Question

A man walks on the edge of a regular pentagon.At any point he decides to move away from the pentagon at any angle.After walking a certain distance he turns back and points himself to the center of the pentagon.What is the probability he would be able to see three edges of the pentagon?(he is allowed to move in the plane of pentagon)

Answer 1

Answer:

1) Markov Chains. Your problem can be modelled by a Markov chain where each vertex is a state. Your Transition Matrix looks like (assuming the center is an absorbing state):

TM=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜01/31/31/31/31/31/3001/30001/301/301/3000001/301/3000001/301/3000001/301/301/30001/30⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

where i'm labeling the first row/column as the center and the other rows/colums as the rest of the vertices clockwise from the top.

Then to compute the average steps to go back to center from each vertex you can compute:

(Id−TM)−1⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜0111111⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜0333333⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

So you need 3 steps to get from any vertex back to the center plus the step to get out from the center so finaly you need 4 steps total.

2) Series. As commented above and observing the simetry in the problem, Markov chains look like overkill.

You can compute

1+∑i=1∞iP(i)

where

P(i)=13(23)i−1

since you have 1 out of three possibilities to return at step i after you have to travelled along the exterior vertices for i−1 steps.

Using that

∑i=1∞iki−1=(∑i=0∞ki)′=1(1−k)2

you get

1+∑i=1∞iP(i)=4

Step-by-step explanation: