# Random straight line walk with conditions

A drunkard is 9 steps away from home, and 1 step behind him is a pit. He will move 1 step forward/backward with probabilities of 0.7/0.3 respectively.What are the probabilities that

(a) he will fall into the pit

(b) reach home

(c) reach 1 step away from home

I had solved a somewhat similar problem, but with no pit and a condition of reaching home in 20 steps or less, by analysing patterns on a 2D lattice, getting the formula

sum of (3/n)*C(2n, n+3)*0.7^(n+3)*0.3^(n-3) for n = 3 to 10 = 0.814401

but i don't know how to tackle this question.

I'm not very savvy mathematically, so the simpler the explanation, the more welcome it will be !

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You can read this page on 1-dimensional random walks mathworld.wolfram.com/RandomWalk1-Dimensional.html – utdiscant Apr 25 '12 at 10:19
I have read upto equation 9, but am unable to figure out how the table can be changed when p is not 1/2 – true blue anil Apr 25 '12 at 16:24

Call $P(n)$ the probability of reaching home if you are $n$ step from the pit. So $P(0)=0$, $P(10)=1$. You are looking for $P(1)$.
For $0<n<10$ $$P(n)=0.7*P(n+1)+0.3*P(n-1)$$ The general solution to this equation is $P(n)=a + b*(0.3/0.7)^n$ (see this link), and using the know value we can write: $P(0)=a+b*1=0$ and $P(10)=a+b*(0.3/0.7)^{10}=1$, and find $$P(n)=a+b*(0.3/0.7)^1=\frac{(0.3/0.7)^n-1}{(0.3/0.7)^{10}-1}$$ $$P(1)=\frac{(0.3/0.7)^1-1}{(0.3/0.7)^{10}-1}$$
The probability of reaching home is $\approx0.5715$ ($P(1)$), the probability of falling $1-P(1)$, and the probability of reaching the last step but don't reaching home is $\frac{(0.3/0.7)^1-1}{(0.3/0.7)^{9}-1}-\frac{(0.3/0.7)^1-1}{(0.3/0.7)^{10}-1}$ because P(reaching the last step AND don't go home) = P(reaching the last step) - P(reaching the last step AND go home) = P(reaching the last step) - P(go home).