Problem: I roll a dice until de sum exceeds 4, the dice is fair & every roll is independent, I'm gonna denote as $\{{X_{i}\}}_{i\geq 1}$. Let $\tau= min_{n\geq 1}(\sum_{i=1}^{n}X_{i}\geq 4)$ and I want to know the distribution of $X_{\tau}$. I wrote the event $\{\tau=k\}=\{\sum_\limits{i=1}^{k}X_i\geq4,\sum_\limits{i=1}^{k-1}X_i\leq3\}$ and $\tau$ only can take values 1, 2, 3 or 4 then $\mathbb{P}(X_\tau=1)= \sum_\limits{k=1}^4 \mathbb{P}(X_\tau=1, \tau=k)=\sum_\limits{k=1}^4 \mathbb{P}(X_k=1,\sum_\limits{i=1}^{k}X_i\geq4,\sum_\limits{i=1}^{k-1}X_i\leq3)= \mathbb{P}(X_1=1,X_1\geq4)+ \mathbb{P}(X_2=1,\sum_\limits{i=1}^{2}X_i\geq4,X_1=3)+ \mathbb{P}(X_3=1,\sum_\limits{i=1}^{3}X_i\geq4,\sum_\limits{i=1}^{2}X_i\leq3)+ \mathbb{P}(X_4=1,\sum_\limits{i=1}^{4}X_i\geq4,\sum_\limits{i=1}^{3}X_i\leq3)$
- Now the first one is zero $\mathbb{P}(X_1=1,X_1\geq4)=0$
- $\mathbb{P}(X_2=1,\sum_\limits{i=1}^{2}X_i\geq4,X_1=3)=\mathbb{P}(X_2=1,X_1=3)=\frac{1}{36}$
- $\mathbb{P}(X_3=1,\sum_\limits{i=1}^{3}X_i\geq4,\sum_\limits{i=1}^{2}X_i\leq3)=\mathbb{P}(X_3=1,X_2=1,X_1=2)+\mathbb{P}(X_3=1,X_1=1,X_2=2)=\frac{2}{6^3}$
- $\mathbb{P}(X_4=1,\sum_\limits{i=1}^{4}X_i\geq4,\sum_\limits{i=1}^{3}X_i\leq3)=\mathbb{P}(X_4=1,X_3=1,X_2=1,X_1=1)$
Questions: am I right? I mean are $\tau$ and $X$ not independent?