Car, gas, crow problem - looking for analytic solution I made up this problem recently.

possible configuration
Imagine you are lost in a car on a long circular road in a hot desert.
You have enough gas to go halfway around the circle. There is a gas
station that is your only hope for survival, but you don’t even know
where you are, let alone which direction will get you to the gas.
With this much information, your best option is to pick a direction
and go that way until you reach the gas or run out. Your chances of
survival are 50%.
Now imagine there is also a dead crow on the side of the road. All
three objects are randomly distributed around the circle with all
locations equally likely. Also, you are told which direction will get
you to the crow before the gas station. Does this help? What should
you do? What are your chances?

I have the sketch of a solution, but it leans on a simulation. Is there a purely analytic solution?

 The gas station and the car divide the road into two segments, and you
 need the smaller segment to survive. The randomly placed crow is more
 likely to be in the larger segment, simply because it’s larger. You
 should go in the other direction.

 coordinate system

 To avoid considering cases
 separately, we define a coordinate system going from the car opposite
 the crow-first direction with circle circumference 1, so g < c. We
 know that 0 ≤ g < 1 but the probability of g → 1 is vanishingly small
 since it would also require that c → 1. It is reasonable to assume
 (and simulation confirms) that the probability density function for g
 is linear so it must be

 probability density function

 The probability of survival is the sum of
 the areas shown, p(g<1/2) = 3/4.

 A: I agree that if you know the direction in which the crow appears before the gas station, then going the opposite direction increases your chances of survival from $1/2$ to $3/4$.
First let's reformulate your problem statement. Suppose your current position is $0$, and use angular displacement from your car to mark points on the road. Then the position of the gas station $X$ and the crow $Y$ are uniformly distributed over $[0,2\pi)$.
If $X<Y$, then with our strategy, we survive iff $X\leq\pi$. We can use Bayes' theorem to compute the survival probability in this case:
$$\mathbb P(X\leq\pi\mid X<Y)=\frac{\mathbb P(X<Y\mid X\leq\pi)\cdot\mathbb P(X\leq\pi)}{\mathbb P(X<Y)}=\mathbb P(X<Y\mid X\leq\pi),$$
since $\mathbb P(X\leq\pi)=\mathbb P(X<Y)=1/2$. But we can break down the last probability by conditioning on the value of $Y$:
\begin{align*}
\mathbb P(X<Y\mid X\leq\pi)&=\mathbb P(X<Y\mid X\leq\pi, Y\leq\pi)\cdot\mathbb P(Y\leq\pi)\\
&\qquad+\mathbb P(X<Y\mid X\leq\pi, Y>\pi)\cdot\mathbb P(Y>\pi)\\
&=\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{2}\\
&=\frac{3}{4}.
\end{align*}
So when $X<Y$ our probability of survival is $3/4$. By an analagous argument, when $X>Y$ our probability of survival is $3/4$, so our new strategy gives us a $3/4$ chance to survive, as you claimed.
