of the time. That's quite often!
And the odds of getting two 14's in a set of 100 rolls is just a binomial random variable. Here's another Matlab script to plot the distribution over a range of possible outcomes (numbers of 14's rolled):
-------
nrolls = 100;
outcomes = 0:20;
p = 1/14;
q = 13/14;
for k = 1:length(outcomes);
v(k) = nchoosek(nrolls,outcomes(k))*(p)^outcomes(k) * (q)^(nrolls-outcomes(k));
end
stem(outcomes,v)
-------
Here's an example of that plot:
https://postimg.org/image/a4m90pjwz/
So at around a 0.02 (2%) probability, rolling only two of one number is a little unlikely. But a 2% chance isn't nothing! And getting a much lower number than you'd expect (0, 1, 2, 3 or 4) has a pretty decent chance of happening (15%).
Anyway we spent a lot of time analyzing data from the game but not a lot of time analyzing our null hypothesis (what would it look like if it was actually random?). So that's worth point out!
-------
nrolls = 100;
nruns = 1000;
data = randi([1, 14], nrolls, nruns);
mu = mean(data);
sigma = std(mu);
-------
In this case the "variable" is the ensemble mean of a bunch of siege rolls, and then we take the standard deviation of all the means to see what the cross-section of a set of 100 rolls looks like. The standard deviation of the set of means converges on ~0.4 the bigger the data set is.
Okay, so, in English? 100 random integers on 1-14 will have a mean that's up to 0.4 off of the expected mean (7.5) 68% of the time. They'll be even further from the expected mean 32% of the time! Which means you'll have an average of 7 or below or 8 or above 32% of the time. That's quite often!
And the odds of getting two 14's in a set of 100 rolls is just a binomial random variable. Here's another Matlab script to plot the distribution over a range of possible outcomes (numbers of 14's rolled):
-------
nrolls = 100;
outcomes = 0:20;
p = 1/14;
q = 13/14;
for k = 1:length(outcomes);
v(k) = nchoosek(nrolls,outcomes(k))*(p)^outcomes(k) * (q)^(nrolls-outcomes(k));
end
stem(outcomes,v)
-------
Here's an example of that plot:
https://postimg.org/image/a4m90pjwz/
So at around a 0.02 (2%) probability, rolling only two of one number is a little unlikely. But a 2% chance isn't nothing! And getting a much lower number than you'd expect (0, 1, 2, 3 or 4) has a pretty decent chance of happening (15%).
Anyway we spent a lot of time analyzing data from the game but not a lot of time analyzing our null hypothesis (what would it look like if it was actually random?). So that's worth point out!
