Question

Suppose that on each play of a game, a gambler either wins 1 with probability p or loses 1 with probability 1–p (or q). The gambler keeps betting until she or he is either up a total of n or down a total of m. What is the probability the gambler will quit an overall winner? You must consider both cases when p = 0.5 and when p ≠ 0.5

Answer 1

Explanation:

From the given information,

Considering both cases when p = 0.5 and when p ≠ 0.5

the probability that the gambler will quit an overall winner is:

`P = (1 - ((1-p)/(p) )^K)/(1- ((1-p)/(p))^N ) \ \ \ is \ p \\eq 0.5 \ and\ K/N = (1)/(2)`

where ;

N.k = n and k = m

Hence, the probability changes to:

`P = (1 -((1-p)/(p))^m)/(1 -((1-p)/(p))^(m+n))` is p ≠ 0.5 and k/N =
`(m)/(m+n)` is P = 0.5