Question

There are 100 equally spaced points around a circle. At 99 of the points, there aresheep, and at 1 point, there is a wolf. At each time step, the wolf randomly moves eitherclockwise or counterclockwise by 1 point. If there is a sheep at that point, he eats it.The sheep don’t move. What is the probability that the sheep who is initially oppositethe wolf is the last one remaining?

Answer 1

Answer:

we have $p_{51} = \frac{1}{100}$

Step-by-step explanation:

considering circle with 100 points on it and then placing the wolf on one and sheep on rest.

pi = P( sheep on i-th spot  is eaten at last}

As in the question we need to find probability of sheep opposite to the wolf, so p_{51}. observe each last spot is replaced by i-1 after eating

Byy LOTP we know thta$pi = \frac{1}{2}pi-1 + \frac{1}{2}pi + 1, i \epsilon {2,... 99}$

also these pi need to satisfy

$\sum_{i=1}^{100} pi =1$

some sheep is eaten at last. Distribution is

$pi = \frac{1}{100}$

satisfies above equation, therefore we have $p_{51} = \frac{1}{100}$