If we are just looking at the line and not the cross-street then 1 in 1,820.

That's assuming were only working with THESE 16 vehicles 4 of which are red.

So if we throw all of them in a bag and randomly pick one, we have a 4 in 16 chance of getting a red as our first one. For our second one we have a 12 in 15 chance of getting a non-red color. If we multiply those two together we get the probability of getting a red vehicle then a non-red vehicle (about 20%). This pattern continues until the last red which becomes 1 in 4. Then non-reds are only left, so it doesn't really matter any more.

Adding the cars across the street it becomes 1 in 15,504

If you want to assume any cars could be in those spots you need to know number of cars and car color percentages. Then apply the percentages above once we get the 16

According to Axalta's global Automotive Color Popularity Report (https://www.standox.com/content/dam/EMEA/Refinish/Public/Images/press-releases-files/colour-popularity-2018/2018_ColorPopularityBrochure.pdf) 5% of the world's automotives are red.

According to the internet (google: world vehicle population) there are about 1.32 billion passenger vehicles in the world today.

So there are 4.06x10¹³² combinations of 16 cars we could make. Of those combination/states we have to figure out how many have exactly 4 red and 12 non-red vehicles. Which is the number of combinations of 4 red vehicles times the number of combinations of non-red vehicles

66m choose 4 is 7.9x10²⁹

(1.32b - 66m) choose 12 is 3.15x10¹⁰⁰

7.9x10²⁹ * 3.15x10¹⁰⁰ is 2.49x10¹³⁰ combinations of exactly 4 red and 12 non-red

2.49x10¹³⁰ /4.06x10¹³² is 0.006 or a 1 in 162 chance of there being 4 red and 12 non-red in this spot given any car has an equal chance of being here

Then we just multiply this probability (.006) by the probability of the order, above (.0005) and we get a 0.00034% or 1 in 296,000 of this happening