Probability - what is the probability that a given tire length will lie in a lenght of interest

I'm going to re-state this questions a little bit, just to settle on the variable names. Let $T$ be the circumference of the tire, let $R$ be the length of the arc on the tire that's painted red, and let $L$ be the length of the road. We'll assume $R>T$, otherwise it's a very strange, short road. (Or an absurdly large tire!)

I'm going to divide this question into two parts. In the first part, we'll imagine that $T$ divides $L$ perfectly. In the second part, we'll see what we can do if that's not the case.

Imagine that $T$ divides $L$ perfectly. That is, if we start on the road segment exactly before the red paint, then we end on the road segment exactly after the red paint. Think about cutting the road up into pieces the size of the tire's circumference $T$, and then overlapping those pieces of road together in your imagination. Now fix a point on the tire in your head. When we roll the tire down on the road, that little point will be in the exact same spot on all those road slices. Our question in this case is: what's the probability that the paint of length $R$ remains within the interval $T$ of the road? The answer there is simply $\frac{T-R}{T}$. This matches well with our intution, too, since a greater painted segment $R$ means there will be a greater likelihood that it will "paint" one of the end points.

The likelihood that $T$ divides $R$ is very low, however it helps us lead into the main point. If we cut up the road into pieces of size $T$, but then have a little bit left over with length $X < T$, it will turn out that the part of the tire that hits the beginning of the last road length will be the same part of the tire that hits the beginning point $A$. This means we can think of the probability that the interval $R$ will be inside $X$.

If $X < R$, then the probability will be zero. For instance, suppose 40cm of a 50cm tire is painted, and the road is 230 cm. If the tire somehow doesn't hit the first point $A$, it will definitely hit the last point $B$, no matter how you try to position the tire at the start.

If $X>R$, then the probability will be $\frac{X - R}{X}$, with similar reasoning as before.

We can get $X$ from $L$ by computing $X = L - T*(\lceil{\frac{L}{T}}\rceil - 1)$. The $\lceil F \rceil$ means we round $F$ up to the nearest integer.

You mention asking what the probability would be if we assigned a random distance $L$ to the road. Without a probability distribution for $L$, I can't really answer that.

I'll give you one more piece of information: what we did in "dividing" the road into similar parts is very similar to examining the quotient space of the road. If you liked what we did in analyzing how the tire hit the road similarly, you should check out quotient spaces a bit more. They're a bit confusing to start off with, but once you get it you'll start finding them everywhere.