# Probability of 15 consecutive green lights

## Introduction

Upon a trip home, my mother and I were noticing a very peculiar occurrence: Traffic lights were almost continuously green. Indeed, exactly fifteen different traffic lights were green consecutively. Now, I am bad at probability, but this seems unlikely.

## Probability Application

I reasoned that since there are three different options for all fifteen traffic lights, the probability of fifteen traffic lights being consecutively green was one in $_{15}P_{3}$. This is because there is only one sequence of traffic light configurations where all of them are green and we must count all of the possible traffic light configurations using a permutation since a sequence such as G,R,Y is not the same as Y,R,G. With that said, the probability of the event comes out to be approximately $0.0366\%$.

## Question

Is this application of probability correct, incorrect, or somewhere in the middle? To make this question very precise, 'somewhere in the middle' means that my application of probability makes many underlying assumptions and ignores many factors. I am not aware what these specific assumptions and factors may be (if they exist), but that's why I ask this question. In what sense am I correct, and in what sense am I incorrect?

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Often, traffic lights are synchronized, meaning that the independence model is not good. So if traffic is not too bad, and you are driving at more or less standard speed, you should hit green fairly often, though $15$ seems on the high side. –  André Nicolas Nov 29 '12 at 21:52
Traffic lights are often timed to produce the effect you noticed, at least on major roads in the direction of the most traffic. Ideally, one would want it on all roads, but that is rather difficult to accomplish. –  Harald Hanche-Olsen Nov 29 '12 at 21:52
Incidentally, the probability of encountering every light being red is much easier to calculate: it is 100% if you are late for something. –  Arkamis Nov 29 '12 at 21:56
By the way, the calculation you made is not the correct one even under the dubious assumption of independence, and the incorrect assumption that all three states are equally likely. The probability of $15$ green in a row would then be $\frac{1}{3^{15}}.$ –  André Nicolas Nov 29 '12 at 22:02
What is $_{15}P_{3}$? The number of permutations of three things from a pool of $15$? I don't see how that could possibly apply here. I think you are trying to compute $3^{15}$: $3$ options that need to be set for $15$ distinguished lights. By your reasoning, this makes the probability of all green much smaller than $0.0366\%$. But as others have noted, the independence and equiprobability (1/3 red, 1/3 yellow, 1/3 red) assumptions are not realistic. –  alex.jordan Nov 29 '12 at 22:02

If red, yellow, and green were equally likely and independent, the probability of getting $15$ greens in a row would be $1$ in $3^{15}$ or about $1$ in $14$ million. The probability of getting $15$ of something in a row would be $1$ in $3^{14}$ or about $1$ in $5$ million. Actually, yellow usually occupies much less time than red or green. Again, if we figure the chance of red and green to be $\frac 12$, you would have $1$ in $2^{15}=32768$ of hitting $15$ greens in a row. I think this is a demonstration that they are not independent. We have a street in San Francisco that I routinely hit six or eight in a row (after maybe hitting a red and waiting for the green) by driving exactly the speed limit.