Examples
1. John starts with $2, and p = 0.6: What is the probability that John obtains a fortune of
N = 4 without going broke?
SOLUTION i = 2, N = 4 and q = 1 − p = 0.4, so q/p = 2/3, and we want
P
2
=
1 − (2/3)
2
1 − (2/3)
4
= 0.91
2. What is the probability that John will becom e infinitely rich?
SOLUTION
1 − (q/p)
i
= 1 − (2/3)
2
= 5/9 = 0.56
3. If John instead started with i = $1, what is the probability that he would go broke?
SOLUTION
The probability he becomes infinitely rich is 1−(q/p)
i
= 1−(q/p) = 1/3, so the probability
of ruin is 2/3.
1.2 Applications
Risk insurance business
Consider an insurance company that earns $1 per day (from interest), but on each day, indepen-
dent of the past, might suffe r a claim against it for the amount $2 with probability q = 1 − p.
Whenever such a claim is suffered, $2 is removed from the reserve of money. Thus on the
n
th
day, the net income for that day is exactly ∆
n
as in the gamblers’ ruin problem: 1 with
probability p, −1 with probability q.
If the insurance company starts off initially w ith a reserve of $i ≥ 1, then what is the
probability it will eventually get ruined (run out of money)?
The answer is given by (4) and (5): If p > 0.5 then the probability is given by (
q
p
)
i
> 0,
whereas if p ≤ 0.5 ruin will always ocurr. This makes intuitive sense because if p > 0.5, then
the average net income per day is E(∆) = p − q > 0, whereas if p ≤ 0.5, then the average net
income per day is E(∆) = p − q ≤ 0. So the c ompany can not expect to stay in business unless
earning (on average) more than is taken away by claims.
1.3 Random walk hitting probabilities
Let a > 0 and b > 0 be integers, and let R
n
denote a simple random walk with R
0
= 0. Let
p(a) = P (R
n
hits level a be fore hitting level −b).
By letting a = N − i and b = i (so that N = a + b), we can imagine a gambler who starts
with i = b and wishes to reach N = a+b before going broke. So we can compute p(a) by casting
the problem into the framework of the gamblers ruin problem: p(a) = P
i
where N = a + b,
i = b . Thus
3