Let us define "fastness" as a characteristic of how fast/agressive a driver is. One aspect of this is the (maximum) time between a traffic light turning red and the driver deciding they have to stop. The assumption here is that a "slow" driver will have low (read: very negative, like minus 5 seconds) value because they start stopping as soon as they notice the orange traffic light. On the contrary, a "fast" driver will have a higher value, like -1 second or even a positive one like 3 seconds, being "a bit" illegal.

(Disclaimer: fast and slow, as words, tend to give a positive "taste" on fast rather than slow. I chose them because they have few letters, and I do not advocate dangerous driving!)

By using this "time from red light to stop" as a random variable X, let's assume a normal distribution of drivers with known parameters m(average) and v(variance).

If that is so, what is the probability distribution that, when being stopped at a red light yourself, the driver at the frontmost of the queue is X-fast?

I guess slow drivers are more probable to appear there because they stop more easily/frequently. The logic behind the solution is more important than formulas here, and that is why I chose to post this here instead of math.se .

  • $\begingroup$ You mention a queue near the end. Are drivers one behind another in a line or are they in parallel lanes independent of each other? Does everyone notice the orange light, or just the brake lights in front? And are crashes permitted? $\endgroup$
    – flinty
    May 13, 2021 at 15:00
  • $\begingroup$ Well, rule of thumb is to create as simple a model as possible. Regarding your questions, let's assume one lane, everyone has full and immediate light knowledge, no crashes, no intersection space (as soon as you pass the light you are on safe zone), and also everyone travels at same speed (so the fastness of the car in front of you does not affect your decision to continue...only your distance from the traffic lights). Feel free to make any other assumption to make calculations easier. $\endgroup$ May 13, 2021 at 17:32
  • $\begingroup$ you may need to define the distribution of red lights. Otherwise if there are no red lights then every driver is the same. $\endgroup$ May 13, 2021 at 23:54
  • $\begingroup$ @Dmitry Kamenetsky I don't understand, could you elaborate? All I know is some roads have traffic lights. $\endgroup$ May 14, 2021 at 6:18
  • $\begingroup$ if there are no traffic lights then there is no distribution. $\endgroup$ May 14, 2021 at 11:24

2 Answers 2


The probability that it's a fast driver in front of you stopped at the light depends on only a few things - the overall prevalence of fast drivers, and the proportion of times fast/slow drivers run red lights. I don't see anything in the question that determines how to relate the stopping time to the likelihood of running the red light, but I'll assume we can calculate it. No matter what the underlying variables are of driver speed, reaction time, and stopping distance, all that really matters is the overall likelihood of someone failing to stop at a red light.

If fast drivers never run a red light, the odds of a fast driver in front of you is simply the proportion of fast drivers on the road. Everyone stops at every red light, so there is nothing different about fast drivers and slow drivers. Only when fast drivers run red lights do they become less likely to be stopped at a red. Simply multiply the proportion of fast drivers by the proportion of fast drivers who stop at the red light, and re-normalize against the remaining proportion of slow drivers (who I assume never run a red light) - this yields the proportion of fast drivers among all the drivers who stopped at the red light.

Example: 60% of drivers are slow and never run a red light, and 40% are fast drivers, 25% of which run red lights. That means that 60% of the population are slow drivers who stop at red lights, 30% are fast drivers who stop at red lights, and 10% are fast drivers who don't stop at red lights. Of the 90% of the population that actually stops at a red light, 2/3 are slow drivers and 1/3 are fast drivers. Despite fast drivers making up 40% of the population, you only stand a 33% chance of having one in front of you at the light, since some of them won't stop at all.


There are two elements to this. First,

The probability of the driver at the frontmost of the queue being X-fast depends directly on the probability of any driver being X-fast.


It also depends on the probability of the driver having reached the light X seconds or more after it turned red.


Consider that, if we set the second element, for example if we knew that the driver reached the light right as it turned red (X=0), the probability of them being X-fast is 0 for every value of X > 0 (since they would have just passed through it) while the rest of the spectrum doesn't get affected. This is because all of them would stop at the light if they were X-fast with X < 0.


The probability of a stopped driver being 0-fast is the probability of any driver being 0-fast times the probability that they reached the light after it turned red (X=0). We can assume that the distribution of any driver to reach the light at a certain moment is uniform. So this multiplication factor is 1 for the moment the light turns orange (say X=-5) since its certain that they reached the light after that or else it would have been green and they would have passed through. And it is 0 for the moment the light turns green again, or at least infinitesimal. It is a straight line between those two points, with a constant negative slope.

So the answer is,

The probability distribution that, when being stopped at a red light yourself, the driver at the frontmost of the queue is X-fast is the distribution of the X-fast drivers, which we assumed normal, times a multiplicative factor that is 1 for the minimum X possible in that specific light (which is the number of seconds the orange light stays on), 0 for the maximum X possible in that specific light (the number of seconds the red light stays on), and decreases constantly.

If you want to consider an additional factor,

It also depends on the moment you, as an observer, reached the light and first saw the frontmost driver there waiting. If you arrive at moment x, the probability of the driver being (X>x)-fast is 0, since they would be going through, while the rest of the distribution remains the same. This is another multiplicative factor that is either 0 or 1. At the very end, you would have to normalize the distribution so that the area below it equals 1.

Final distribution

Sorry for the lame drawing. I think this is at least a partial solution, but I would be happy to be corrected if there is a flaw in my logic.


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