Question

Suppose that you are gambling against an infinitely rich adversary and at each stage you either win or lose 1 unit with respective probabilities p and 1 − p. Show that the probability that you eventually go broke is 1 if $p\le \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and${\left(\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$p$}\right.\right)}^i$if $p>\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$where q = 1 − p and i is your initial fortune.

Short answer

This proves the statement.

Step-by-step solution

Given information

We know from Example $4j$it's clear that

$P_{n,m}=p{P}_{n-1,m}+(1-p)P_{n,m-1}$

where symbols have their usual meaning.

Step 2

Finally we obtain -

$P_{n,m}=\sum _{k=n}^{m+n-1}\left(\begin{array}{c}m+n-1\\ k\end{array}\right)p^k{(1-p)}^{m+n-1-k}$

Therefore we have, as the no. of trial tends to infinity, we get

$P=\left\{\begin{array}{l}1;p\le \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ {\left(\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$p$}\right.\right)}^i;p>\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}\right.$

where$(m+n)\to \infty$