No Stopping At Red Lights

2023/07/01

There’s something about red lights that I find irritating, and recently, they’ve become an unavoidable part of my daily routine.

Each day, I undertake a 20-minute stroll from Broadtec Royal International to my office situated near Beijing Jinhui Mansion, both framed in blue. Sounds idyllic, doesn’t it? What’s even better is the simplicity of the route. It’s a straight path along a gentle curve that eliminates any risk of me losing my way.

Despite these pleasures, my journey is regularly interrupted by the nuisance of five intersections (highlighted in red) with traffic lights along my route. More often than not, I find myself halted, at least once, before the red signal. It’s particularly frustrating when the sun is shining mercilessly overhead, and I’m rushing to reach my destination. I can’t help but wonder about the possibility of making this journey without a single stop at a red light, and I’m determined to find the odds.

Model

I recreated the route as seen below. In this simplified illustration, each hatched square symbolizes an intersection, and each rectangle signifies the stretch of road between these intersections. On my journey back home, I start from point $P_0$ and end at $P_{4xy}$. Needless to say, the reverse journey from home to my office follows the same pattern.

I also need the following assumptions to complete the model:

• Each traffic light shows only red and green (ignore yellow lights), and they last for an equal length of time
• Each time when I have 2 options to choose
  • On points like $P_0$, I choose whichever way I can see a green light (so, each way shares an equal probability)
  • On points like $P_{0y}$, I choose $P_{0y}\rightarrow P_{0xy}$ when there is a green light, otherwise I choose $P_{0y}\rightarrow P_1$ (also, each way shares an equal probability)
• Each time when I have only 1 option, either on point $P_{0x}$ ($P_{0x}\rightarrow P_{0xy}$) or $P_{0xy}$ ($P_{0xy}\rightarrow P_{1x}$), I have to wait whenever there is a red light

Based on these assumptions, I know the following initial probabilities
$$\begin{aligned} P_0 = 1, \\ P_{0x} = 0, \\ P_{0y} = 0, \end{aligned}$$

where $P_{0x} = 0$ and $P_{0y} = 0$ indicates I am still at $P_0$.

This is then changed to
$$\begin{aligned} P_0' = 0, \\ P_{0x}' = \frac{1}{2}, \\ P_{0y}' = \frac{1}{2}. \end{aligned}$$

For the routes $P_{0y}\rightarrow P_{0xy}$ and $P_{0x}\rightarrow P_{0xy}$, there is still a possibility that I come across a red light after a green light on the routes $P_{0}\rightarrow P_{0y}$ and $P_{0}\rightarrow P_{0x}$, which happens when the previous green light has not finished. For simplicity, I also assume

• I only need half of the green light time to finish $P_{0}\rightarrow P_{0y}$ or $P_{0}\rightarrow P_{0x}$, then I have to wait at the red light in such cases

So, the probability of $P_{0y}\rightarrow P_{0xy}$ and $P_{0x}\rightarrow P_{0xy}$ both become half of their previous ones, meaning that the probability of reaching the upper-right corner is
$$\begin{aligned} P_{0xy} &= \frac{1}{2}P_{0x}' + \frac{1}{2}P_{0y}', \\ &= \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}, \\ &= \frac{1}{2}. \end{aligned}$$

Solution

Let’s take a closer look at a general model unit as below, where $i=0,1,2,\cdots$.

For each iteration, we follow the same pattern, getting $P_{i+1}$ and $P_{(i+1)x}$ from $P_i$ and $P_{ix}$. After $(i+1)$th iteration,
$$\begin{aligned} P_{i+1} &= \frac{1}{4}P_{i}, \\ P_{(i+1)x} &= P_{ixy} \\ &= \frac{1}{2}P_{ix}' + \frac{1}{2}P_{iy}' + \frac{1}{2}P_{ix} \\ &= \frac{1}{2}P_i + \frac{1}{2}P_{ix}. \end{aligned}$$

So, for $n\geq1$, knowing the initial probabilities are $P_0 = 1, P_{0x} = 0$, we have
$$\begin{aligned} P_n &= \left( \frac{1}{4} \right)^n P_0, \\ P_{nx} &= \frac{1}{2}P_{n-1} + \frac{1}{2}P_{(n-1)x} \\ &= \frac{1}{2}\left( \frac{1}{4} \right)^{n-1} P_0 + \frac{1}{2}P_{(n-1)x} \\ &= \frac{1}{2}\left( \frac{1}{4} \right)^{n-1} + \frac{1}{2}P_{(n-1)x}. \end{aligned}$$

Noticing that I need to compute the probability of $P _{4y}\rightarrow P_{4xy}$ and add it to $P_{4xy}$, which is not included in the first 4 iterations, the final answer is actually $P_{5x}$. Then we have
$$\begin{aligned} P_{5x} &= \frac{1}{2}\cdot\left( \frac{1}{4} \right)^{4} + \left( \frac{1}{2} \right)^{2}\left( \frac{1}{4} \right)^{3} + \left( \frac{1}{2} \right)^{3}\left( \frac{1}{4} \right)^{2} + \left( \frac{1}{2} \right)^{4}\left( \frac{1}{4} \right) + \left( \frac{1}{2} \right)^5 + \left( \frac{1}{2} \right)^5 \cdot 0\\ &= \frac{1+2+2^2+2^3+2^4}{2^9} \\ &= \frac{2^5-1}{2^9} \\ &= \frac{31}{512} \\ &\approx 6.05\%, \end{aligned}$$

which is surprisingly low!

Let that sink in for a moment. The chance that I can walk my entire route without hitting a single red light is a slender 6.05%! The odds are slim, indeed, but isn’t there a certain allure to this unpredictability? Each day presents a new roll of the dice, a fresh challenge to conquer.

So the next time you find yourself stopping at a red light, perhaps you’ll share in my curiosity and wonderment. What are the odds of your journey being uninterrupted? While it might seem trivial, these are the sorts of mysteries that bring a bit of excitement to our daily routines. And who knows, maybe one magical day, the traffic lights will all align in my favor, and I’ll stroll through my route as if the red lights were merely illusions! Until then, I’ll keep tracking my progress and embracing the adventure that each new day brings.