Hey caters.
I'm sorry but I don't understand your question: could you be more specific?
Lets say that you have 2 cups and in 1 of them there is an equal number of white and black say 100 white and 100 black. Lets say that in the other cup you add another 100 black to the 100 white and 100 black you put in there.
Now lets say the black is 0 and white is 1.
These are the transformations:
white from 50/50 back to the same cup
black from 50/50 go to 2/3 black 1/3 white
black from 2/3 black back to same cup
white from 2/3 black go to 50/50
Now what is the probability of each of these outcomes?
With 3 colors base 3
with 4 colors base 4 and so on we can express bases to any number as long as we have that many colors.
We would have for 3 these outcomes:
white from 33/33/33 back to same cup
black from 33/33/33 to 25/50/25
red from 33/33/33 to 25/25/50
black from 25/50/25 back to same cup
red from 25/50/25 to 25/25/50
white from 25/50/25 to 33/33/33
red from 25/25/50 back to same cup
white from 25/25/50 to 33/33/33
black from 25/25/50 to 25/50/25
Now for any number of colors we are going to have x^2 outcomes
How would we figure out the probability of each outcome in the binary black/white, ternary black/white/red, quarternary black/white/red/orange, quinary black/white/red/orange/yellow, base 6 black/white/red/orange/yellow/green, base 7 black/white/red/orange/yellow/green/blue, octal black/white/red/orange/yellow/green/blue/indigo, nonary black/white/red/orange/yellow/green/blue/indigo/violet, and base 10 black/white/red/orange/yellow/green/blue/indigo/violet/gray when we have the colors assigned to these digits:
black = 0
white = 1
red = 2
orange = 3
yellow = 4
green = 5
blue = 6
indigo = 7
violet = 8
and
gray = 9
and for any base we get x^2 possible outcomes?
red from 20/20/20/40 go to 20/20/40/20
Notice the first number is % black, the second number is % white, the third number is % red and the fourth number is % orange. There are 4 possible outcomes per cup because 1 cup has equal amounts of black,white,red,and orange but the others have 40 % of 1 color and 20% of the others, and the fact that you can draw any of the 4 colors. There are 4 cups because we put more beads of a certain color in the other 3. That gives us x^2 outcomes possible for any number of colors.
I will give you another example.
red from 25/25/25/25 go to 20/20/40/20
A transition matrix capture all of the probabilities. If you want the probabilities at the nth time step then all you have to do is calculate T^n where T is the transition matrix.
You can do this symbolically as well if you don't have actual constant numbers. By calculating T^n you get all the probabilities from going to state 0 to state n.
So basically once you get T, and if you want all probabilities at a particular time spot, then you calculate T^n (symbolically or numerically) and then just use that to get them.
A simple 2x2 example (2 states) is the following:
[0.75 0.25]
[0.15 0.85]
where P(A1=1|A0=1) = 0.75, P(A1=2|A1=1) = 0.15 and P(A1=2|A0=1) = 0.25.