When you take money out to pay bills, do you generally take out the same dollar amount each unit of time, or do you take out the same percentage of your bankroll each time? If it is the latter, then you can easily incorporate that into the Kelly system. If Kelly tells you to wager a fraction f, then you should wager f(1 - r), where r is the fraction you are removing with each wager.
If, however, you are removing the same dollar amount each time, then it is quite complicated. Suppose you start with a bankroll of B_0. Let B_n be your bankroll after n wagers. Without removing anything, your bankroll changes as
B_n = X_n B_{n-1},
where X_n is the proportional change in your bankroll due to the n-th wager. Solving this recursion,
B_n = B_0 e^{S_n},
where S_n = Y_1 + ... + Y_n, and Y_j = log(X_j). For large n, S_n ~ un, where u = E[log(X_j)]. This is why the Kelly formula tells you to maximize the expected logarithm. In the long run, your bankroll will look like B_n ~ B_0 e^{un}, so hopefully u is positive. (It will be if you have an edge, and if you follow the Kelly system.)
Now, if you are taking out C dollars with each wager, then the recursion changes to
B_n = X_n B_{n-1} - C.
Solving this,
B_n = (B_0 - CZ_n)e^{S_n},
where Z_n = \sum_1^n e^{-S_j}. If u > 0, then this sum will converge as n goes to infinity to some random limit Z. In the long run, your bankroll will look like
B_n ~ (B_0 - CZ)e^{un}.
Two things to notice. First, if Z > B_0/C, then you will go broke. So there is a risk of ruin. Second, if you survive, and Z < B_0/C, then your bankroll will grow at the same rate as before, and you do
not need to adjust the original Kelly recommendation.
As for risk of ruin, it is P(Z > B_0/C), and this looks like a very difficult probability to compute. But here is a (possibly crude) approximation.
Approximate Z by the integral from 0 to infinity of e^{-S(t)}, where S(t) is some continuous interpolation of S_n. Now, S_n is a random walk whose steps have mean u and variance s^2, where u and s are the mean and standard deviation of log(X_j). (The parameters u and s must be estimated from your tournament results.) Therefore, we might try using
S(t) = sB(t) + ut,
where B is a Brownian motion. In that case, it is known that the integral from 0 to infinity of e^{-S(t)} has the same distribution as 2/(Gs^2), where G has the gamma distribution with scale parameter 1 and shape parameter a = 2u/s^2. Using this, we can estimate that your risk of ruin is less than
[(2C/B_0)^a]/(a*Gamma(a)*s^{2a}).
By the way, Gamma(a) is the
Gamma function.