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There's a great question/answer at: Calculating probabilities over different time intervals

This is an awesome answer, but I'd like to ask a related question:

What if the period goes the other direction, for example, the probability is determined for a year, but you want to see the probability of it happening over 50 years?

For example, let's say there's a 5% chance of a fire during the course of a month. How likely would this be over the course of a year? What about over 30 years?

And, what if there's a 5% chance during 5 months of the year, and 10% chance during 7 months of the year. What would be the chance of a fire during the year? What about over 30 years?

desertnaut
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Neil Ticktin
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1 Answers1

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In all calculations, we will assume independence. That may not be reasonable in the case of forest fires.

Suppose that the probability of a fire in the course of a month is $0.05$, that is, $5\%$, which is very high for any individual structure.

Then the probability of no fire in the month is $0.95$.

The probability of no fire for $12$ months in a row is then $(0.95)^{12}$.

It follows that the probability of at least one fire in a year is $1-(0.95)^{12}$.

This is about $0.45964$.

For $30$ years, the same reasoning gives $1-(0.95)^{360}$. This is very close to $1$. That may feel counterintuitive. However, as mentioned earlier, the probability that a house has a fire in a given month is very much smaller than $0.05$.

Now we look at the problem where we have probability $0.05$ each month for $5$ months, and $0.10$ each month for $7$ months. Then the probability of no fire in the $5$ months is $(0.95)^5$, and the probability of no fire in the other $7$ months is $(0.90)^7$. So the probability of no fire in a year is $(0.95)^5(0.90)^7$. It follows that the probability of at least one fire in the year is $1-(0.95)^5(0.90)^7$.
Over $30$ years, in the $5$-$7$ scenario, the probability of no fire is $((0.95)^5(0.90)^7)^{30}=(0.95)^{150}(0.90)^{210}$. So the probability of at least one fire is $1- (0.95)^{150}(0.90)^{210}$. This is nearly $1$.

André Nicolas
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  • Excellent.

    To be clear, and since I know just enough to be dangerous, when you say a probability of "nearly 1" -- that means (in layman terms), you should expect a fire to happen (i.e., nearly a 100% chance of happening). Right?

    One more variant then.

    Assuming the above is still all good -- what if we add a second and third factor (e.g., all three happening simultaneously). For example, chances of a fire happening at the same time as a water supply cutoff, as well as on a hot day. What would the formula for look like if there's a second and third factor? The three multiplied together?

     – Neil Ticktin Sep 11 '13 at 22:55
  • By nearly $1$ in that case I mean that the probability of no fire is about $10^{-13}$. That is about one one-millionth of the probability of winning the grand prize in the standard "big" lottery if you buy $1$ ticket. – André Nicolas Sep 11 '13 at 23:06
  • Perfect.

    And, the second part: what if we add a second and third factor (e.g., all three happening simultaneously). For example, chances of a fire happening at the same time as a water supply cutoff, as well as on a hot day. What would the formula for look like if there's a second and third factor? The three multiplied together?

     – Neil Ticktin Sep 12 '13 at 02:52
  • We need to be careful. Multiplication is fine if events are independent. But with forest fires, hot day and forest fire are not independent. Same is true for house fires, probably. I imagine in that case fires are more likely on a cold day. – André Nicolas Sep 12 '13 at 02:58
  • Understood. Because these are NOT forest fires, but residential/business fires, I believe they are independent (like you said, may be more in winter than summer). Thanks! – Neil Ticktin Sep 12 '13 at 07:31