Probability vs. Expectation: A Case Study

A common claim goes like this:

"If your external dependency has a 1% chance of failing each day, that is a 365% chance of failure in a year."

This is mathematically incorrect. Probabilities cannot exceed 100%. The statement confuses expected values with probabilities. Let us carefully derive the correct formulation.

Daily vs. Annual Probability

Let the daily failure probability be

p = 0.01

and the number of days be

n = 365

On any given day, the probability of no failure is

1 - p

Assuming independence between days, the probability of no failures over the entire year is

(1 - p)^{n}

Therefore, the probability of at least one failure in the year is

1 - (1 - p)^{n}

Numerical Evaluation

Substituting the values,

(1 - 0.01)^{365} = (0.99)^{365} \approx 0.0255

So the chance of zero failures is about 2.55%, and the chance of at least one failure is

1 - (0.99)^{365} \approx 0.9745 (97.45%)

This is the correct answer: with a 1% daily failure probability, the annual probability of at least one failure is about 97%.

Why "365%" Is Wrong

The incorrect intuition arises from multiplying the daily failure probability by the number of days:

365 \times 0.01 = 3.65

This quantity is not a probability. It is the expected number of failures per year, i.e. the mean of a Binomial distribution with parameters

X \sim Binomial (n = 365, p = 0.01)

For this distribution,

E [X] = n p = 3.65

Poisson Approximation

For large n and small p, the Binomial distribution can be approximated by a Poisson distribution with parameter

λ = n p = 3.65

Then

Pr (X = k) \approx \frac{e ^{- λ} λ ^{k}}{k !}

In particular, the probability of no failures is

Pr (X = 0) \approx e^{- λ} = e^{- 3.65} \approx 0.026

and the probability of at least one failure is

1 - e^{- 3.65} \approx 0.974

which closely matches the exact calculation.

Rigorous Interpretation

The expression np gives the expected count of failures, not a probability.
The correct probability of at least one failure is obtained by computing the complement of zero failures.
In the independent-trial model, this probability grows rapidly with n and approaches 1 as n goes to infinity.

Formally,

n \to \infty lim Pr (at least one failure) = 1

for any fixed p > 0.

Conclusion

The so called "365% chance of failure" is a textbook case of conflating probability with expectation. The rigorous calculation shows:

Expected number of failures: 3.65
Probability of at least one failure in a year: approximately 97.45%

This fallacy is a useful teaching example in discrete probability: additive reasoning about risks can mislead, while multiplicative reasoning through complements yields the correct result.