
Problem You are given a graph with n nodes (states) and the edges between them have probabilities associated with it. You are initially in state "start" and asked to calculate the probability/expectation of a "end" state
Solution
1) Build state graph (states as nodes and transitions as edges) from the description of problem statement. Identify the start states and end states.
2)
A) If the graph is a DAG (Directed Acyclic Graph) use Dynamic Programming to calculate the probability of values associated with required state.
B) If the graph is not a DAG and has cycles, build a set of equations describing the state transitions and use Gaussian Elimination to solve for the probability/expectation
Another approach is simulation. If desired accuracy of result is less for eg 1e6 or so, simulate above transitions large number of times. The values should converge to actual values. This will work only if desired accuracy of result is less. For most accurate answer use Equation solving. Description
Example 1 (Using Gaussian Elimination / Simulation) A manufacturer of a Chutes And Ladders game (also called Snakes and Ladders) wants to know what is the expected number of dice throws (6 sided ) to finish the game by a single person. If expected value is high user will get bored as it takes lot of time to finish. If it is less user will finish game quickly and lose interest. The board is given in input as an array t[] which denotes where a player will end if he reaches the cell i. If there is no chute or ladder at cell i, then t[i] = i , else it will be the cell where chute or ladder points to. Note : Here we are not looking at game as 2 player, but trying to evaluate how much time a single user will take to finish.  States  Position of player position 0 .. n1 where n is number of cells. Start state is 0 and end state is >= n1 ( when user goes beyond last cell it is considered a finish) Transitions  Dice throw probability = p = 1/6. Additionally with the presence of chutes and ladders, user will move to other cells.
 This graph has cycles as based on chutes and ladders user can go back to previous cells. So build a set of equations and solve , say using Gaussian elimination
 Let e[i] be expected number of throws to finish game from state i.
e[i] = 0 i >= n1
= 1 + p * e[i0] + p * e[i1] + p * e[i2] + p * e[i3] + p * e[i4] + p * e[i5]
where i0 .. i5 are 6 states calculated from where player goes when dice shows 1 .. 6 as well as combining the fact where the resulting position has a chute or ladder While buidling the equation you might need to bring the nonconstant terms of RHS of last relation to LHS (to make it a equation in standard form)
 Writing relation for each state above you will have n variables and n equations  Solve using Gaussian elimination and e[0] is the answer
int m=6;
double p =1/m;
double e[n,n+1]; // equation for n x n+1 matrix
For(i,0,n1)
{
e[i,i] = 1.0;
if(i==n1)
continue;
For(j,1,m)
{
e[i,n] += 1;
int a = i+j;
a=min(a,n1);
int b=t[a];
e[i,b] = p;
}
}
GaussianSolve(e)
answer = e[0,n];
Practice : CoinGame, BranchPrediction
Simulation  Set initial values of e[i] to 0 and simulate above relation large number of times. The values should converge to actual values.
double e[n];
For(u,1,1000000) // simulate 1000000 times
{
For(i,0,n1)
{
if(i == n1)
{
e[i]=0;
continue;
}
e[i] += 1;
For(j,1,m)
{
int a = i+j;
a=min(a,n1);
int b=t[a];
e[i] += p * e[b];
}
}
}
answer=e[0];
Practice : SchoolTrip,
Example 2 (using DP)  You are throwing a n sided fair dice and noting the face which comes up. What is the expected number of throws before which you will be able to see all n faces at least once. State: i , the number of faces seen till now. Start state is 0 and end state is n. Transitions : Each with probability 1/n to other states, as to how many are seen newly after a throw.
Let e[i] be the expected number of throws required to see all n faces assuming you have already seen i faces (i 0 .. n)
e[i] = 0 i == n // seen all faces
= 1 + (i/n) * e[i] + (ni)/n * e[i+1]
 (i/n) probability that you will see already seen faces and (ni)/n probability that you will see a unseen face at which seen count goes up by 1.  Since RHS has e[i], we can bring it to LHS and regrouping
e[i] = (n/(ni) + e[i+1])
This now is a DAG and hence use DP to solve ie find all e[i], i in 0 .. n. Answer is e[0]
e[n]=0;
For(i,n1,0)
e[i] = (n/(ni) + e[i+1])
answer = e[0];
Practice : Favdice , PartyGame 