Chance of 1/n event having happened after n tries approaching 63% I've recently been wondering about this phenomenon. If you have an event that occurs 1/n times and you try to trigger this event for n times, with n being a large enough number, the chance of the event having occurred at least once seems to always be around 63% ( chance = 1-(1-1/n)^n). However I thought such chances would follow a bell curve with the event occuring before n to 50% of people and after to the other 50%. This would mean that there would only be a 50% chance of it having happened at n tries. Could someone tell me where the mistake in my thought process is here? 
 A: Say you have some procedure whose result is "success" with probability $\frac 1n.$
You can do the procedure as many times as you need, and each time the probability of "success" is $\frac 1n,$ independent of the results of any other iteration.
You do the procedure repeatedly until the result is "success",
and then set $X$ to the number of times you did the procedure.
So if you get "success" the very first time, $X=1$, but if you first get "success" the tenth time you do the procedure, $X=10.$
It seems that you imagine the probability distribution of $X$ is somewhat like a Gaussian or normal distribution, the usual "bell curve" that people talk about.
And indeed if this were true, then you would find that $X < n$ with a probability around $50\%$ and $X > n$ with a probability around $50\%.$
(These can't be exact because there is a non-zero probability that $X=n$ exactly.)
But in this process the most likely value for $X$ is $X=1.$
It doesn't matter how large $n$ is, $X=1$ is still the most likely outcome.
Specifically,
\begin{align}
P(X = 1) &= \frac 1n, \\
P(X = 2) &= \left(1 - \frac1n\right) \frac 1n, \\
P(X = 3) &= \left(1 - \frac1n\right)^2 \frac 1n, \\
P(X = 4) &= \left(1 - \frac1n\right)^3 \frac 1n, \\
\end{align}
and in general for any possible outcome $X= k,$
$$
P(X = k) = \left(1 - \frac1n\right)^{k-1} \frac 1n.
$$
Since $1 - \frac 1n < 1,$ each time you multiply by $1 - \frac1n$ the result gets smaller.
So this distribution is highly asymmetrical with the largest probabilities all piled up at the low end of the range of outcomes, near $X=1,$ even when the expected value is very far from $1.$
These probabilities near $1$ are balanced by the long, thin tail on the right-hand side of the distribution, where the probabilities get very small but also are still non-zero even for numbers that are very, very far above the expected value.
(In fact there is no limit to how large $X$ can get with non-zero probability.)
This is completely unlike the normal distribution, which is perfectly symmetric around its mean with the highest density of probability at the mean.
If $n = 2,$ for example, the expected value of $X$ is $2.$
But $X$ is twice as likely to be less than $2$ as to be greater than $2.$
Of course it's also possible that $X=2$ exactly, and this is somewhat likely to happen (probability $\frac14$); the $63\%$ approximate probability is really meant for larger values of $n.$
But for larger $n$, even though there's only a $37\%$ probability that $X$ will be greater than $n$, it's possible for $X$ to be much, much larger than $n$.
Even though those very, very large values of $X$ are very unlikely, they're still very, very large, and this makes up for their low probability of occurring.
That's how the $37\%$ portion of the distribution above the expected value can balance out the $63\%$ portion below the expected value.
