Question

Consider the unfavourable game in Example 2.4.2. This time, suppose that the initial fortune of gambler A is i dollars with i ≤ 98. Suppose that the initial fortune of gambler B is 100 − i dollars. Show that the probability is greater than 1/2 that gambler A losses i dollars before winning 100 − i dollars.

Short answer

The probability is greater than 1/2 that gambler A losses i dollars before winning

100 − i dollars.

Step-by-step solution

Given information

initial fortune of gambler A is i dollars with i ≤ 98

initial fortune of gambler B is 100 − i dollars

Total fortune\(k = 100\)

\(p = 0.4\)

Computing the probability

Assuming that it is an unfair play

Required probability is given by\({a_i}\)

Let\({a_i}\)denote the probability that the fortune of gambler A will reach k dollars before it reaches 0 dollars, given that his initial fortune is i dollars.

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

Let\(\left( {\frac{{1 - p}}{p}} \right) = r\)

Then\({a_i} = \frac{{{r^i} - 1}}{{{r^k} - 1}}\)where\(i = 1,2,...,k - 1\)

Here p=0.4

So

\(\begin{aligned}{c}r = \frac{{1 - p}}{p}\\ = \frac{3}{2}\end{aligned}\)

Which is greater than 1 and\(k > i\)

Then from the above formula

\(\begin{aligned}{l}{a_i} = \frac{{{{\left( {\frac{3}{2}} \right)}^i} - 1}}{{{{\left( {\frac{3}{2}} \right)}^k} - 1}}\\ = \frac{{{{\left( {1.5} \right)}^i} - 1}}{{{{\left( {1.5} \right)}^{100}} - 1}}\end{aligned}\)

For\(i = 98\),\({a_i} = 0.44\)

For\(i = 97\),\({a_i} = 0.30\)

And so on

So the total probability that gambler A losses i dollars before winning 100 − i dollars

=0.44+0.30+…

Which is >0.5

Hence,

\({a_i} > \frac{1}{2}\)