MATH 367, fall 98, CHAPTER 8 Random Walks

1. Peter has unlimited credit. The probability that Peter wins in each game is p. Then the probability that Peter is ever ahead by k games is:

\{ \begin{matrix}1 & if p ≥ q \\ (\frac{p}{q})^{k} & if p < q \end{matrix}

2. Peter has s dollars and Paul has t − s dollars. Between both of them there are t dollars in play. Let p_{s} = the probability that Peter ends up with all the money (if he starts with s dollars). Then

p_{s} = \frac{1 − (q/p)^{s}}{1 − (q/p)^{t}} ifp ≠ q

and p_{s} = s/t if p = q = 1/2.

3. Duration of the play: X = net amount that Peter wins overall, Y the number of games played. Then

E(X) = E(Y)(p − q)

If Peter starts with s and t is the total amount of money in play,

E(X) = (t − s)p_{s} − s(1 − p_{s}) = tp_{s} − s

So,

E(Y) = \frac{tp_{s} − s}{p − q} if p ≠ q

E(Y) = s(t − s) if p = q = 1/2

If Peter has unlimited credit and Peter plays till he is k dollars ahead, then E(X) = k. So

E(Y) = \frac{k}{p − q} if p ≠ q

otherwise (if p = 1/2) they play forever.

Homework: read chapter 7 and do problems 1, 2, 3, 4, 5, 9, 11, 12, 25–29, 31, 33.