Discussion Problem (Expected Number)

[Amit Goyal]

A token is placed on the corner square of a 3×3 chess board. The token is
moved either up, down, left, or right, with equal probability. If the token
would move off the edge of the board it "wraps around'' the board and moves
to the opposite side instead. For example, if the token was in the bottom
right corner and moves down, it would move to the top right corner. What is
the expected number of moves it will take for the token to end up on the
center square of the board for the first time?

[Sinistral]

suppose the token takes mean "t" moves to reach to the centre if it starts from the corner.
suppose the token takes mean "a" moves to reach to the centre if it starts from anywhere else except the corners or centre (ie if it starts from the middle of the end row/column): lets call this position as side middle.

starting from corner, it can either go to corner (from 2 sides) or side middle (from 2 sides).
so t = .25t + .25t + .25(1+a) + .25(1+a)
where 1+a is the no. of moves it takes to reach to centre from corner via side middle.

starting from side middle it can either go to corner (from 2 sides) or centre or side middle.
So a = .25*1 + (1+t)*.25 + (1+t)*.25 + (1+a)*.25
where 1+t is the no of moves it takes to reach to centre from side middle via corner

solving above two equations give t=7 moves

[anon_econ]

I'm getting 10 :/
Sir, what's the correct answer?

[anon_econ]

In terms of your notation I had got t=1+0.5*a+0.5*t and a=1+0.5*t+0.25*a+0.25*0

[anon_econ]

Hmm okay there's a difference only in the first equation. I was kind of lazy to read your solution first. Don't you think another .5 should be added to the rhs of the first one?

[Amit Goyal]

Well done Vasudha.

Using symmetry, it is easy to see that Expected no. of steps is
same from boxes a11, a13, a31 and a33. Lets call this x.
And again by symmetry, Expected no. of steps is same from boxes
a12, a21, a23 and a32. Lets call this y.
We want to find x.
Note that
x = E(no. of steps from box a11)
   = 1 + 0.25 E(no. of steps from box a12) + 0.25 E(no. of steps from
box a21) + 0.25 E(no. of steps from box a13) + + 0.25 E(no. of steps
from box a31)
   = 1 + 0.25y + 0.25y + 0.25x + 0.25x
   = 1 + 0.5y + 0.5x
Thus, x = 2 + y
y = E(no. of steps same from box a12)
   = 1 + 0.25 E(no. of steps from box a22) + 0.25 E(no. of steps from
box a32) + 0.25 E(no. of steps from box a11) + + 0.25 E(no. of steps
from box a13)
   = 1 + 0 + 0.25y + 0.25x + 0.25x
Thus, 3y = 4 + 2x
Solving above two for x and y we get,
x = 10, y = 8.
Thus, expected number of moves it will take for the token to end up on
the
center square of the board for the first time is 10.

[Sinistral]

@Vasudha: oh yes . my 2 ts in the RHS of first equation need to replaced by (1+t). it will bring back 0.5 , as pointed by you.

@Amit Goyal:
Thank you so much for the detailed solution. :)