Probability of consecutive disjoint events

Question

Let be $A$ and $B$ two disjoint events which represent outcomes of a stochastic experiment. We assume $P(A)+P(B)>0$. Show that if you repeat the experiment, then the probability that $A$ appears before $B$ is given by $\frac{P(A)}{P(A)+P(B)}$.

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I don't quite understand how to come up with this solution. In my opinion we have to distinguish between the outcomes of the different stages, i.e. $(A_1,A_2),(A_1,B_2), (B_1,A_2) (B_1,B_2)$. But we have no information about how those events of different stages are related to each other.

Actually we are looking for $$ P(\{(A_1,B_2)\})=P(A_1\mid A_1\cup B_1)\cdot P(B_2\mid A_1\cap (A_1\cup B_1)). $$ Rearranging and using the definition of the conditional probability delivers: $$ \begin{align*} &P(A_1\mid A_1\cup B_1)P(B_2\mid A_1\cap (A_1\cup B_1))=\frac{P(A_1)}{P(A_1\cup B_1)}\frac{P(B_2\cap A_1)}{P(A_1)}\\ &=\frac{P(A_1\cap B_2)}{P(A_1)+P(B_1)}=\dots\overset{???}{=}\frac{P(A)}{P(A)+P(B)}. \end{align*} $$ Is there any chance to prove the claim without any further assumptions? Maybe I am missing something?

Answer 1

Let the experiment be dice throw. A is that 1 appears; and B is that 2 appears. Let C be the outcome that 3,4,5,6 appear.

You have to repeat the experiment till either one of A or B appears. You may have to do multiple dice throws (may be >2).

Answer 2

You repeat untill you hit something in $A \cup B$ not just twice, if it were twice you have to guarantee that either $A$ or $B$ appears in every try , so in that case you should have: $P(A \cup B ) = 1 $ , but that is not the case. If you repeat untill you are in $A \cup B$ the chance of getting $A$ is just:

$$ \frac{P(A|A \cup B)}{P(A \cup B)} = \frac{P(A)}{P(A)+P(B)} $$

The way to think about this is that you don't care about other outcomes (not in $A \cup B$) thus the experiment is just modeled by a Bernoulli process in essence, as you truncate everything else.