1
Gambler’s Ruin Problem
Consider a gambler who starts with an initial fortune of $1 and then on each successive gamble
either wins $1 or loses $1 independent of the past with probabilities
p
and
q
= 1

p
respectively.
Let
R
n
denote the total fortune after the
n
th
gamble. The gambler’s objective is to reach a total
fortune of $
N
, without ﬁrst getting
ruined
(running out of money). If the gambler succeeds,
then the gambler is said to
win
the game. In any case, the gambler stops playing after winning
or getting ruined, whichever happens ﬁrst. There is nothing special about starting with $1,
more generally the gambler starts with $
i
where 0
< i < N
.
While the game proceeds,
{
R
n
:
n
≥
0
}
forms a simple random walk
R
n
= Δ
1
+
···
+ Δ
n
, R
0
=
i,
where
{
Δ
n
}
forms an i.i.d. sequence of r.v.s. distributed as
P
(Δ = 1) =
p, P
(Δ =

1) =
q
=
1

p
, and represents the earnings on the succesive gambles.
Since the game stops when either
R
n
= 0 or
R
n
=
N
, let
τ
i
= min
{
n
≥
0 :
R
n
∈ {
0
,N
}
R
0
=
i
}
,
denote the time at which the game stops when
R
0
=
i
. If
R
τ
i
=
N
, then the gambler wins, if
R
τ
i
= 0, then the gambler is ruined.
Let
P
i
=
P
(
R
τ
i
=
N
) denote the probability that the gambler wins when
R
0
=
i
. Clearly
P
0
= 0 and
P
N
= 1 by deﬁnition, and we next proceed to compute
P
i
,
1
≤
i
≤
N

1.
The key idea is to condition on the outcome of the ﬁrst gamble, Δ
1
= 1 or Δ
1
=

1, yielding
P
i
=
pP
i
+1
+
qP
i

1
.
(1)
The derivation of this recursion is as follows: If Δ
1
= 1, then the gambler’s total fortune
increases to
R
1
=
i
+1 and so by the Markov property the gambler will now win with probability
P
i
+1
. Similarly, if Δ
1
=

1, then the gambler’s fortune decreases to
R
1
=
i

1 and so
by the Markov property the gambler will now win with probability
P
i

1
. The probabilities