# What is the probability of an independent event occurring after repeated attempts?

Let's suppose that I have an event whose probability of occurring is $\frac{44}{1000}$.

Let's also assume that I can make multiple independent attempts at observing the event.

I want to know what the compounded probability is that the event would occur after a certain number of tries. My intuition is that each event has the same chance of occurring as each attempt is completely independent.

Since I couldn't reason my way out of it, I thought instead about the probability of the event NOT occurring. I asked myself "What is the probability of the event NOT occurring x times in a row?"

Since these attempts are independent and the probability of 2 independent events occurring in sequence is $P=P(A)*P(B)$, I computed $P=(1-\frac{44}{1000})^x$

With $x=100$, $P\approx 1.1\%$. Applying the same logic again this means that the probability of not not seeing the event in 100 attempts is $1-P(100)=98.9\%$ Following this reasoning, at x=323 Excel returns 0, meaning near $100\%$ probability (this is of course an artifact of finite precision, it approaches but not reaches 0).

This seems to contradict the earlier statement that each event had an independent probability of $\frac{44}{1000}$.

What is the correct way of reasoning about this that resolves the seeming contradiction between the odds of the event occurring and the odds of it not occurring after x attempts?

Your reasoning for computing the probability of the event not occurring after x attempts is correct.

The reason why Excel returns 0 for x=323, is one of numerical precision in the computation. Since numbers used in Excel to perform computations have limited precision, not only do you incur in numerical errors, but they can't represent numbers which are sufficiently small.

If you use arbitrary precision to compute the probability (www.wolframalpha.com for example), you will see that it is not 0.

If you want to know the probability of observing the event for the first time on the x-th attempt, you simply need to compute the probability of not seeing it x-1 times and then seeing it:

$$\left(1 - \frac{44}{1000}\right)^{x-1} \cdot \frac{44}{1000}$$

This is known as the geometric distribution.

• "this is of course an artifact of finite precision, it approaches but not reaches 0" ... I see now that I was missing the last term, which resolves the contradiction as it clearly illustrates that I'm asking two separate questions which are allowed to have two distinct probabilities. – Stephen Nov 2 '12 at 0:26
• @Stephen I have computed (1-44/1000)^323 and it's approximately 5*10^-7. That should be no problem for Excel's numerical precision unless you are formatting that cell to contain numbers with a limited number of decimal places (i.e. in Excel's general format number it displays the correct results). – pedrosorio Nov 2 '12 at 0:41
• Since I was just doing a quick approximation to get an idea of how quickly it decreased I only used 6 decimal places, so your answer makes perfect sense. Thanks. – Stephen Nov 2 '12 at 0:44