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

Answer:

Step-by-step 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:

$P = \dfrac{1 - (\dfrac{1-p}{p} )^K}{1- (\dfrac{1-p}{p})^N } \ \ \ is \ p \neq 0.5 \ and\ K/N = \dfrac{1}{2}$

where ;

N.k = n  and k  = m

Hence, the probability changes to:

$P = \dfrac{1 -(\dfrac{1-p}{p})^m}{1 -(\dfrac{1-p}{p})^{m+n}}$ is  p ≠ 0.5   and k/N = $\dfrac{m}{m+n}$  is P = 0.5