I have a maze as shown above(use the link) and state 3 contains prize while state 7 contains shock. a mouse can be placed in any state from 1 to 9 randomly and it move through the maze uniformly at random
Pi denote the probability that mouse reaches state 3 before state 7, given that AIM started in compartment i.
how to compute Pi for ∈ {1,2,3,4,5,6,7,8,9}.
Intuitively, it seems that the probabilities of landing on either square would be even, as long as the starting point is random.
+---+---+---+
| 1 | 2 | 3 |
+---+---+---+
| 4 | 5 | 6 |
+---+---+---+
| 7 | 8 | 9 |
+---+---+---+
If the mouse starts in position 3 or 7, the game is over.
If the mouse starts in position 1, there is a 1/3 chance of it ending up in position 2, 5 or 4, and so on.
IF we start with each cell containing a 1/9 probability, we can compute the frequency distribution for the next generation by multiplying the current value by the probability of a mouse moving in from another location. For example, in the second generation. Cell 1 will have 1/5 of the mice from cell 2 + 1/5 of the mice from cell 4 and 1/8 the mice from cell 5. So, the next generation of cell 1 is (1/9)(1/5)+(1/9)(1/5)+(1/9)(1/8), or 21/360, or 0.0583. We can then compute the probabilities for all the remaining cells.
Here are the first five generations in a terribly formatted table
1 2 3 4 5
1 0.111 0.058 0.059 0.046 0.039
2 0.111 0.095 0.078 0.065 0.054
3 0.111 0.169 0.228 0.274 0.312
4 0.111 0.095 0.078 0.065 0.054
5 0.111 0.163 0.115 0.101 0.082
6 0.111 0.095 0.078 0.065 0.054
7 0.111 0.169 0.228 0.274 0.312
8 0.111 0.095 0.078 0.065 0.054
9 0.111 0.058 0.059 0.046 0.039
Let Px be the probability that the game ends in position 3 if it starts in position x.
We know that P3=1 and P7=0
If you start in any other cell, then after you move you are essentially beginning the game again in the new cell. The probabilities for the other 7 cells can therefore be calculated from the probabilities for their neighbors that they can move to:
P5 = P2/4 + P4/4 + P6/4 + P8/4
P2 = P1/3 + P5/3 + P3/3
P1 = P2/2 + P4/2
... etc.
For each cell you have a linear equation -- 9 equations for 9 cells. Use Gaussian elimination or similar technique to solve the system of equations for the 9 probabilities.
