# Calculating probabilities over longer period of time

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?

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 $1$.

• 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 2:52
• 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 2:58
• 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 7:31