# Please explain me the difference between two ways to calculate a time when something with a probability of P will happen

Let's take a look at an example:(taken from the brilliant.org course on probability ) Given a chance for failure to happen in any month is 12.6% There are 2 ways to calculate the T (time, or months amount to be more precise)

1. The first way is presented on brilliant.org. they suggest the formula T=1/p meaning that 1/0.126 = 7.9 months

2. I knew another way, but maybe I was wrong. The way I was familiar with is:

The chance the failure does not happen in the first month is 1-0.126=0.87

The chance it will ALSO not happen in the second month is 0.87 * 0.87 =0.87^2 = 0.76

The chance it will ALSO not happen in the third month is 0.87 * 0.87 * 0.87 =0.87^3 = 0.67

and so on.(see the image attached)

Yes, I know that the second method instead gives a probability something to happen in any month than giving the specific month when the failure is likely to happen, which does the 1/p formula but I will be glad to understand why the formula gives me month 8 where the probability not to happen is 30.4%. It sounds much better if the formula would return a month where the probability is, let us say, below 5%.

Thank you very much.

• $~T = 1/p~$ is equal to the expected time before a success occurs. This means that if you are considering a time interval of $~n,~$ then the expected number of successes in this interval is $~\dfrac{n}{T} = np.~$ Your alternate computation expresses the idea that the probability that there is at least one success in the first $~n~$ time intervals is $~1 - (1-p)^n.$ There is no contradiction here. ...see next comment Apr 7, 2023 at 9:07
• This is because computing the expected number of successes in a time interval of $~n~$ is different from computing the probability of at least one success in $~n~$ time intervals. That is, the phrase "at least one success" includes the events when there is more than one success. Apr 7, 2023 at 9:08
• As a clarifying illustration, suppose that you are given the following information. In a certain time interval, there will either be $~0,1,$ or $~2~$ successes. Further suppose that the probability of these mutually exclusive events is $~0.4, 0.3, ~$ and $~0.3,~$ respectively. Then, the probability of at least one success in this interval is $~1 - 0.4 = 0.6.~$ Further, the expected number of successes in this interval will be $$(0 \times 0.4) + (1 \times 0.3) + (2 \times 0.3) = 0.9.$$ Apr 7, 2023 at 9:16
• "giving the specific month when the failure is likely to happen" Warning! The failure is not "more likely" to happen in the month given by "Brilliant". The failure is equally (un)likely to happen in any month. This number only means the following: Suppose you have a very large amount of factories/people doing this thing. Then for each factory, write down the month that failure happens. Then average all these month numbers. You end up with "Brilliant"s answer. But this is not "more likely", only the average. Apr 7, 2023 at 10:28

Let's say chance of failure is $$p$$.

We want to calculate the expected value of the month that failure happens (Call this $$F_i$$ for failure happening in month $$i$$).

This is given by:

$$1\cdot P(F_1)+2\cdot P(F_2)+3\cdot P(F_3)+\dots$$$$=1\cdot p+2\cdot(1-p)\cdot p+3\cdot(1-p)^2\cdot p+\dots$$$$=p\left(\sum_{k=1}^\infty k(1-p)^{k-1}\right)$$

Solving this sum (e.g. using wolfram or wikipedia) we obtain that $$\sum_{k=1}^\infty k(1-p)^{k-1}=\frac1{p^2}$$ and thus this average is equal to $$\frac1p$$. This includes both your formulas and the equation that "Brilliant" grabbed from thin air.

It sounds much better if the formula would return a month where the probability is, let us say, below 5%.

If the probability is below 5%, say day 25, then it is "very likely" that failure will have happened. This means that if I do this experiment many times (Say I buy some equipment every time it breaks and write down how many months that takes each time), then it is almost always before day 25. So day 25 is an over-estimate of the average number of days it takes.