STATISTICAL ANALYSIS probability

Question

Here is a modification of the players’ ruin problem in L2.

Two players A and B plays a series of independent games, with A’s probability of winning for each game = p and B’s probability of winning = 1 – p. When A wins, B pays A $1, and when B wins, A pays B $1. A starts with $M, and B starts with an infinite amount of money (i.e. B will never ruin). The whole game will end only when A runs out of money (i.e. “ruins”). (We do not specify A’s maximum amount of money for the game to end; or equivalently, we assume that the end will continue whenever A has a positive amount of money.)


Show that A’s probability of getting ruined = (not A’s probability of winning)

  1     if t = q/p ≥ 1 (i.e. p ≤ ½); 

t^M   if t = q/p < 1 (i.e. p > ½).


[Hint: You need to consider 3 different cases: t > 1; t = 1; and t < 1.]

Answer 1

Consider the gambler’s capital after each game as a random walk. Let P(k) be the probability that A will eventually be ruined given that he currently has k dollars. We know

• P(0) = 1, because once A has no money he is already ruined.

• For k ≥ 1, after one game his capital is k+1 with probability p (he wins $1) and k-1 with probability q=1‑p (he loses $1). Conditioning on the first game gives a difference equation

P(k) = p·P(k+1) + q·P(k‑1).

To solve, we look for a solution of the form P(k) = C·t^k + D, where t = q/p (with t =1 when p = q = 1/2). Substituting into the recurrence shows this is indeed the general solution.

**Case 1: t > 1 (that is p ≤ q).** Here the random walk drifts downwards. Because t > 1, the term C·t^k would grow without bound as k increases. Since ruin probabilities are bounded by 1, we must take C = 0. Using the boundary condition P(0)=1 gives D = 1. Thus P(k)=1 for every starting capital. In words, if A is not as likely to win as to lose (p≤5⁄½), the walk is transient to the left, and no matter how much capital A starts with he will eventually reach 0 with probability 1.

**Case 2: t = 1 (p=q=1/2).** When the games are fair the recurrence reduces to P(k) = 1/2(P(k+1)+P(k‑1)). The general solution is linear: P(k) = C k + D. To keep the probabilities bounded as k→∞ (there is no upper boundary), we must set C=0. Using P(0)=1 again forces D=1. Hence P(k)=1 for all k. A symmetric random walk in one dimension is recurrent, so with probability 1 it will eventually hit 0.

**Case 3: t < 1 (p > q).** When A’s chance of winning each game is larger than B’s, the random walk has a positive drift. The t^k term now decays to zero as k increases, so we let D=0. Applying the boundary condition P(0)=1 determines C=1. Therefore P(k)=t^k. Since t=q/p<1, the probability of eventual ruin decreases geometrically in the starting capital. For example, if p=2/3 and q=1/3 then t=1/2 and a player beginning with $M has ruin probability (1/2)^M.

Putting these cases together we obtain the result you were asked to show:

\[

P(\text{ruin})=\begin{cases}

1,&\text{if }t=\frac{q}{p}\ge 1\;(p\le \frac{1}{2}),\\[6pt]

\left(\frac{q}{p}\right)^M,&\text{if }t=\frac{q}{p}<1\;(p>\frac{1}{2}).

\end{cases}

\]

(The middle case p=q also yields 1.) This formula captures the intuition that if A wins less often than B, ruin is certain, whereas if A wins more often his probability of ruin decays exponentially with his initial capital.