You owe me a dime.
I’m not sure what the average of the deviation from 50-50 would be,
In N independent trials of an experiment with a probability of success (heads) equal to p, the average number of successes would be p*N and the variance would be p*(1-p)*N. (Standard deviation sqrt(p*(1-p)*N))). In a hundred trials with a fair coin, the mean would be 50, and the standard deviation would be 5. About two thirds of the time, you would expect to get between 45 and 55 heads (or tails.)
As the number of trials grows arbitrarily large the distribution approaches a normal distribution (DeMoivre-LaPlace theorem, a special case of the central limit theorem.) For most purposes, one can make practial predictions about likelihoods of outcomes by assuming the population conforms to a normal distribution, but care needs to be taken for tail probabilities. See the binomial distribution for all the gory details.
The ratio of to the standard deviation to the mean decreases as 1/sqrt(N), regardless of p, for p not equal to zero or one. (In which cases, the std. dev. is zero.)
I have regular plain old BASIC somewhere. Suppose if I got ambitious, I could run it through ten thousand trials or so.
Then convert the program to do roulette...
;-)