Thread: A "Random Walk" Problem

Peter and Paul bet one dollar each on each game. Each is willing to allow the other unlimited credit.Use a calculator to make a table showing, to four decimal places, for each of p =1/10, 1/3, 0.49, .499, 0.501, 0.51, 2/3, 9/10 the probabilities that Peter is ever ahead by $10, by $100, and by $1000.

Im not too sure what to do with this problem. Can someone help me get started with this?

Attempt. For P = 1/10 and Peter is ahead by $10.
Q = 9/10

I tried using this formula
P* = (1-r^s)/(1 - r^t)
where
P* is the probability that Peter wins all the money.
r = q/p (q not equal p)
Peter starts with s dollars
Paul starts with t - s dollars.

But the problem does not give the initial amount for each person.

Okay, another attempt. The probability that peter is ever ahead by k games is
{1 if p>=q
{ (p/q)^k if p < q


Im doing the case where Peter is ahead by $10.
p =1/10, 1/3, 0.49, .499, 0.501, 0.51, 2/3, 9/10

by above since p >=q 0.501, 0.51, 2/3, 9/10.
The probability that Peter is ever ahead is 1.

For p = 1/10 , q = 9/10 , k =10
I got 2.867E-10

For p = 1/3 , q = 2/3 , k =10
I got 1/1024

For p = 0.49 , q = 0.51 , k =10
I got 0.6703

For p = .499 , q = .501 , k =10
0.960789

I definitely not too sure about this method. Please leave a comment if you have any ideas.