Q. 3.19 - A First Course in Probability

Question

Consider the gambler’s ruin problem, with the exception that $A$ and $B$ agree to play no more than n games. Let $P_{n,i}$denote the probability that $A$ winds up with all the money when $A$ starts with $i$ and $B$ starts with $N-i$. Derive an equation for $P_{n,i}$ in terms of $P_{n-1,i+1}$ and $P_{n-1,i-1}$, and compute $P_{7,3}$, $N=5$. 

Step-by-Step Solution

Answer

Value of $P_{7,3}$ in terms of $P_{n-1,i+1}$ and $P_{n-1,i-1}$ is $P_{7,3}=p^2+2p^{3}q+5p^4{q}^2$

Step1: Recursive probability

Computing factorials could be famous example of recursive programming. 

A number's factorial is calculated by multiplying it by all the numbers below it, up to and including one.

Step2: Find P 7 , 3

The recursive relation is like this,

$P_{n,i}=0$,  $i=0$ or $n<N-i$

       $=1$,     $i=N$

       $=p{P}_{n-1,i+1}+q{P}_{n-1,i-1}$     other cases

where $q=1-p$ is that the chance that $A$ will win the round, and $p$ is that the likelihood that $A$ will win the round?

Find $P_{7,3}$ with $N=5$,

$P_{7,3}=p{P}_{6,4}+q{P}_{6,2}$

$=p^2+pq{P}_{5,3}+pq{P}_{5,3}+q^2{P}_{5,1}$

$=p^2+2pq{P}_{5,3}+q^2{P}_{5,1}$

$=p^2+2p^{2}q{P}_{4,4}+2p{q}^2{P}_{4,2}+p{q}^2{P}_{4,2}$

$=p^2+2p^{2}q{P}_{4,4}+3p{q}^2{P}_{4,2}$

$=p^2+2p^{3}q+2p^2{q}^2{P}_{3,3}+3p^2{q}^2{P}_{3,3}$

$=p^2+2p^{3}q+5p^3{q}^2{P}_{2,4}$

$=p^2+2p^{3}q+5p^4{q}^2$