Nines of safety: a proposed unit of measurement of risk

What's new 2021-10-05

In everyday usage, we rely heavily on percentages to quantify probabilities and proportions: we might say that a prediction is ${50\%}$ accurate or ${80\%}$ accurate, that there is a ${2\%}$ chance of dying from some disease, and so forth. However, for those without extensive mathematical training, it can sometimes be difficult to assess whether a given percentage amounts to a “good” or “bad” outcome, because this depends very much on the context of how the percentage is used. For instance:

(i) In a two-party election, an outcome of say ${51\%}$ to ${49\%}$ might be considered close, but ${55\%}$ to ${45\%}$ would probably be viewed as a convincing mandate, and ${60\%}$ to ${40\%}$ would likely be viewed as a landslide.
(ii) Similarly, if one were to poll an upcoming election, a poll of ${51\%}$ to ${49\%}$ would be too close to call, ${55\%}$ to ${45\%}$ would be an extremely favorable result for the candidate, and ${60\%}$ to ${40\%}$ would mean that it would be a major upset if the candidate lost the election.
(iii) On the other hand, a medical operation that only had a ${51\%}$ , ${55\%}$ , or ${60\%}$ chance of success would be viewed as being incredibly risky, especially if failure meant death or permanent injury to the patient. Even an operation that was ${90\%}$ or ${95\%}$ likely to be non-fatal (i.e., a ${10\%}$ or ${5\%}$ chance of death) would not be conducted lightly.
(iv) A weather prediction of, say, ${30\%}$ chance of rain during a vacation trip might be sufficient cause to pack an umbrella, even though it is more likely than not that rain would not occur. On the other hand, if the prediction was for an ${80\%}$ chance of rain, and it ended up that the skies remained clear, this does not seriously damage the accuracy of the prediction – indeed, such an outcome would be expected in one out of every five such predictions.
(v) Even extremely tiny percentages of toxic chemicals in everyday products can be considered unacceptable. For instance, EPA rules require action to be taken when the percentage of lead in drinking water exceeds ${0.0000015\%}$ (15 parts per billion). At the opposite extreme, recycling contamination rates as high as ${10\%}$ are often considered acceptable.

Because of all the very different ways in which percentages could be used, I think it may make sense to propose an alternate system of units to measure one class of probabilities, namely the probabilities of avoiding some highly undesirable outcome, such as death, accident or illness. The units I propose are that of “nines“, which are already commonly used to measure availability of some service or purity of a material, but can be equally used to measure the safety (i.e., lack of risk) of some activity. Informally, nines measure how many consecutive appearances of the digit ${9}$ are in the probability of successfully avoiding the negative outcome, thus

${90\%}$ success = one nine of safety
${99\%}$ success = two nines of safety
${99.9\%}$ success = three nines of safety

and so forth. Using the mathematical device of logarithms, one can also assign a fractional number of nines of safety to a general probability:

Definition 1 (Nines of safety) An activity (affecting one or more persons, over some given period of time) that has a probability ${p}$ of the “safe” outcome and probability ${1-p}$ of the “unsafe” outcome will have ${k}$ nines of safety against the unsafe outcome, where ${k}$ is defined by the formula
$\displaystyle k = -\log_{10}(1-p) \ \ \ \ \ (1)$
(where ${\log_{10}}$ is the logarithm to base ten), or equivalently
$\displaystyle p = 1 - 10^{-k}. \ \ \ \ \ (2)$

Remark 2 Because of the various uncertainties in measuring probabilities, as well as the inaccuracies in some of the assumptions and approximations we will be making later, we will not attempt to measure the number of nines of safety beyond the first decimal point; thus we will round to the nearest tenth of a nine of safety throughout this post.

Here is a conversion table between percentage rates of success (the safe outcome), failure (the unsafe outcome), and the number of nines of safety one has:

Success rate ${p}$ Failure rate ${1-p}$ Number of nines ${k}$ ${0\%}$ ${100\%}$ ${0.0}$ ${50\%}$ ${50\%}$ ${0.3}$ ${75\%}$ ${25\%}$ ${0.6}$ ${80\%}$ ${20\%}$ ${0.7}$ ${90\%}$ ${10\%}$ ${1.0}$ ${95\%}$ ${5\%}$ ${1.3}$ ${97.5\%}$ ${2.5\%}$ ${1.6}$ ${98\%}$ ${2\%}$ ${1.7}$ ${99\%}$ ${1\%}$ ${2.0}$ ${99.5\%}$ ${0.5\%}$ ${2.3}$ ${99.75\%}$ ${0.25\%}$ ${2.6}$ ${99.8\%}$ ${0.2\%}$ ${2.7}$ ${99.9\%}$ ${0.1\%}$ ${3.0}$ ${99.95\%}$ ${0.05\%}$ ${3.3}$ ${99.975\%}$ ${0.025\%}$ ${3.6}$ ${99.98\%}$ ${0.02\%}$ ${3.7}$ ${99.99\%}$ ${0.01\%}$ ${4.0}$ ${100\%}$ ${0\%}$ infinite

Thus, if one has no nines of safety whatsoever, one is guaranteed to fail; but each nine of safety one has reduces the failure rate by a factor of ${10}$ . In an ideal world, one would have infinitely many nines of safety against any risk, but in practice there are no ${100\%}$ guarantees against failure, and so one can only expect a finite amount of nines of safety in any given situation. Realistically, one should thus aim to have as many nines of safety as one can reasonably expect to have, but not to demand an infinite amount.

Remark 3 The number of nines of safety against a certain risk is not absolute; it will depend not only on the risk itself, but (a) the number of people exposed to the risk, and (b) the length of time one is exposed to the risk. Exposing more people or increasing the duration of exposure will reduce the number of nines, and conversely exposing fewer people or reducing the duration will increase the number of nines; see Proposition 7 below for a rough rule of thumb in this regard.

Remark 4 Nines of safety are a logarithmic scale of measurement, rather than a linear scale. Other familiar examples of logarithmic scales of measurement include the Richter scale of earthquake magnitude, the pH scale of acidity, the decibel scale of sound level, octaves in music, and the magnitude scale for stars.

Remark 5 One way to think about nines of safety is via the Swiss cheese model that was created recently to describe pandemic risk management. In this model, each nine of safety can be thought of as a slice of Swiss cheese, with holes occupying ${10\%}$ of that slice. Having ${k}$ nines of safety is then analogous to standing behind ${k}$ such slices of Swiss cheese. In order for a risk to actually impact you, it must pass through each of these ${k}$ slices. A fractional nine of safety corresponds to a fractional slice of Swiss cheese that covers the amount of space given by the above table. For instance, ${0.6}$ nines of safety corresponds to a fractional slice that covers about ${75\%}$ of the given area (leaving ${25\%}$ uncovered).

Now to give some real-world examples of nines of safety. Using data for deaths in the US in 2019 (without attempting to account for factors such as age and gender), a random US citizen will have had the following amount of safety from dying from some selected causes in that year:

Cause of death Mortality rate per ${100,\! 000}$ (approx.) Nines of safety All causes ${870}$ ${2.0}$ Heart disease ${200}$ ${2.7}$ Cancer ${180}$ ${2.7}$ Accidents ${52}$ ${3.3}$ Drug overdose ${22}$ ${3.7}$ Influenza/Pneumonia ${15}$ ${3.8}$ Suicide ${14}$ ${3.8}$ Gun violence ${12}$ ${3.9}$ Car accident ${11}$ ${4.0}$ Murder ${5}$ ${4.3}$ Airplane crash ${0.14}$ ${5.9}$ Lightning strike ${0.006}$ ${7.2}$

The safety of air travel is particularly remarkable: a given hour of flying in general aviation has a fatality rate of ${0.00001}$ , or about ${5}$ nines of safety, while for the major carriers the fatality rate drops down to ${0.0000005}$ , or about ${7.3}$ nines of safety.

Of course, in 2020, COVID-19 deaths became significant. In this year in the US, the mortality rate for COVID-19 (as the underlying or contributing cause of death) was ${91.5}$ per ${100,\! 000}$ , corresponding to ${3.0}$ nines of safety, which was less safe than all other causes of death except for heart disease and cancer. At this time of writing, data for all of 2021 is of course not yet available, but it seems likely that the safety level would be even lower for this year.

Some further illustrations of the concept of nines of safety:

Each round of Russian roulette has a success rate of ${5/6}$ , providing only ${0.8}$ nines of safety. Of course, the safety will decrease with each additional round: one has only ${0.5}$ nines of safety after two rounds, ${0.4}$ nines after three rounds, and so forth. (See also Proposition 7 below.)
The ancient Roman punishment of decimation, by definition, provided exactly one nine of safety to each soldier being punished.
Rolling a ${1}$ on a ${20}$ -sided die is a risk that carries about ${1.3}$ nines of safety.
Rolling a double one (“snake eyes“) from two six-sided dice carries about ${1.6}$ nines of safety.
One has about ${2.6}$ nines of safety against the risk of someone randomly guessing your birthday on the first attempt.
A null hypothesis has ${1.3}$ nines of safety against producing a ${p = 0.05}$ statistically significant result, and ${2.0}$ nines against producing a ${p=0.01}$ statistically significant result. (However, one has to be careful when reversing the conditional; a ${p=0.01}$ statistically significant result does not necessarily have ${2.0}$ nines of safety against the null hypothesis. In Bayesian statistics, the precise relationship between the two risks is given by Bayes’ theorem.)
If a poker opponent is dealt a five-card hand, one has ${5.8}$ nines of safety against that opponent being dealt a royal flush, ${4.8}$ against a straight flush or higher, ${3.6}$ against four-of-a-kind or higher, ${2.8}$ against a full house or higher, ${2.4}$ against a flush or higher, ${2.1}$ against a straight or higher, ${1.5}$ against three-of-a-kind or higher, ${1.1}$ against two pairs or higher, and just ${0.3}$ against one pair or higher. (This data was converted from this Wikipedia table.)
A ${k}$ -digit PIN number (or a ${k}$ -digit combination lock) carries ${k}$ nines of safety against each attempt to randomly guess the PIN. A length ${k}$ password that allows for numbers, upper and lower case letters, and punctuation carries about ${2k}$ nines of safety against a single guess. (For the reduction in safety caused by multiple guesses, see Proposition 7 below.)

Here is another way to think about nines of safety:

Proposition 6 (Nines of safety extend expected onset of risk) Suppose a certain risky activity has ${k}$ nines of safety. If one repeatedly indulges in this activity until the risk occurs, then the expected number of trials before the risk occurs is ${10^k}$ .

Proof: The probability that the risk is activated after exactly ${n}$ trials is ${(1-10^{-k})^{n-1} 10^{-k}}$ , which is a geometric distribution of parameter ${10^k}$ . The claim then follows from the standard properties of that distribution. $\Box$

Thus, for instance, if one performs some risky activity daily, then the expected length of time before the risk occurs is given by the following table:

Daily nines of safety Expected onset of risk ${0}$ One day ${0.8}$ One week ${1.5}$ One month ${2.6}$ One year ${2.9}$ Two years ${3.3}$ Five years ${3.6}$ Ten years ${3.9}$ Twenty years ${4.3}$ Fifty years ${4.6}$ A century

Or, if one wants to convert the yearly risks of dying from a specific cause into expected years before that cause of death would occur (assuming for sake of discussion that no other cause of death exists):

Yearly nines of safety Expected onset of risk ${0}$ One year ${0.3}$ Two years ${0.7}$ Five years ${1}$ Ten years ${1.3}$ Twenty years ${1.7}$ Fifty years ${2.0}$ A century

These tables suggest a relationship between the amount of safety one would have in a short timeframe, such as a day, and a longer time frame, such as a year. Here is an approximate formalisation of that relationship:

Proposition 7 (Repeated exposure reduces nines of safety) If a risky activity with ${k}$ nines of safety is (independently) repeated ${m}$ times, then (assuming ${k}$ is large enough depending on ${m}$ ), the repeated activity will have approximately ${k - \log_{10} m}$ nines of safety. Conversely: if the repeated activity has ${k'}$ nines of safety, the individual activity will have approximately ${k' + \log_{10} m}$ nines of safety.

Proof: An activity with ${k}$ nines of safety will be safe with probability ${1-10^{-k}}$ , hence safe with probability ${(1-10^{-k})^m}$ if repeated independently ${m}$ times. For ${k}$ large, we can approximate

$\displaystyle (1 - 10^{-k})^m \approx 1 - m 10^{-k} = 1 - 10^{-(k - \log_{10} m)}$

giving the former claim. The latter claim follows from inverting the former. $\Box$

Remark 8 The hypothesis of independence here is key. If there is a lot of correlation between the risks between different repetitions of the activity, then there can be much less reduction in safety caused by that repetition. As a simple example, suppose that ${90\%}$ of a workforce are trained to perform some task flawlessly no matter how many times they repeat the task, but the remaining ${10\%}$ are untrained and will always fail at that task. If one selects a random worker and asks them to perform the task, one has ${1.0}$ nines of safety against the task failing. If one took that same random worker and asked them to perform the task ${m}$ times, the above proposition might suggest that the number of nines of safety would drop to approximately ${1.0 - \log_{10} m}$ ; but in this case there is perfect correlation, and in fact the number of nines of safety remains steady at ${1.0}$ since it is the same ${10\%}$ of the workforce that would fail each time.
Because of this caveat, one should view the above proposition as only a crude first approximation that can be used as a simple rule of thumb, but should not be relied upon for more precise calculations.

One can repeat a risk either in time (extending the time of exposure to the risk, say from a day to a year), or in space (by exposing the risk to more people). The above proposition then gives an additive conversion law for nines of safety in either case. Here are some conversion tables for time:

From/to Daily Weekly Monthly Yearly Daily 0 -0.8 -1.5 -2.6 Weekly +0.8 0 -0.6 -1.7 Monthly +1.5 +0.6 0 -1.1 Yearly +2.6 +1.7 +1.1 0

From/to Yearly Per 5 yr Per decade Per century Yearly 0 -0.7 -1.0 -2.0 Per 5 yr +0.7 0 -0.3 -1.3 Per decade +1.0 + -0.3 0 -1.0 Per century +2.0 +1.3 +1.0 0

For instance, as mentioned before, the yearly amount of safety against cancer is about ${2.7}$ . Using the above table (and making the somewhat unrealistic hypothesis of independence), we then predict the daily amount of safety against cancer to be about ${2.7 + 2.6 = 5.3}$ nines, the weekly amount to be about ${2.7 + 1.7 = 4.4}$ nines, and the amount of safety over five years to drop to about ${2.7 - 0.7 = 2.0}$ nines.

Now we turn to conversions in space. If one knows the level of safety against a certain risk for an individual, and then one (independently) exposes a group of such individuals to that risk, then the reduction in nines of safety when considering the possibility that at least one group member experiences this risk is given by the following table:

Group Reduction in safety You ( ${1}$ person) ${0}$ You and your partner ( ${2}$ people) ${-0.3}$ You and your parents ( ${3}$ people) ${-0.5}$ You, your partner, and three children ( ${5}$ people) ${-0.7}$ An extended family of ${10}$ people ${-1.0}$ A class of ${30}$ people ${-1.5}$ A workplace of ${100}$ people ${-2.0}$ A school of ${1,\! 000}$ people ${-3.0}$ A university of ${10,\! 000}$ people ${-4.0}$ A town of ${100,\! 000}$ people ${-5.0}$ A city of ${1}$ million people ${-6.0}$ A state of ${10}$ million people ${-7.0}$ A country of ${100}$ million people ${-8.0}$ A continent of ${1}$ billion people ${-9.0}$ The entire planet ${-9.8}$

For instance, in a given year (and making the somewhat implausible assumption of independence), you might have ${2.7}$ nines of safety against cancer, but you and your partner collectively only have about ${2.7 - 0.3 = 2.4}$ nines of safety against this risk, your family of five might only have about ${2.7 - 0.7 = 2}$ nines of safety, and so forth. By the time one gets to a group of ${1,\! 000}$ people, it actually becomes very likely that at least one member of the group will die of cancer in that year. (Here the precise conversion table breaks down, because a negative number of nines such as ${2.7 - 3.0 = -0.3}$ is not possible, but one should interpret a prediction of a negative number of nines as an assertion that failure is very likely to happen. Also, in practice the reduction in safety is less than this rule predicts, due to correlations such as risk factors that are common to the group being considered that are incompatible with the assumption of independence.)

In the opposite direction, any reduction in exposure (either in time or space) to a risk will increase one’s safety level, as per the following table:

Reduction in exposure Additional nines of safety ${\div 1}$ ${0}$ ${\div 2}$ ${+0.3}$ ${\div 3}$ ${+0.5}$ ${\div 5}$ ${+0.7}$ ${\div 10}$ ${+1.0}$ ${\div 100}$ ${+2.0}$

For instance, a five-fold reduction in exposure will reclaim about ${0.7}$ additional nines of safety.

Here is a slightly different way to view nines of safety:

Proposition 9 Suppose that a group of ${m}$ people are independently exposed to a given risk. If there are at most
$\displaystyle \log_{10} \frac{1}{1-2^{-1/m}}$
nines of individual safety against that risk, then there is at least a ${50\%}$ chance that one member of the group is affected by the risk.

Proof: If individually there are ${k}$ nines of safety, then the probability that all the members of the group avoid the risk is ${(1-10^{-k})^m}$ . Since the inequality

$\displaystyle (1-10^{-k})^m \leq \frac{1}{2}$

is equivalent to

$\displaystyle k \leq \log_{10} \frac{1}{1-2^{-1/m}},$

the claim follows. $\Box$

Thus, for a group to collectively avoid a risk with at least a ${50\%}$ chance, one needs the following level of individual safety:

Group Individual safety level required You ( ${1}$ person) ${0.3}$ You and your partner ( ${2}$ people) ${0.5}$ You and your parents ( ${3}$ people) ${0.7}$ You, your partner, and three children ( ${5}$ people) ${0.9}$ An extended family of ${10}$ people ${1.2}$ A class of ${30}$ people ${1.6}$ A workplace of ${100}$ people ${2.2}$ A school of ${1,\! 000}$ people ${3.2}$ A university of ${10,\! 000}$ people ${4.2}$ A town of ${100,\! 000}$ people ${5.2}$ A city of ${1}$ million people ${6.2}$ A state of ${10}$ million people ${7.2}$ A country of ${100}$ million people ${8.2}$ A continent of ${1}$ billion people ${9.2}$ The entire planet ${10.0}$

For large ${m}$ , the level ${k}$ of nines of individual safety required to protect a group of size ${m}$ with probability at least ${50\%}$ is approximately ${\log_{10} \frac{m}{\ln 2} \approx (\log_{10} m) + 0.2}$ .

Precautions that can work to prevent a certain risk from occurring will add additional nines of safety against that risk, even if the precaution is not ${100\%}$ effective. Here is the precise rule:

Proposition 10 (Precautions add nines of safety) Suppose an activity carries ${k}$ nines of safety against a certain risk, and a separate precaution can independently protect against that risk with ${l}$ nines of safety (that is to say, the probability that the protection is effective is ${1 - 10^{-l}}$ ). Then applying that precaution increases the number of nines in the activity from ${k}$ to ${k+l}$ .

Proof: The probability that the precaution fails and the risk then occurs is ${10^{-l} \times 10^{-k} = 10^{-(k+l)}}$ . The claim now follows from Definition 1. $\Box$

In particular, we can repurpose the table at the start of this post as a conversion chart for effectiveness of a precaution:

Effectiveness Failure rate Additional nines provided ${0\%}$ ${100\%}$ ${+0.0}$ ${50\%}$ ${50\%}$ ${+0.3}$ ${75\%}$ ${25\%}$ ${+0.6}$ ${80\%}$ ${20\%}$ ${+0.7}$ ${90\%}$ ${10\%}$ ${+1.0}$ ${95\%}$ ${5\%}$ ${+1.3}$ ${97.5\%}$ ${2.5\%}$ ${+1.6}$ ${98\%}$ ${2\%}$ ${+1.7}$ ${99\%}$ ${1\%}$ ${+2.0}$ ${99.5\%}$ ${0.5\%}$ ${+2.3}$ ${99.75\%}$ ${0.25\%}$ ${+2.6}$ ${99.8\%}$ ${0.2\%}$ ${+2.7}$ ${99.9\%}$ ${0.1\%}$ ${+3.0}$ ${99.95\%}$ ${0.05\%}$ ${+3.3}$ ${99.975\%}$ ${0.025\%}$ ${+3.6}$ ${99.98\%}$ ${0.02\%}$ ${+3.7}$ ${99.99\%}$ ${0.01\%}$ ${+4.0}$ ${100\%}$ ${0\%}$ infinite

Thus for instance a precaution that is ${80\%}$ effective will add ${0.7}$ nines of safety, a precaution that is ${99.8\%}$ effective will add ${2.7}$ nines of safety, and so forth. The mRNA COVID vaccines by Pfizer and Moderna have somewhere between ${88\% - 96\%}$ effectiveness against symptomatic COVID illness, providing about ${0.9-1.4}$ nines of safety against that risk, and over ${95\%}$ effectiveness against severe illness, thus adding at least ${1.3}$ nines of safety in this regard.

A slight variant of the above rule can be stated using the concept of relative risk:

Proposition 11 (Relative risk and nines of safety) Suppose an activity carries ${k}$ nines of safety against a certain risk, and an action multiplies the chance of failure by some relative risk ${R}$ . Then the action removes ${\log_{10} R}$ nines of safety (if ${R > 1}$ ) or adds ${-\log_{10} R}$ nines of safety (if ${R<1}$ ) to the original activity.

Proof: The additional action adjusts the probability of failure from ${10^{-k}}$ to ${R \times 10^{-k} = 10^{-(k - \log_{10} R)}}$ . The claim now follows from Definition 1. $\Box$

Here is a conversion chart between relative risk and change in nines of safety:

Relative risk Change in nines of safety ${0.01}$ ${+2.0}$ ${0.02}$ ${+1.7}$ ${0.05}$ ${+1.3}$ ${0.1}$ ${+1.0}$ ${0.2}$ ${+0.7}$ ${0.5}$ ${+0.3}$ ${1}$ ${0}$ ${2}$ ${-0.3}$ ${5}$ ${-0.7}$ ${10}$ ${-1.0}$ ${20}$ ${-1.3}$ ${50}$ ${-1.7}$ ${100}$ ${-2.0}$

Some examples:

Smoking increases the fatality rate of lung cancer by a factor of about ${20}$ , thus removing about ${1.3}$ nines of safety from this particular risk; it also increases the fatality rates of several other diseases, though not quite as dramatically an extent.
Seatbelts reduce the fatality rate in car accidents by a factor of about two, adding about ${0.3}$ nines of safety. Airbags achieve a reduction of about ${30-50\%}$ , adding about ${0.2-0.3}$ additional nines of safety.
As far as transmission of COVID is concerned, it seems that constant use of face masks reduces transmission by a factor of about five (thus adding about ${0.7}$ nines of safety), and similarly for constant adherence to social distancing; whereas for instance a ${30\%}$ compliance with mask usage reduced transmission by about ${10\%}$ (adding only ${0.05}$ or so nines of safety).

The effect of combining multiple (independent) precautions together is cumulative; one can achieve quite a high level of safety by stacking together several precautions that individually have relatively low levels of effectiveness. Again, see the “swiss cheese model” referred to in Remark 5. For instance, if face masks add ${0.7}$ nines of safety against contracting COVID, social distancing adds another ${0.7}$ nines, and the vaccine provide another ${1.0}$ nine of safety, implementing all three mitigation methods would (assuming independence) add a net of ${2.4}$ nines of safety against contracting COVID.

In summary, when debating the value of a given risk mitigation measure, the correct question to ask is not quite “Is it certain to work” or “Can it fail?”, but rather “How many extra nines of safety does it add?”.

As one final comparison between nines of safety and other standard risk measures, we give the following proposition regarding large deviations from the mean.

Proposition 12 Let ${X}$ be a normally distributed random variable of standard deviation ${\sigma}$ , and let ${\lambda > 0}$ . Then the “one-sided risk” of ${X}$ exceeding its mean ${{\bf E} X}$ by at least ${\lambda \sigma}$ (i.e., ${X \geq {\bf E} X + \lambda \sigma}$ ) carries
$\displaystyle -\log_{10} \frac{1 - \mathrm{erf}(\lambda/\sqrt{2})}{2}$
nines of safety, the “two-sided risk” of ${X}$ deviating (in either direction) from its mean by at least ${\lambda \sigma}$ (i.e., ${|X-{\bf E} X| \geq \lambda \sigma}$ ) carries
$\displaystyle -\log_{10} (1 - \mathrm{erf}(\lambda/\sqrt{2}))$
nines of safety, where ${\mathrm{erf}}$ is the error function.

Proof: This is a routine calculation using the cumulative distribution function of the normal distribution. $\Box$

Here is a short table illustrating this proposition:

Number ${\lambda}$ of deviations from the mean One-sided nines of safety Two-sided nines of safety ${0}$ ${0.3}$ ${0.0}$ ${1}$ ${0.8}$ ${0.5}$ ${2}$ ${1.6}$ ${1.3}$ ${3}$ ${2.9}$ ${2.6}$ ${4}$ ${4.5}$ ${4.2}$ ${5}$ ${6.5}$ ${6.2}$ ${6}$ ${9.0}$ ${8.7}$

Thus, for instance, the risk of a five sigma event (deviating by more than five standard deviations from the mean in either direction) should carry ${6.2}$ nines of safety assuming a normal distribution, and so one would ordinarily feel extremely safe against the possibility of such an event, unless one started doing hundreds of thousands of trials. (However, we caution that this conclusion relies heavily on the assumption that one has a normal distribution!)

See also this older essay I wrote on anonymity on the internet, using bits as a measure of anonymity in much the same way that nines are used here as a measure of safety.