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 $P_{7,3}$,$N=5$.

Short answer

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

Step-by-step solution

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.

Find P7,3

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$N=5$,

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

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

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

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

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

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

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

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