Probability cut-off Is there a probability cut-off point?
e.g. if the probability of something keeps halving do you stop at a certain point and say the rest of the possibilities are negligible?
What do you do then?
Ignore the rest of the possibilities in all following calculations? e.g. percentages, averages, ratios, etc.
What is an event rate and is it related?
 A: If you're  trying to do an exact calculation, then no, you should not ignore any of the possibilities.  Of course, if you just want an approximate result, ignoring low-probability events may be justified, as long as the effect of those events on what you're trying to calculate is not disproportionately large.
For example, say you're playing a game where you toss a fair coin repeatedly.  Every time the coin comes up heads, you win \$1; the first time it comes up tails, the game ends.  Then your expected winnings from the game are
$$\frac{\$1}2 + \frac{\$1}4 + \frac{\$1}8 + \frac{\$1}{16} + \dotsb = \sum_{k=1}^\infty \frac{\$1}{2^k} = \$1.$$
If you truncate this series after $n$ terms, the sum will not be exactly \$1.  But it won't be off by more than $\$1/2^n$, so the difference becomes negligible (less that one cent) after only 7 terms.
However, let's say you're playing a different game, where you win \$0.01 on the first heads, \$0.02 on the second, \$0.04 on the third, \$0.08 on the fourth and so on.  You might think that this game doesn't pay as well as the previous one, since the rewards are in cents instead of dollars, and since the game is very likely to end before they grow much larger.  But if you calculate the expected winnings from playing this game, they are $$\frac{\$0.01}{2} + \frac{\$0.02}{4} + \frac{\$0.04}{8} + \dotsb = \sum_{k=1}^\infty \frac{2^{k-1}}{2^k}\cdot\$0.01 = \sum_{k=1}^\infty \frac{\$0.01}{2} = \infty.$$
This despite the fact that, if you only play up to $n$ rounds of this game, your expected winnings are only $n/2$ cents, and that the odds of your game lasting more than, say, 20 rounds are literally less than one in a million!  However, if you do manage to toss 20 heads in a row, you'll already have won over \$10,000, and the prizes just keep going up from there.
The example I used above is known as the St. Petersburg paradox.  You can find a lot of interesting discussion about the implications of this paradox, but the message to take home from this example is that you shouldn't neglect even extremely unlikely events if the effects of those events, if and when they do happen, is equally extreme.
A: From the center for evidence based medicine, the event rate is defined as: 

The proportion of patients in a group in whom the event is observed. Thus, if out of 100 patients, the event is observed in 27, the event rate is 0.27. Control event rate (CER) and experimental event rate (EER) are used to refer to this in control and experimental groups of patients, respectively. The patient expected event rate (PEER) refers to the rate of events we'd expect in a patient who received no treatment or conventional treatment.

Ilmari is completely correct that, in general, neglecting low probability events is a bad idea.  However, how does this relate to the event rate?  Take as an example some disease with probability of $p=.5$ of manifesting in any member of the population.  Then, take a sample of the population of size $n=500$.  How many people would we expect to have the disease?  Under the strong assumption that the disease is an independent random process, the number of cases in the same is a binomial RV - lets call it $\mathbb{X}$.  We use the usual form of expected value to find the expected number of cases:$$E\left[ \mathbb{X} \right] = \sum_{k=0}^{n} k \binom{n}{k} p^k (1-p)^{n-k} = np$$ 
Which essentially gives use the event rate for the sample.  For the n and p given above, that expected event rate would be $.5*500=250$.
Now consider what would happen if you had some cut-off level m below which you equated the probability to 0.  You'd still run the same equation from above, but only pick off terms with a probability greater than a certain value. Essentially, you'd be doing the calculation:$$E\left[ \mathbb{X} \right] = \sum_{k=0}^nk{n \choose k}p^k(1-p)^{n-k}H\left({n \choose k}p^k(1-p)^{n-k}-m\right)$$ Where H(x) is 0 if $x\leq 0$ and 1 otherwise (per Ilmari's suggestion). 
For $m=\frac{1}{1000000}, E\left[ \mathbb{X} \right]= 249.9733$.  For $m=\frac{1}{1000}, E\left[ \mathbb{X} \right]= 233.184.$  If n were larger, the underestimations would be more significant.  
The upshot is that it is rarely a good idea to neglect probabilities close to 0, regardless of bound.  Effects might not be present in small samples or with very small bounds, but they will be there.  Use the fullest precision possible, whenever possible, at least while working out a problem.   Round only at the end if the situation demands it.
