# Probability of Vehicles and Pedestrians meeting

#### Stone2594

##### New Member
Hi,

I need to produce a probability for a vehicle and a pedestrian meeting along a 1km road.

Here is what I know:
•cars take 3 minutes to travel
•pedestrians take 12 minutes to travel
•30 cars an hour
•20 pedestrians an hour

The cars and pedestrians may also be grouped or evening distributed on the road as well.

This is a simplified version of what I’m trying to achieve, but i need it simple so I can explain this to my non-mathematical boss. It’s so we can make an argument that safety is at risk with a proposed new development which will increase pedestrians and vehicles on this road.

I already have a method, I just want others views on this.

Thanks

#### katxt

##### Well-Known Member
The problem sounds interesting but it's not clear exactly what you want. How about you explain your method for a start, and we can discuss it. kat

#### Stone2594

##### New Member
I’m working on a lane with no walkway, and need to know the probability that a pedestrian will come across a car.

https://www.dropbox.com/s/2kf7agr86l5m3ev/Probability of Car meeting Pedestrian.xlsx?dl=0

Explanation is below.

Worst Case - Evenly Spaced
If all cars and peds were evenly spaced, then the probability of a car on that road during an hour is 100% (60mins/30cars = 2mins/car, 2mins/car<3min to travel) similarly for the peds. So 100% chance a ped will come across a car.

Best case - single group
If all cars were in 1 group (assume side by side for simplicity), then it would take the cars 3 minutes to drive on that road, or 5% of the hour (3mins/60mins)
And peds 12 minutes to walk, or 20% of the hour (12mins/60mins).
So the probability that they meet is 5% x 20% =1%.

Average
The result above cover a massive range and not ideal when presenting. So I want a mid-point and this isn’t simply finding the middle of these probabilities.
I approached it like this,
Car:
The max number of groups of cars per hour to equal the 3 min travel time is 20 (60mins/3mins).
This is 66.7% of the total 30 cars.
So if there are 20 groups or more, a car will 100% be on the lane at any time in that hour.
As for below 20 groups, I have simply found assumed on average this will be 10 groups (50% of 20 groups).
So 100%*33.3% + 50%*66.7% = 66.7% of a car on this lane at any point in the hour.
Peds:
Do the same approach for peds, and this results in 87.5%.
66.7%*87.5% = 58.3% of a pedestrian and car meeting on a lane.

Please let me know what you think.

Thanks

#### katxt

##### Well-Known Member
Here are my initial thoughts. It looks like you want the probability that any particular pedestrian (or group of peds) will meet at least one car somewhere along their 12 min walk.
I think that you can assume that cars are independent. If you know that in this part of the world, cars sometimes travel in groups, then use the number of groups rather than the number of cars (27 groups for instance, rather than 30 cars.)

A reasonable approach would be to first find the expected number of meets per walk. Split the 30 cars/hr into 15/hr each way and consider two situations - the cars going against the walker and the cars going with the walker.
First, 15 cars/hr going against the walker. With cars arriving at 1 per 4 minutes and taking 3 min to travel, we have an expected value of ¾ of a car already on the 1 km road. In the 12 minutes the walker takes, 3 more cars have entered the lane from the far end. The walker meets these three cars + the ¾ already on the road - an expected value of 3¾ cars met.
Next, 15 cars going with the walker. 3 cars enter the lane from behind during the 12 minute walk. All of these overtake the walker except ¾ still remaining on the road behind the walker at the 1 km mark. Expected meets 2¼.
In total the walker meets 6 cars on the 1 km walk. (I didn't anticipate that that would cancel so nicely so there is probably an easy way to show the average is 6.)

Now, assume that all walks and drives and meetings occur at random with an average rate of 6 per walk. This is the classic Poisson situation. The probability of 0 meets is given by exp(-6) = 0.25% near enough. So the probability of at least one meeting is 99.75%. You can now also calculate the probability of exactly 1, 2, ... meetings if that's of interest.

(Edited after a sudden thought.)

An alternative is to do a simulation which might appeal to your boss. Cheers. kat

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