Why is this method incorrect?

This is similar to the question, "does $\displaystyle \int \frac{f(x)}{g(x)} dx= \frac{\int f(x) dx}{\int g(x) dx}?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = \frac{\int_0^1 (\theta/2)d\theta}{\int_0^1 (\theta/2 + 1/2)d \theta}=\frac{1/2}{1 + 1/2} = \boxed{1/3}.$$

The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.

Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.

Your ${\theta \over \theta + 1} = P(L | \theta, S)$. So if you integrate $\int_0^1 {\theta \over \theta + 1}\; d\theta = \int_0^1 P(L | \theta, S)\; d \theta$, since $\theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!

You needed something akin to the pdf of $\theta$, conditioned on $S$, shown in red below:

• $P(L|S) = \int_0^1 p(L, \theta | S)\; d \theta = \int_0^1 P(L | \theta, S) \color{red}{p(\theta | S)}\; d \theta$

Frankly, finding $p(\theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:

• $p(L, \theta | S) = p(L, \theta, S) / P(S)$

• $p(L, \theta, S) = \theta / 2$

• So, $P(L|S) = \int_0^1 {\theta / 2 \over P(S)} \; d\theta = {\int_0^1 (\theta /2) d\theta \over P(S)}$ and you are back to the answer by @paulinho