Peoples Physics Book Version-2

http://www.ck12.org Chapter 31. Random Walks 1

Above, we have answered the question posed at the beginning of this section for the three-step case. That is, we have
completely determined what the likelihood of the walker being in any possible location is at the end of this walk.

Question

Find the probability mass function for a single roll of a fair die.

Answer

The outcomes are simply the possible numbers, since it is a fair die, they have equal probability: 1/6.

Therefore the PMF is

( 1 , 2 , 3 , 4 , 5 , 6 );( 1 / 6 , 1 / 6 , 1 / 6 , 1 / 6 , 1 / 6 , 1 / 6 )

Question

Why are particular outcomes (sequences of steps, not end locations) equally likely in fair-coin random walks

with any number of steps?

Answer

Since steps right and left are equally likely, any particular sequence ofNsteps has a probability of( 1 / 2 )−N.

(Why? Think of the two-step example given in the beginning of the chapter: ifp= 1 / 2

P^2 = ( 1 −P)^2 =P( 1 −P) = ( 1 −P)(P) = 1 / 4

.

Fair Coin General Case

Now let us try to generalize these results to to a random walk withP= 1 /2 andNsteps. The intuition we obtained
from considering the simple case above can be summarized as follows: to find the probability mass function of the
end location of a random walker, one should: