Deceptive mathematics: why 99 percent, repeated, is not safe enough

Key idea

Making go decisions at a 99 percent probability of success, we will probably see a long string of successes. And yet after only 18 such decisions, the chance that at least one has failed has risen to 17 percent, or one in six: the same odds as a game of Russian roulette.

By Mike Meier · Edition 1.0, November 10, 2026 · Adapted from Lessons Learned and Murphy’s Corollary, SETP, 2015

A retelling of the paper; Mike’s exact words are quoted.

High odds, long strings of success 

Let’s imagine we’re making go decisions when we have a 99 percent probability of success. Chances are, we’ll see a long string of successful outcomes. Each one feels like confirmation.

And yet, after only 18 such decisions, our overall cumulative probability of success has fallen to 83 percent. In other words, our cumulative probability of failure has risen to 17 percent, or one in six: the same level as in the game of Russian roulette.

I don’t know how many of you took me up on my offer of ten million dollars with one chance in six of a catastrophic outcome, but my guess is not many.

The arithmetic 

* This assumes each decision is independent and carries the same risk. In practice, as standards slip with each success (Factor II), the risk on later decisions tends to rise.

Raise the probability of an outcome to the power of the number of trials to get the probability of that outcome for every event in the series. At 99 percent, 18 times: 0.9918 = 0.83. The chance that at least one fails is what’s left: 1 − 0.83 = 0.17.*

Try the numbers 

A go decision that succeeds % of the time, made times, fails at least once 17% of the time: about one chance in 6.

Failure becomes more likely than not after 69 decisions at this level of confidence.

100 crews, each making the same series of decisions

On average: 17 of 100 crews had at least one failure.

A six-chamber revolver cylinder with one chamber loaded: one chance in six.
99 percent, 18 times: the same odds as one round in a six-chamber cylinder.

The 99.9 percent for top landings is an illustration, not Mike’s figure. The Rhino Jump is explained in Three Random Thoughts On Safety.

Try 99.9, then 99.99. Each extra nine buys about ten times as many decisions before the odds turn against you. It never buys an unlimited number.

The numbers in a table 

Chance that at least one go decision fails
Decisionsat 99%at 99.9%at 99.99%
11.0%0.10%0.010%
109.6%1.0%0.10%
1817%1.8%0.18%
5039%4.9%0.50%
10063%9.5%1.0%
50099%39%4.9%
1,000>99.9%63%9.5%

More likely than not to fail after 69 decisions at 99%, 693 at 99.9%, and 6,932 at 99.99%.

The data set is too small 

A long string of successes is exactly what a 99 percent decision produces, so the string tells us very little. Our data set is too small, at least at the beginning of any such process. A landing that fails one time in a thousand averages about a thousand successes before the first crash, and runs of 1,500 or more happen a little more than one time in five. Fifteen hundred top landings without a crash could not show that the landing was safe.

Which raises a hard question, one Mike was asked at the SETP symposium: if we can’t reliably tell 99.9 percent from 99.99 percent before the fact, don’t we need 100 percent? Is 100% safety possible?

Cite this page

Meier, M. (2026, November 10). Deceptive mathematics: Why 99 percent, repeated, is not safe enough (Edition 1.0). meiersafe.com. https://meiersafe.com/deceptive-mathematics/

Chicago and BibTeX

Meier, Mike. “Deceptive Mathematics: Why 99 Percent, Repeated, Is Not Safe Enough.” Edition 1.0. meiersafe.com, November 10, 2026. https://meiersafe.com/deceptive-mathematics/.

@misc{meier2026deceptivemathematics,
  author = {Meier, Mike},
  title  = {Deceptive Mathematics: Why 99 Percent, Repeated, Is Not Safe Enough},
  year   = {2026},
  note   = {Edition 1.0},
  url    = {https://meiersafe.com/deceptive-mathematics/}
}

Deceptive mathematics by Mike Meier, meiersafe.com/deceptive-mathematics/, CC BY-SA 4.0.