1. Randome Walk Problem

Peter has probability 1/4 of winning each game. Peter and Paul each bet $100 on each game. They each start with$400 and play until one of them goes broke. What is the probability that Paul goes broke?

How do you do this problem>?

I used this formula:
Suppose Peter starts with s dollars and Paul starts with t-s dollars. Let P* be the probability that Peter wins all the money. Then
P* = (1 -r^s) /(1-r^t)
where r = q/p (p not equal q).

Attempt:
s = 400 (Given)
t = 800 (from t-400= 400)
p = 1/4 (Given)
q = 3/4 (q=1-p) = (1-1/4)
r = 3 (q/p)

After that I substituted each value into the equation
P* = (1 -r^s) /(1-r^t)
P* = (1 -3^400) /(1-3^800)
P* = 1.41742 E-191 (This is wrong)

But when I tried s = 4 and t = 8
P* = (1 -r^s) /(1-r^t)
P* = (1 -3^4) /(1-3^8)
P* = 1/82 = 0.012195. (which is correct).

Can someone explain this to me? Why does my first attempt gives the wrong solution? When I tried another problem just like this (with s and t being single digits), it works well with direct substitution.

2. Originally Posted by jjbrian
Peter has probability 1/4 of winning each game. Peter and Paul each bet $100 on each game. They each start with$400 and play until one of them goes broke. What is the probability that Paul goes broke?

Ans: 0.01220
How do you do this problem>?

I used this formula:
Suppose Peter starts with s dollars and Paul starts with t-s dollars. Let P* be the probability that Peter wins all the money. Then
P* = (1 -r^s) /(1-r^t)
where r = q/p (p not equal q).

Attempt:
s = 400 (Given)
t = 800 (from t-400= 400)
p = 1/4 (Given)
q = 3/4 (q=1-p) = (1-1/4)
r = 3 (q/p)
When you do this you assume that they start with $400 and play$1 each game. Obviously the longer the game is, the less probability that Peter has of winning.

3. the information given (Peter and Paul each bet $100 on each game.They each start with$400 and play until one of them goes broke.) in the question is just for you to figure out the state space which is 1,2,3,4 in this case