Consider a gambler whose initial fortune is $2$ dollars, and who repeatedly makes fair bets in which his fortune either increases or decreases by $1$ dollar. He keeps betting until he runs out of money. Given that the gambler got broke on his $10th$ bet, what is the probability that his wealth trajectory went upto $6$ dollars and then crashed down to zero?
From what I understand, isn't there just one possible case in which this can happen? $+2, +3... +6, +5, ...0$? So shouldn't the probability just be $\frac{1}{1024}$? ($1/2$ for each state change?) I know I am missing something but I am unable to figure it out. It has something to do with conditional probability I think but I don't know what exactly is wrong here.
You are asked for the conditional probability that the one case you mentioned occurred, given that the gambler went broke on the $10$'th bet. By definition, $$ \mathbb P(A \mid B) = \dfrac{\mathbb P(A \cap B)}{\mathbb P(B)}$$ You have found $\mathbb P(A \cap B)$, but you need to divide by $\mathbb P(B)$.
EDIT: Let $f(n,b)$ be the probability that it takes exactly $n$ bets to go broke starting with $b$ dollars. Thus $f(0,0)=1$, $f(n,0) = 0$ if $n \ge 1$, $f(n, b) = 0$ if $b > n$, $f(n,b) = (f(n-1,b-1) + f(n-1,b+1))/2$ for $1 \le b \le n$. You want $f(10, 2)$. Compute $f(1,1)$, $f(2,2)$, $f(3,1)$, $f(3,3)$, etc.