Expected value and recursion lll

We'll use symmetry and name classify same types of squares on basis on possible choices at the square, then, we'll use conditional probability. Choices:

V -> 24

W -> 19

X -> 14

Y -> 11

Z -> 15

Let $E(i)$ be the expected number of jumps taken from a type of square by toad to reach trap.

If we are on any of Z squares then we have 15 choices:

3Z,4W,4Y,2X,1V and 1 trap.

If we go to any of 3Z,4W,4Y,2X,1V our $E(Z)$ will be set to respective $E(i)$ +1, as we have made a jump to reach the square.

Similarly following this for all types of square we get following Equations:

$E(Z)$ = $(3/15$ ) $(1+E(Z))$ + $(4/15$ ) $(1+E(W))$ + $(4/15$ ) $(1+E(Y))$ + $(2/15$ ) $(1+E(X))$ + $(1/15$ ) $(1+E(V))$ + $(1/15$ ) $(1)$

$E(W)$ = $(1/19$ ) $(1+E(V))$ + $(3/19$ ) $(1+E(W))$ + $(4/19$ ) $(1+E(Z))$ + $(3/19$ ) $(1+E(X))$ + $(6/19$ ) $(1+E(Y))$ + $(2/19$ ) $(1)$

$E(V)$ = $(0/24$ ) $(1+E(V))$ + $(4/24$ ) $(1+E(W))$ + $(4/24$ ) $(1+E(Z))$ + $(4/24$ ) $(1+E(X))$ + $(8/24$ ) $(1+E(Y))$ + $(4/24$ ) $(1)$

$E(Y)$ = $(1/11$ ) $(1+E(V))$ + $(3/11$ ) $(1+E(W))$ + $(2/11$ ) $(1+E(Z))$ + $(2/11$ ) $(1+E(X))$ + $(2/11$ ) $(1+E(Y))$ + $(1/11$ ) $(1)$

$E(X)$ = $(1/14$ ) $(1+E(V))$ + $(3/14$ ) $(1+E(W))$ + $(2/14$ ) $(1+E(Z))$ + $(2/14$ ) $(1+E(X))$ + $(4/14$ ) $(1+E(Y))$ + $(2/14$ ) $(1)$

The needed expected value is:

$E$ = $(1/25$ ) $(E(V))$ + $(4/25$ ) $(E(W))$ + $(4/25$ ) $(E(Z))$ + $(4/25$ ) $(E(X))$ + $(8/25$ ) $(E(Y))$ + $(4/25$ ) $(0)$

Solving these equations we get: $E$ = $\frac{2251163}{279075}$