Likelihood of 10000:1 probability happening exactly once in 10,000 tries I am interested in understanding the difference between "likelihood" of a random event with a particular probability actually occurring the exact probability it is said to be likely.  i.e. if an event has a 1 in 10000 probabilty, what is the likelihood that in 10000 trials it will occur exactly 1 time, not 2 times, not 0 times, not 3 times, etc. and how does one express (and account for) the deviation?  
If an event has a probability of 1:10,000, therefore in 100,000 trials it would then be likely to occur 10 times; in 1,000,000 trials, it would be likely to occur 100 times, but would it not be also just as likely that it occur in any given set of 1,000,000 trials any number of times, for example: 98 times, 99 times, 101 times, 96 times, 102 times, etc.
Statistically speaking how many trials must be averaged and accounted for to approach a statistical certainty that a particular result is actually 1:10000, and not 1:9999 or 1:10001 or 1:10000.5, etc.?
 A: I came up to this question based on its title, while hoping to find the probability of an event with $p = \frac{1}{n}$ happening at least once in $n$ iterations. I know your question was about exactly once but I guess it's somehow related.
It looks like for $n$ sufficiently large, this likelihood tends to $1 / e ≃ 0.632$ and is (quite surprisingly) almost independent of $n$.
Explanation:
Suppose I roll a dice 6 times. The probability of getting 1 at least once out of those 6 tries is:

Probability of not getting '1' for each try:
$p = \frac{5}{6}$
Probability of not getting any '1' in 6 tries:
$p = \frac{5}{6}^{6}$
Probability of getting '1' at least once in 6 tries:
$p = 1 - \frac{5}{6}^{6} \approx 0.665$

Similarly, suppose an event has a probability of 1/10000. The probability of this event happening at least once out of 10000 tries is:

$p = 1 - \frac{9999}{10000}^{10000} \approx 0.634$

We can extrapolate this for any n and get:

Probability of event with $p = \frac{1}{n}$ occurring at least once out of $n$ tries:
$p = 1 - (\frac{n-1}{n})^{n}$
And since:
$\lim\limits_{n \rightarrow +\infty} \frac{n-1}{n}^{n} = \lim\limits_{n \rightarrow +\infty} (1 - \frac{1}{n})^{n} = \frac{1}{e} \approx 0.368$
We can say that:
$\lim\limits_{n \rightarrow +\infty} 1 - \frac{n-1}{n}^{n} \approx 0.632$

Plotting this equation in Grapher, we get something like this:

Conclusion: although it makes perfect sense, I was actually quite surprised by the fact that the probability of an event having $p = \frac{1}{n}$ happening at least once out of $n$ tries is almost independent of $n$, for $n$ as little as $3$ already.
A: Let establish on simpler problem on dice. Lets calculate the likelihood probability that on 6 throws of dice, score will be 1 exactly once.
How many ways can this happen [and their respective probabilities]:
1 is scored in first throw but not in any other throws[1/6*5/6*5/6*...] [=3125/46656]
1 is scored in second throw but not in any other throw [5/6*1/6*5/6*...] [=3125/46656]
...
...

so total probability that 1 is scored only once in 6 throws is (3125/46656)*6 = 3125/7776
You can extend same development for events with probability 1/n. Probability of event occurring only once in n trials would be
((n-1)/n)^(n-1)

This might look a bit familiar when I rearrange it:
(1-1/n)^(n-1)

Other part of your question: reducing deviation as number of samples increases, is already well explained in another answer.
