Probability of an event based on percentage in fixed lapse of time.

Question

I am a software engineer. I am also a former triathlete that rides with a large group of friends every time we have a chance.

i am trying to come up with a little software to distribute among us riders, where it "sort of" predict the probability that you may fell. You know what they say... two kind of riders, those who fell, those who will.

So, here the numbers.

We are $98$ riders (give and take). Last year $4$ of them fell. Rough eye calculation is $4$%. Let's assume that we ride every Sunday of the year ($52$).

What are the probabilities that you will fall this year? (Given these numbers).

I tried to apply some formulas that I found in Quora, Mathpage and other math resources but the resulting numbers don't make sense to me. At least not to be able to say in plain english... "you have $1$ in $30$ chances to fall this year"

Does it apply the same in February than November?

Thank you so much.

Answer 1

Without more information, your best estimate of a rider falling in one year is $\frac{4}{98}=0.0408...$.

Assuming a constant probability of falling on any ride (and constant for all riders), this works out to a probability of about 0.0008 of falling on any given ride.

Of course, depending on the fall, one may be so injured that one does not ride the next ride, or the next, etc., so there is likely some interdependence here.

As well, if a rider falls on one ride, people may be apt to be more cautious on the next few rides, which would make these probabilities non-constant.

Further, some conditions may cause more falls. In particular, ice leads to falls, so probabilities of falling, for a given length ride, may be higher in winter. On the other hand, people may ride shorter distances in winter, so it is not clear how the probability changes. (I fell on ice a few weeks ago!)

In short, you'll need a lot more (hard to get) information to do anything besides just a rough 0.008 probability of a rider falling on any ride.

Edit: Here are a few technical details. Suppose the probability of falling on a ride is $p$. Then the probability of not falling on a ride is $1-p$, and the probability of not falling in year of rides is $(1-p)^{52}$ (assuming each ride is independent of every other ride). Thus the probability of falling is $$1-(1-p)^{52}.$$
Since out best estimate for this value is $\frac{4}{98}$, we solve the equation $$ 1-(1-p)^{52}=\frac{4}{98}$$ to find $p=0.0008010769740...$, or about $0.0008$.

Answer 2

Given the single data point that $4$ out of $98$ riders fell in a year, all you can assume is that all riders have the same chance to fall in a year, which is $\frac 4{98} \approx 4\%$. If you then know that some riders ride more than others, you might choose to assume a rate of falling per mile. You would then have to compute the total number of miles ridden by the riders in a year. The result will be a rider who rides twice as much as another will have twice the chance of falling. If your weather is like mine, it only rains in the winter. You might think that there is more chance of falling in the winter than the summer. To assess this, you need to have data on when those falls occurred. Maybe there is zero chance of a fall when the roads are dry and all the falls happen when the roads are wet.

[TL,DR] All of which is a long way of saying that without a lot more detailed data you cannot give a reasonable estimate.