This just encourages massive army blobs instead of multiple parallel armies.
You're right, but that varies depending on the advantage conferred by differential army size. Besides, multiple armies will always have the advantage of being able to assault multiple locations simultaneously.
exponents are your friend
Yup, I was thinking something along the lines of:
y = -1/(n^(x-1)) + m
where:
y is the ratio of number of people that can fight for the larger army versus the smaller army.
x is the ratio of the number of tickets of the larger army versus the smaller army.
n is a constant that affects the advantage conferred by differential army size. (Increasing n means increasing advantage)
m is the maximum advantage
I would suggest an m at 2 or lower, and an n at 1.5 or lower.
For example, at m=2 and n=1.2:
Army A has 120 tickets and Army B has 100 tickets (20% advantage, x = 1.2), Army A could field 4% (y=1.04) more men at one time than Army B at one time. Roughly 61 v 59.
Army A has 1000 tickets and Army B has 500 tickets (100% advantage, x = 2), Army A could field 17% (y=1.17) more men at one time than Army B. Roughly 65 v 55.
Army A will never be able to field more than m times more men than Army B, in this case, the maximum advantage is twice as many men. The advantage conferred by differential ticket army size experiences diminishing returns.
The numbers can be tweaked as seen fit, but I think this is at least proof of concept.
EDIT: Urgh, just realized my formula only really works for m=2. Any tips on how to generalize this, such that at x=1, y=1 for all m?
DOUBLE EDIT: The real general formula is much messier, it is:
y = -1/(n^(x-p)) + m
where p = 1 - log
n(1/(m-1))