91Ó°ÊÓ

The probability that the person will win one dollar is 1/3 and the probability that the person will lose one dollar is 2/3.

By playing the game the person wants to win two dollars.

Need to prove the probability that before she loses her initial fortune she will achieve her goal is less than 1/4.

Here, given that:

\(\begin{aligned}{l}p = \frac{1}{3}\\(1 - p) = \frac{2}{3}\\k = i + 2\end{aligned}\)

It is know that,

\({a_i} = \frac{{{{\left( {\frac{{1 - p}}{p}} \right)}^i} - 1}}{{{{\left( {\frac{{1 - p}}{p}} \right)}^k} - 1}};\forall i = 1,2,...,k - 1\)

The probability of achieving the goal in the given criteria is

\(\begin{aligned}{c}{a_i} = \frac{{{{\left( {\frac{{1 - p}}{p}} \right)}^i} - 1}}{{{{\left( {\frac{{1 - p}}{p}} \right)}^k} - 1}}\\ = \frac{{{{\left( {\frac{{\frac{2}{3}}}{{\frac{1}{3}}}} \right)}^i} - 1}}{{{{\left( {\frac{{\frac{2}{3}}}{{\frac{1}{3}}}} \right)}^k} - 1}}\\ = \frac{{{2^i} - 1}}{{{2^k} - 1}}\\ = \frac{{{2^i} - 1}}{{{2^{\left( {i + 2} \right)}} - 1}}\\ = \frac{{1 - \left( {\frac{1}{{{2^i}}}} \right)}}{{4 - \left( {\frac{1}{{{2^i}}}} \right)}}\end{aligned}\)

Since,\(\frac{{1 - x}}{{4 - x}} < \frac{1}{4};\forall x\left( {0 < x < 1} \right)\)

Therefore, \({a_i} < \frac{1}{4};\forall i \ge 0\)[Proved]