# Skydiving Risk Assessment

## Introduction

This post gives a simple risk assessment of skydiving and other such activities that have added risk of death. This post is not a recommendation; I am not a doctor.

Let’s start by quantifying risk of death in terms of micromorts, where 1 micromort is equal to $10^{-6}$ probability of death (micro is $10^{-6}$, mort is spanish for death). From wikipedia, below are micromort numbers associated with travel:

- Travelling 1000 mi by motorcycle: ~170
- Travlling 1000 mi by walking: 60
- Travelling 1000 mi by car: 4
- Travelling 1000 mi by jet: 1

By comparison, below are the numbers for some extreme sports:

- Skiing: 0.7 per day
- Marathons: 7 per run
- Skydiving: 8 per jump
- BASE Jumping: 430 per jump
- Mountaineering (Ascent to Matterhorn): ~2800 per ascent attempt
- Mountaineering (Ascent to Everest): ~38,000 per ascent attempt

Finally, for perspective here are some miscellaneous risks:

- Ecstasy: 0.5 per tablet
- All causes: ~20 per day
- Non-natural causes (all minus natural): 1.6 per day

## Analysis

The last figure is interesting as it suggests that risk on the order of magnitude of $10^{-6}$ is unavoidable in our every day lives. If we look closer into how these figure are calculated for a particular classification, they just divided the total deaths per day for that classification by the total population. For instance, for “all causes”, they divided total mortalities per day in the US by the population of the US.

\[2*10^{-5} \approx (2.5 M / 365)/(300M)\]These risk numbers are empirical averages of risk of death gathered over a sample of millions. To gauge how accurate these averages are, we can model each person as a random variable that takes values 0 and 1 (Bernoulli) and consider the variance of these estimates.

The variance of a bernoulli random variable \(X_i\) with true parameter \(p\) is \(p(1 - p)\), and the variance of an empirical average \(\hat{p} = \sum_{i = 1}^n \frac{1}{n} X_i\) is \(Var(\hat{p}) = \frac{p (1 - p)}{n} \leq \frac{p}{n}\). By making some mild assumptions about \(p\), such as \(p < 0.01\), we can compute that the square root of the variance (“standard deviation”) is also on the order of \(10^{-6}\).

From these mean and stdev estimates, we can construct a high-probability upper and lower bound for the true value of these statistics as \(mean + c * std\) and \(mean - c * std\), where c is a small constant; With high probability (“confidence”), the true value is between these two bounds.

Using this upper confidence bound analysis, it seems that the background risk is on the order of \(10^{-6}\) with high probability. In contrast, the high probability upper bound for risk of skydiving is

\[8*10^{-6} + 2 * \sqrt{\frac{0.01}{50M}} = 36 * 10^{-6}\]Thus, with high confidence, the risk of skydiving is one order of magnitude higher than the “background” risk we encounter every day. Going on one skydive equals the risk of living 10-30 days.

From these numbers, it seems that the average risk of skydiving is fairly mild. Just going by the averages listed (without any confidence analysis), one skydive is about the same risk as of driving from Seattle to LA and back.

## Final Remarks

I think it’s useful to estimate high confidence upper bounds on risk of unfamiliar (and inherently risky) activities. However, ultimately you’d still have to decide whether that risk is tolerable, and it’s not clear what that threshold should be (or even if you should live according to a strict threshold). But the calculated risk probably should not be too much higher than the background risk if it’s for a recreational activity.