An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant?
Pick one:
A) 20
B) 43
C) 38
D) 10
E) 55
The answer is 10.   If the spider started at a corner diagonally on the same face as the ant the answer would be 9, and if the spider started at an adjacent corner the answer would be 7.   Let x=number of turns to reach ant from starting point.   Let y=number of turns to reach ant from diagonal corner on same face as ant.   Let z=number of turns to reach ant from an adjacent corner to ant.   After one turn the spider will be on a diagonal corner of a common face as the ant. So the mean number of turns from the x position is one more than the mean number from the y position:   E(x)=1+E(y).   Once at a y position there is a 2/3 chance it will then move to a z position, and a 1/3 chance back to an x position:   E(y)=(2/3)*(1+E(z))+(1/3)*(1+E(x)).   If the spider arrives at a z position there is a 1/3 chance it will move to the ant, and a 2/3 chance it will move back to a y position:   E(z)=(1/3)*1+(2/3)*(1+E(y)).   With these three equations and three unknowns it is not difficult to solve for E(x), E(y), and E(z).   correct answer 10
<result>D</result>
An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant?